Free example problems + complete solutions for typical related rates problems. Learn our 4-step problem solving strategy to solve any problem, and practice it using the problems below so you’ll be ready for your exam!

Related Rates: Overall Problem Solving Strategy

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We developed the general strategy we will use to solve these problems in our blog post 4 Steps to Solve Any Related Rates Problem. (Link will open in a new tab.) You might find it helpful to read that post before proceeding if you haven’t already. The following steps are not a recipe for you simply to follow, but rather a sequence to help guide your thinking and lead you to be able to solve each problem you encounter.

In the subsections below, we’ll solve each problem using the strategy outlined above, step by step each time. We’ll look at problems that require each of the approaches listed in 2.B above: Be sure to practice some of each — but if you need to focus your time, put it on **similar triangles** and the **Pythagorean theorem** since those problems appear most frequently.

**Draw a picture of the physical situation.****Write an equation that relates the quantities of interest.**- Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
- To develop your equation, you will probably use:
- a simple geometric fact (like the relation between a sphere’s volume and its radius, or the relation between the volume of a cylinder and its height); or
- a trigonometric function (like $\tan{\theta}$ = opposite/adjacent); or
- similar triangles; or
- the Pythagorean theorem.

**Take the derivative with respect to time of both sides of your equation. Remember the chain rule.****Solve for the quantity you’re after.**

In the subsections below, we’ll solve each problem using the strategy outlined above, step by step each time. We’ll look at problems that require each of the approaches listed in 2.B above:

i. Geometric fact. Typical problems: A circle’s radius changes, a snowball melts, a rectangle’s height and/or width changes.

ii. Trig function. Typical problems: A searchlight rotates, a rocket takes off, a kite travels horizontally.

iii. Similar triangles. Water fills a cone or trough, sand falls onto a conical pile, person walks away from a light pole that casts a shadow.

iv. Pythagorean theorem. Typical problems: Cars/ships/joggers move along 90 degree paths, baseball players run along the diamond, boat is pulled toward a dock.

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Related Rates: Using a Simple Geometric Fact

Many related rates problems make use of a simple geometric fact. For example, and so forth.

The following problems illustrate.

- the area of a circle is $A = \pi r^2$
- the volume of a sphere is $V = \dfrac{4}{3} \pi r^3$
- the area of a rectangle is $A = l w$

The following problems illustrate.

Related Rates: Spherical snowball melts

A spherical snowball is melting at the rate of $2 \pi$ cm$^3$/hr. It melts symmetrically such that it is always a sphere. How fast is its radius changing at the instant $r = 10$ cm?

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Answer: $-\dfrac{1}{200} \text{ cm/hr}$. [If you are using a web-based homework system, see the “Caution” at the end of this solution.]

Let’s unpack the question statement:

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

*B. To develop your equation, you will probably use. . . a simple geometric fact. *

In this problem we’re given the value for $\dfrac{dV}{dt}$, and we’re after the value of $\dfrac{dr}{dt}$. Hence we need to start with an expression that relates*V* to *r*. Since the snowball is a sphere, we know that relationship:

$$V = \frac{4}{3}\pi r^3$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*} \frac{d}{dt}V & = \frac{d}{dt} \left( \frac{4}{3}\pi r^3\right) \\[12px] \frac{dV}{dt} &= \frac{4}{3}\pi \frac{d}{dt} \left( r^3\right) \\[12px] &= \frac{4}{3}\pi \left( 3r^2 \frac{dr}{dt} \right) \\[12px] &= 4 \pi r^2 \frac{dr}{dt} \end{align*} **4. Solve for the quantity you’re after.**

Solving the equation above for $\dfrac{dr}{dt}$: \begin{align*} \frac{dV}{dt} &= 4 \pi r^2 \frac{dr}{dt} \\[12px] \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] \end{align*} Now we just have to substitute values. Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr, and the problem asks about when $r=10$ cm: \begin{align*} \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] &= \frac{1}{4 \pi (10)^2} (-2 \pi) \\[12px] &= -\frac{1}{200} \text{ cm/hr} \quad \cmark \end{align*} The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

**Caution**: IF you are using a web-based homework system and the question asks,

Let’s unpack the question statement:

- We’re told that the snowball’s volume
*V*is changing at the rate of $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr. (We must insert the negative sign “by hand” since we are told that the snowball is melting, and hence its volume is*decreasing*.) - As a result, its radius is changing, at the rate $\dfrac{dr}{dt}$, which is the quantity we’re after.
- The snowball always remains a sphere.
- Toward the end of our solution, we’ll need to remember that the problem is asking us about $\dfrac{dr}{dt}$ at a particular instant, when $r = 10$ cm.

See the figure.

In this problem we’re given the value for $\dfrac{dV}{dt}$, and we’re after the value of $\dfrac{dr}{dt}$. Hence we need to start with an expression that relates

$$V = \frac{4}{3}\pi r^3$$

\begin{align*} \frac{d}{dt}V & = \frac{d}{dt} \left( \frac{4}{3}\pi r^3\right) \\[12px] \frac{dV}{dt} &= \frac{4}{3}\pi \frac{d}{dt} \left( r^3\right) \\[12px] &= \frac{4}{3}\pi \left( 3r^2 \frac{dr}{dt} \right) \\[12px] &= 4 \pi r^2 \frac{dr}{dt} \end{align*}

Open to read why that dr/dt is there.

Are you wondering why that $\dfrac{dr}{dt}$ appears? The answer is the Chain Rule. While the derivative of $r^3$ with respect to *r* is $\dfrac{d}{dr}r^3 = 3r^2$, the derivative of $r^3$ with respect to *time t* is $\dfrac{d}{dt}r^3 = 3r^2\dfrac{dr}{dt}$. Remember that *r* is a function of time *t*: the radius *changes* as time passes and the snowball melts. We could have captured this time-dependence explicitly by writing our relation as
$$V(t) = \frac{4}{3}\pi [r(t)]^3$$
to remind ourselves that both *V* and *r* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}V(t) &= \frac{d}{dt}\left[ \frac{4}{3}\pi [r(t)]^3\right] \\ \\
\frac{dV(t)}{dt} &= \frac{4}{3} \pi 3[r(t)]^2 \left[\frac{d}{dt}r(t)\right]\\ \\
&= 4\pi [r(t)]^2 \left[\frac{dr}{dt}\right]
\end{align*}

[Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr in this problem, and we’re looking for $\dfrac{dr}{dt}$.]
Most people find that writing the explicit time-dependence *V(t)* and *r(t)* annoying, and so just write *V* and *r* instead. Regardless, you *must* remember that *r* depends on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dr}{dt}$ term.

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Solving the equation above for $\dfrac{dr}{dt}$: \begin{align*} \frac{dV}{dt} &= 4 \pi r^2 \frac{dr}{dt} \\[12px] \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] \end{align*} Now we just have to substitute values. Recall $\dfrac{dV}{dt} = -2 \pi$ cm$^3$/hr, and the problem asks about when $r=10$ cm: \begin{align*} \frac{dr}{dt} &= \frac{1}{4 \pi r^2} \frac{dV}{dt} \\[12px] &= \frac{1}{4 \pi (10)^2} (-2 \pi) \\[12px] &= -\frac{1}{200} \text{ cm/hr} \quad \cmark \end{align*} The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

At what rate does the radius *decrease*?

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Related Rates: Circle expands

What is the radius of an expanding circle at a moment when the rate of change of its area is numerically half as big as the rate of change of its radius?

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Answer: $\dfrac{1}{4 \pi}$

When a circle expands, obviously its area changes; the*rate* at which its area changes is $\dfrac{dA}{dt}$. Its radius changes as well, at the rate $\dfrac{dr}{dt}$. Those two quantities are related, and that recognition is the crux of this problem.

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

B. To develop your equation, you will probably use . . . a simple geometric fact.

Because we ultimately need a relation between $\dfrac{dA}{dt}$ and $\dfrac{dr}{dt}$, we start with a relation between*A* and *r*. Fortunately, that’s a particularly well-known relation:

$$A = \pi r^2$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*} \frac{d}{dt}(A) &= \frac{d}{dt} \left(\pi r^2 \right) \\[12px] \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] \end{align*}**4. Solve for the quantity you’re after.**

The problem is asking us to find*r* at the instant when $\dfrac{dA}{dt} = \dfrac{1}{2} \dfrac{dr}{dt}$, or when $ \dfrac{\left(\dfrac{dA}{dt}\right)}{\left(\dfrac{dr}{dt}\right)} = \dfrac{1}{2}$.

Starting from the preceding equation:

\begin{align*} \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] r &= \frac{1}{2 \pi} \dfrac{\left(\dfrac{dA}{dt}\right)}{\left(\dfrac{dr}{dt}\right)} \\[12px] &= \frac{1}{2 \pi} \left(\frac{1}{2}\right) = \frac{1}{4 \pi} \quad \cmark \end{align*}

When a circle expands, obviously its area changes; the

See the figure.

B. To develop your equation, you will probably use . . . a simple geometric fact.

Because we ultimately need a relation between $\dfrac{dA}{dt}$ and $\dfrac{dr}{dt}$, we start with a relation between

$$A = \pi r^2$$

\begin{align*} \frac{d}{dt}(A) &= \frac{d}{dt} \left(\pi r^2 \right) \\[12px] \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] \end{align*}

The problem is asking us to find

Starting from the preceding equation:

\begin{align*} \frac{dA}{dt} &= 2 \pi r \frac{dr}{dt} \\[12px] r &= \frac{1}{2 \pi} \dfrac{\left(\dfrac{dA}{dt}\right)}{\left(\dfrac{dr}{dt}\right)} \\[12px] &= \frac{1}{2 \pi} \left(\frac{1}{2}\right) = \frac{1}{4 \pi} \quad \cmark \end{align*}

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Related Rates: Rectangle, find *dA/dt* given *dl/dt* and *dw/dt*

The length of a rectangle is increasing at the rate of 8 cm/s, and its width is increasing at the rate of 5 cm/s. Consider the moment when its length is 3 cm and its width is 4 cm. How fast is the area of the rectangle increasing?

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Answer: $\dfrac{dA}{dt} = 47\, \tfrac{\text{cm}^2}{\text{s}} $

Let’s first unpack this question:

We’re given that the rectangle’s length is changing at the rate $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s. We’re looking for $\dfrac{dA}{dt}$ at the particular instant specified in the problem.

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. *

The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.

*B. To develop your equation, you will probably use . . .a simple geometric fact.*

We’re looking for $\dfrac{dA}{dt}$ given values for both $\dfrac{dl}{dt}$ and $\dfrac{dw}{dt}$. That suggests we want to start with a relationship among the rectangle’s area*A* and its length *l* and width *w*. Fortunately, that’s one we immediately know:

$$A = lw$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

Both $w$ and $l$ are changing with time, and so when we take the derivative we must use the Product Rule and include both $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}A &= \frac{d}{dt}(lw) \\[8px] \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \end{align*}**4. Solve for the quantity you’re after.**

We want $\dfrac{dA}{dt}$ at the moment when $l = 3$ cm and $w = 4$ cm. Furthermore, recall that the problem tells us $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s.

Hence \begin{align*} \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \\[8px] &= \left( 8 \, \tfrac{\text{cm}}{\text{s}}\right) (4 \, \text{cm}) + (3 \, \text{cm}) \left(5 \, \tfrac{\text{cm}}{\text{s}} \right) \\[8px] &= 32 \, \tfrac{\text{cm}^2}{\text{s}} + 15 \, \tfrac{\text{cm}^2}{\text{s}} \\[8px] &= 47 \, \tfrac{\text{cm}^2}{\text{s}} \quad \cmark \end{align*}

Let’s first unpack this question:

We’re given that the rectangle’s length is changing at the rate $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s. We’re looking for $\dfrac{dA}{dt}$ at the particular instant specified in the problem.

See the figure.

The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.

We’re looking for $\dfrac{dA}{dt}$ given values for both $\dfrac{dl}{dt}$ and $\dfrac{dw}{dt}$. That suggests we want to start with a relationship among the rectangle’s area

$$A = lw$$

Both $w$ and $l$ are changing with time, and so when we take the derivative we must use the Product Rule and include both $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}A &= \frac{d}{dt}(lw) \\[8px] \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \end{align*}

We want $\dfrac{dA}{dt}$ at the moment when $l = 3$ cm and $w = 4$ cm. Furthermore, recall that the problem tells us $\dfrac{dl}{dt} = 8$ cm/s, and its width is changing at the rate $\dfrac{dw}{dt} = 5$ cm/s.

Hence \begin{align*} \frac{dA}{dt} &= \frac{dl}{dt}w + l \frac{dw}{dt} \\[8px] &= \left( 8 \, \tfrac{\text{cm}}{\text{s}}\right) (4 \, \text{cm}) + (3 \, \text{cm}) \left(5 \, \tfrac{\text{cm}}{\text{s}} \right) \\[8px] &= 32 \, \tfrac{\text{cm}^2}{\text{s}} + 15 \, \tfrac{\text{cm}^2}{\text{s}} \\[8px] &= 47 \, \tfrac{\text{cm}^2}{\text{s}} \quad \cmark \end{align*}

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Related Rates: Rectangle, constant area, length increases

A rectangle has constant area 500 square centimeters. Its length is increasing at the rate of 8 centimeters per second. What is its width at the moment the width is decreasing at the rate of 1 centimeter per second?

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Answer: $w = \sqrt{\dfrac{500}{8}} \text{ cm}$

This rectangle has constant area; you might imagine it as a flat piece of rubber that has constant area. As you stretch it out in one direction, it must shrink in the other to compensate. The faster you stretch it out, the faster its width must decrease to keep the area constant.

The problem tells us that the length is changing at the constant rate $\dfrac{dl}{dt} = 8$ cm/s, a positive value since the length is increasing with time. We’re after something about when the rate that the width changes, $\dfrac{dw}{dt}$ , has a certain value (-1 cm/s). So we need to relate the width*w* to the length *l* at every moment. Fortunately we can do that easily since the area remains constant.

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. *

The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.

*B. To develop your equation, you will probably use . . .a simple geometric fact.*

Here we use the relation between the area of a rectangle and its width and length. Remember that the problem says that the rectangle’s area stays constant at 500 cm$^2$:

$$A = 500 = wl$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

Both $w$ and $l$ are changing with time, and so when we take the derivative we must include $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(wl) \\ \\ 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \end{align*}**4. Solve for the quantity you’re after.**

We want to find $w$ at the instant $\dfrac{dw}{dt} = -1$ cm/s. (Note that we must insert that negative sign “by hand” since the problem states that the width is*decreasing*.) We also know from the problem statement that $\dfrac{dl}{dt} = 8$ cm/s (a positive value since the length is increasing with time).

Looking at the preceding equation, we see that we currently have two unknowns, $l$ and $w$. Since we’re looking for $w$, let’s eliminate $l$ by noting that the rectangle has constant area: $A = 500 = lw$. Hence $l = \dfrac{500}{w}$.

Then from the preceding equation: \begin{align*} 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \\ \\ 0 &= \frac{dw}{dt}\frac{500}{w} + w\frac{dl}{dt} \\ \\ w\frac{dl}{dt} &= – \frac{500}{w}\frac{dw}{dt} \\ \\ w^2 &= – 500 \dfrac{\left(\dfrac{dw}{dt}\right)}{\left(\dfrac{dl}{dt}\right)} \end{align*} Recall $\dfrac{dw}{dt} = -1$ cm/s and $\dfrac{dl}{dt} = 8$ cm/s:

\begin{align*} w^2 &= -(500)\frac{(-1)}{8} = \frac{500}{8} \\ w &= \sqrt{\dfrac{500}{8}} \text{ cm} \end{align*}

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This rectangle has constant area; you might imagine it as a flat piece of rubber that has constant area. As you stretch it out in one direction, it must shrink in the other to compensate. The faster you stretch it out, the faster its width must decrease to keep the area constant.

The problem tells us that the length is changing at the constant rate $\dfrac{dl}{dt} = 8$ cm/s, a positive value since the length is increasing with time. We’re after something about when the rate that the width changes, $\dfrac{dw}{dt}$ , has a certain value (-1 cm/s). So we need to relate the width

See the figure.

The rectangle’s width and the length are both changing, so we’ll leave them each as a variable.

Here we use the relation between the area of a rectangle and its width and length. Remember that the problem says that the rectangle’s area stays constant at 500 cm$^2$:

$$A = 500 = wl$$

Both $w$ and $l$ are changing with time, and so when we take the derivative we must include $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(wl) \\ \\ 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \end{align*}

We want to find $w$ at the instant $\dfrac{dw}{dt} = -1$ cm/s. (Note that we must insert that negative sign “by hand” since the problem states that the width is

Looking at the preceding equation, we see that we currently have two unknowns, $l$ and $w$. Since we’re looking for $w$, let’s eliminate $l$ by noting that the rectangle has constant area: $A = 500 = lw$. Hence $l = \dfrac{500}{w}$.

Then from the preceding equation: \begin{align*} 0 &= \frac{dw}{dt}l + w\frac{dl}{dt} \\ \\ 0 &= \frac{dw}{dt}\frac{500}{w} + w\frac{dl}{dt} \\ \\ w\frac{dl}{dt} &= – \frac{500}{w}\frac{dw}{dt} \\ \\ w^2 &= – 500 \dfrac{\left(\dfrac{dw}{dt}\right)}{\left(\dfrac{dl}{dt}\right)} \end{align*} Recall $\dfrac{dw}{dt} = -1$ cm/s and $\dfrac{dl}{dt} = 8$ cm/s:

\begin{align*} w^2 &= -(500)\frac{(-1)}{8} = \frac{500}{8} \\ w &= \sqrt{\dfrac{500}{8}} \text{ cm} \end{align*}

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Related Rates: Square, sides grow

A square has side-length *x.* Each side increases at the rate of 0.5 meters each second.**(a)** Find the rate at which the square's perimeter is increasing.**(b)** Find the rate at which the square's area increasing at the moment the area is $25 \text{m}^2.$**(a)** $2.0 \tfrac{\text{m}}{\text{s}}$**(b)** $5.0 \, \tfrac{\text{m}^2}{\text{s}}$

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It’s all free: our only aim is to support your learning via our community-supported and ad-free site. **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the length of each side of the square*x.* Recall that each side increases at the rate $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}} $. We’ll use that value at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The square’s side-length changes as the situation progresses, so we’re calling that the variable*x.*

*B. To develop your equation, you will probably use . . . a simple geometric fact. *

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dP}{dt},$ where*P* is the square’s perimeter. Hence we need to start with an expression that relates *P* to *x.* Since the object is a square, we know that relationship:
$$ P = 4x $$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\[ \begin{align*} \dfrac{d}{dt}P &= \dfrac{d}{dt}(4x) \\[8px] \dfrac{dP}{dt} &= 4 \dfrac{dx}{dt} \end{align*} \]

**4. Solve for the quantity you’re after.**

This step is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$, so \[ \begin{align*} \dfrac{dP}{dt} &= 4 \dfrac{dx}{dt} \\[8px] &= 4 \left( 0.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[8px] &= 2.0 \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \] This result makes sense conceptually: the square has four sides, each of which is growing at the rate of $0.5 \, \tfrac{\text{m}}{\text{s}}$. Hence the perimeter grows at the rate $\dfrac{dP}{dt} = 4 \left(0.5 \, \tfrac{\text{m}}{\text{s}} \right) = 2.0 \, \tfrac{\text{m}}{\text{s}}.$ **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the length of each side of the square*x.* Recall that each side increases at the rate $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}} $, and we’re interested in the moment when $A = 25 \text{ m}^2.$ We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The square’s side-length changes as the situation progresses, so we’re calling that the variable*x.*

*B. To develop your equation, you will probably use . . . a simple geometric fact. *

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dA}{dt},$ where*A* is the square’s area. Hence we need to start with an expression that relates *A* to *x.* Since the object is a square, we know that relationship:
$$ A = x^2 $$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}A &= \dfrac{d}{dt}x^2 \\[8px]
\dfrac{dA}{dt} &= 2x \dfrac{dx}{dt}
\end{align*} \]

**4. Solve for the quantity you’re after.**

We know that $$ \frac{dA}{dt}= 2x \dfrac{dx}{dt} $$ and we know that $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$. We also know that we’re interested in the moment when $A = 25 \text{ m}^2;$ to complete our calculuation, we need to know the value of*x* at that moment.

*Begin subproblem to find x at the moment of interest.*

Since $A = x^2,$ when $A = 25 \text{ m}^2:$ \[ \begin{align*} x^2 &= 25 \text{ m}^2\\[8px] x &= 5.0 \text{ m} \end{align*} \]*End subproblem.*

We can now complete our calculation: \[ \begin{align*} \frac{dA}{dt}&= 2x \dfrac{dx}{dt} \\[8px] &= 2(5.0 \text{ m}) \left(0.5 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 5.0 \, \tfrac{\text{m}^2}{\text{s}} \quad \cmark \end{align*} \]

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Solution SummarySolution (a) DetailSolution (b) Detail

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See the figure. We’ve labeled the length of each side of the square

The square’s side-length changes as the situation progresses, so we’re calling that the variable

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dP}{dt},$ where

\[ \begin{align*} \dfrac{d}{dt}P &= \dfrac{d}{dt}(4x) \\[8px] \dfrac{dP}{dt} &= 4 \dfrac{dx}{dt} \end{align*} \]

Open to read why the dx/dt is there.

Are you wondering why $\dfrac{dx}{dt}$ appears? The answer is the Chain Rule.
While the derivative of $4x$ with respect to *x* is
$$\dfrac{d}{dx}4x = 4,$$
the derivative of $4x$ with respect to *time t* is
$$\dfrac{d}{dt}4x = 4\dfrac{d}{dt}x = 4\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that*x* is a function of time *t*: the side-length *x* *changes* as time passes. We could have captured this time-dependence explicitly by writing our relation as
$$ P(t) = 4x(t)$$
to remind ourselves that both *P* and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}P(t) &= \frac{d}{dt} \big[ 4x(t)\big]\\[8px]
\frac{dP(t)}{dt}&= 4 \frac{d}{dt}\big[ x(t)\big] \\[8px]
&= 4 \dfrac{dx(t)}{dt}
\end{align*}

Remember that

[Recall $\dfrac{dx(t)}{dt} = .5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dP(t)}{dt}.$]
Most people find that writing the explicit time-dependence $P(t)$ and *x(t)* annoying, and so just write *P* and *x* instead. Regardless, you *must* remember that both *P* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.

[collapse]

This step is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$, so \[ \begin{align*} \dfrac{dP}{dt} &= 4 \dfrac{dx}{dt} \\[8px] &= 4 \left( 0.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[8px] &= 2.0 \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \] This result makes sense conceptually: the square has four sides, each of which is growing at the rate of $0.5 \, \tfrac{\text{m}}{\text{s}}$. Hence the perimeter grows at the rate $\dfrac{dP}{dt} = 4 \left(0.5 \, \tfrac{\text{m}}{\text{s}} \right) = 2.0 \, \tfrac{\text{m}}{\text{s}}.$

See the figure. We’ve labeled the length of each side of the square

The square’s side-length changes as the situation progresses, so we’re calling that the variable

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dA}{dt},$ where

Open to read why the dx/dt is there.

Are you wondering why $\dfrac{dx}{dt}$ appears? The answer is the Chain Rule.
While the derivative of $x^2$ with respect to *x* is
$$\dfrac{d}{dx}x^2 = 2x,$$
the derivative of $x^2$ with respect to *time t* is
$$\dfrac{d}{dt}x^2 = 2x\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that*x* is a function of time *t*: the side-length *x* *changes* as time passes. We could have captured this time-dependence explicitly by writing our relation as
$$ A(t) = \big[ x(t)\big]^2$$
to remind ourselves that both *A* and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}A(t) &= \frac{d}{dt} \big[ x(t)\big]^2\\[8px]
\frac{dA(t)}{dt}&= 2 \big[ x(t)\big] \dfrac{dx(t)}{dt}
\end{align*}

Remember that

[Recall $\dfrac{dx(t)}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dA(t)}{dt}$ at the moment when $A(t) = 25 \text{ m}^2.$]
Most people find that writing the explicit time-dependence $A(t)$ and *x(t)* annoying, and so just write *A* and *x* instead. Regardless, you *must* remember that both *A* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.

[collapse]

We know that $$ \frac{dA}{dt}= 2x \dfrac{dx}{dt} $$ and we know that $\dfrac{dx}{dt} = 0.5 \, \tfrac{\text{m}}{\text{s}}$. We also know that we’re interested in the moment when $A = 25 \text{ m}^2;$ to complete our calculuation, we need to know the value of

Since $A = x^2,$ when $A = 25 \text{ m}^2:$ \[ \begin{align*} x^2 &= 25 \text{ m}^2\\[8px] x &= 5.0 \text{ m} \end{align*} \]

We can now complete our calculation: \[ \begin{align*} \frac{dA}{dt}&= 2x \dfrac{dx}{dt} \\[8px] &= 2(5.0 \text{ m}) \left(0.5 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 5.0 \, \tfrac{\text{m}^2}{\text{s}} \quad \cmark \end{align*} \]

[hide solution]

Related Rates: Cube, sides grow

A cube has side-length *x*. Each side increases at the rate of $0.05$ m/s.**(a)** At what rate is its volume increasing at the moment when $x = 0.4$ m?**(b)** At what rate is its surface area increasing at the moment when $x = 0.6$ m?**(a)** $0.024 \, \tfrac{\text{m}^3}{\text{s}}$**(b)** $0.36 \, \tfrac{\text{m}^2}{\text{s}}$

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It’s all free: our only aim is to support your learning via our community-supported and ad-free site. **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the length of each side of the cube*x.* Recall that each side increases at the rate $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}} $, and we’re interested in the moment when $x = 0.4$ m. We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The cube’s side-length changes as the situation progresses, so we’re calling that the variable*x.*

*B. To develop your equation, you will probably use . . . a simple geometric fact. *

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dV}{dt}.$ Hence we need to start with an expression that relates*V* to *x.* Since the object is a cube, we know that relationship:
$$ V = x^3 $$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}V &= \dfrac{d}{dt}x^3 \\[8px]
\dfrac{dV}{dt} &= 3x^2 \dfrac{dx}{dt}
\end{align*} \]

**4. Solve for the quantity you’re after.**

This is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ and are interested in the moment when $x(t) = 0.4$ m: \[ \begin{align*} \dfrac{dV}{dt} &= 3x^2 \dfrac{dx}{dt} \\[8px] &= 3(0.4 \text{ m})^2 \left(0.05 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 0.024 \, \tfrac{\text{m}^3}{\text{s}} \quad \cmark \end{align*} \] **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the length of each side of the cube*x.* Recall that each side increases at the rate $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}} $, and we’re interested in the moment when $x = 0.6$ m. We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The cube’s side-length changes as the situation progresses, so we’re calling that the variable*x.*

*B. To develop your equation, you will probably use . . . a simple geometric fact. *

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dA}{dt}.$ Hence we need to start with an expression that relates the cube’s area*A* to *x.* Since we have a cube, we know that relationship: the cube has 6 sides, each with area $x^2.$ Hence
$$ A = 6x^2 $$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}A &= \dfrac{d}{dt}\left( 6x^2\right) \\[8px]
\dfrac{dV}{dt} &= 12x \dfrac{dx}{dt}
\end{align*} \]

**4. Solve for the quantity you’re after.**

This is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ and are interested in the moment when $x(t) = 0.6$ m: \[ \begin{align*} \dfrac{dA}{dt} &= 2x \dfrac{dx}{dt} \\[8px] &= 12(0.6 \text{ m}) \left(0.05 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 0.36 \, \tfrac{\text{m}^2}{\text{s}} \quad \cmark \end{align*} \]

Show/Hide Solution

Solution SummarySolution (a) DetailSolution (b) Detail

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See the figure. We’ve labeled the length of each side of the cube

The cube’s side-length changes as the situation progresses, so we’re calling that the variable

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dV}{dt}.$ Hence we need to start with an expression that relates

Open to read why the dx/dt is there.

Are you wondering why $\dfrac{dx}{dt}$ appears? The answer is the Chain Rule.
While the derivative of $x^3$ with respect to *x* is
$$\dfrac{d}{dx}x^3 = 3x^2,$$
the derivative of $x^3$ with respect to *time t* is
$$\dfrac{d}{dt}x^3 = 3x^2\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that*x* is a function of time *t*: the side-length *x* *changes* as time passes. We could have captured this time-dependence explicitly by writing our relation as
$$ V(t) = \big[ x(t)\big]^3$$
to remind ourselves that both *V* and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}V(t) &= \frac{d}{dt} \big[ x(t)\big]^3\\[8px]
\frac{dV(t)}{dt}&= 3 \big[ x(t)\big]^2 \dfrac{dx}{dt}
\end{align*}

Remember that

[Recall $\dfrac{dx(t)}{dt} = .05 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dV(t)}{dt}$ at the moment when $x(t) = 0.4$ m.]
Most people find that writing the explicit time-dependence $V(t)$ and *x(t)* annoying, and so just write *V* and *x* instead. Regardless, you *must* remember that both *V* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.

[collapse]

This is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ and are interested in the moment when $x(t) = 0.4$ m: \[ \begin{align*} \dfrac{dV}{dt} &= 3x^2 \dfrac{dx}{dt} \\[8px] &= 3(0.4 \text{ m})^2 \left(0.05 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 0.024 \, \tfrac{\text{m}^3}{\text{s}} \quad \cmark \end{align*} \]

See the figure. We’ve labeled the length of each side of the cube

The cube’s side-length changes as the situation progresses, so we’re calling that the variable

In this problem, we’re given the value for $\dfrac{dx}{dt},$ and we’re looking for the value of $\dfrac{dA}{dt}.$ Hence we need to start with an expression that relates the cube’s area

Open to read why the dx/dt is there.

Are you wondering why $\dfrac{dx}{dt}$ appears? The answer is the Chain Rule.
While the derivative of $x^2$ with respect to *x* is
$$\dfrac{d}{dx}x^2 = 2x,$$
the derivative of $x^2$ with respect to *time t* is
$$\dfrac{d}{dt}x^2 = 2x\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that*x* is a function of time *t*: the side-length *x* *changes* as time passes. We could have captured this time-dependence explicitly by writing our relation as
$$ A(t) = 6\big[ x(t)\big]^2$$
to remind ourselves that both *V* and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}A(t) &= \frac{d}{dt} 6\big[ x(t)\big]^2\\[8px]
\frac{dV(t)}{dt}&= 12 [x(t)] \dfrac{dx}{dt}
\end{align*}

Remember that

[Recall $\dfrac{dx(t)}{dt} = .05 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dA(t)}{dt}$ at the moment when $x(t) = 0.6$ m.]
Most people find that writing the explicit time-dependence $A(t)$ and *x(t)* annoying, and so just write *A* and *x* instead. Regardless, you *must* remember that both *A* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.

[collapse]

This is straightforward in this problem. We have $\dfrac{dx}{dt} = 0.05 \, \tfrac{\text{m}}{\text{s}}$ and are interested in the moment when $x(t) = 0.6$ m: \[ \begin{align*} \dfrac{dA}{dt} &= 2x \dfrac{dx}{dt} \\[8px] &= 12(0.6 \text{ m}) \left(0.05 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px] &= 0.36 \, \tfrac{\text{m}^2}{\text{s}} \quad \cmark \end{align*} \]

[hide solution]

Related Rates: Cylinder drains

A cylinder filled with water has a 3.0-foot radius and 10-foot height. It is drained such that the depth of the water is decreasing at 0.1 feet per second. How fast is the water draining from the tank? **1. Draw a picture of the physical situation.**

See the figure. Let’s call the height (or depth) of the water at any given moment*y*, as shown.

We are told that the water level in the cup is decreasing at the rate of $0.1\, \tfrac{\text{ft}}{\text{s}}$, so $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$. Remember that we have to insert that negative sign “by hand” since the water’s height is*decreasing*.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The height of the water changes as time passes, so we’re calling that the variable*y*.

*B. To develop your equation, you will probably use . . . a simple geometric fact.*

The volume*V* of any cylinder is its circular cross-sectional area $\left(\pi r^2 \right)$ times its height. Here, at any moment the water’s height is *y*, and so the volume of water in the cylinder is:
$$V = \pi r^2y$$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

Note that the cylinder’s radius,*r*, is constant (*r* = 3.0 ft), so we’ll treat it as a constant when we take the derivative. By contrast, the water’s height *y* is *not* constant; instead it changes, and indeed, it changes at the rate $\dfrac{dy}{dt}$.
\begin{align*}
\frac{dV}{dt} &= \frac{d}{dt}\left(\pi r^2 y \right) \\ \\
&= \pi r^2 \frac{d}{dt}(y) \\ \\
&= \pi r^2 \frac{dy}{dt} \\ \\
\end{align*}

**4. Solve for the quantity you’re after.**

At this point we’re just substituting values: the problem told us $r = 3.0 \, \text{ft}$ and $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for $\dfrac{dV}{dt}$. Starting from our last expression above: \begin{align*} \frac{dV}{dt} &= \pi r^2 \frac{dy}{dt} \\ \\ &= \pi (3.0\, \text{ft})^2 \left(-0.1\, \tfrac{\text{ft}}{\text{s}}\right) \\ \\ &= -2.8 \, \tfrac{\text{ft}^3}{\text{s}} \quad \cmark \end{align*} That is, water is draining from the tank at the rate $\dfrac{dV}{dt} = -2.8\, \tfrac{\text{ft}^3}{\text{s}}$. The negative value indicates that the water’s volume in the tank*V* is decreasing, which is correct.

**Caution**: IF you are using a web-based homework system and the question asks,

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Show/Hide Solution

See the figure. Let’s call the height (or depth) of the water at any given moment

We are told that the water level in the cup is decreasing at the rate of $0.1\, \tfrac{\text{ft}}{\text{s}}$, so $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$. Remember that we have to insert that negative sign “by hand” since the water’s height is

The height of the water changes as time passes, so we’re calling that the variable

The volume

Note that the cylinder’s radius,

Open to read why the dy/dt is there.

Are you wondering why $\dfrac{dy}{dt}$ appears? The answer is the Chain Rule. While the derivative of $y$ with respect to *y* is
$$\dfrac{d}{dy}y = 1,$$
the derivative of $y$ with respect to *time t* is
$$\dfrac{d}{dt}y = \dfrac{dy}{dt}.$$
(Recall that that rate is $\dfrac{dy}{dt} = -0.1 \, \tfrac{\text{ft}}{\text{s}}$ in this problem.) Remember that *y* is a function of time *t*: the water’s depth *y* *changes* as time passes. We could have captured this time-dependence explicitly by writing our relation as
$$ V(t) = \pi r^2y(t)$$
to remind ourselves that both *V* and *y* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}V(t) &= \frac{d}{dt} \big[\pi r^2 y(t)\big]\\[8px]
\dfrac{dV(t)}{dt} &= \pi r^2\frac{d}{dt} \big[ y(t)\big]\\[8px]
&= \pi r^2 \dfrac{dy(t)}{dt}
\end{align*}

[Recall $\dfrac{dy(t)}{dt} = -0.1 \, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for $\dfrac{dV(t)}{dt}.$]
Most people find that writing the explicit time-dependence $V(t)$ and *y(t)* annoying, and so just write *V* and *y* instead. Regardless, you *must* remember that both *V* and *y* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dy}{dt}$ term.

[collapse]

At this point we’re just substituting values: the problem told us $r = 3.0 \, \text{ft}$ and $\dfrac{dy}{dt} = -0.1\, \tfrac{\text{ft}}{\text{s}}$, and we’re looking for $\dfrac{dV}{dt}$. Starting from our last expression above: \begin{align*} \frac{dV}{dt} &= \pi r^2 \frac{dy}{dt} \\ \\ &= \pi (3.0\, \text{ft})^2 \left(-0.1\, \tfrac{\text{ft}}{\text{s}}\right) \\ \\ &= -2.8 \, \tfrac{\text{ft}^3}{\text{s}} \quad \cmark \end{align*} That is, water is draining from the tank at the rate $\dfrac{dV}{dt} = -2.8\, \tfrac{\text{ft}^3}{\text{s}}$. The negative value indicates that the water’s volume in the tank

At what rate does the volume *decrease*?

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[hide solution]

Related Rates: Using a Trig Function

If a problem asks you how fast an angle is changing, you *must* use a trigonometric relationship to related the angle to other changing quantities. That is, you'll use one of these

\begin{align*} \tan{\theta} &= \frac{\text{opposite}}{\text{adjacent}} \\[12px] \cos{\theta} &= \frac{\text{adjacent}}{\text{hypotenuse}} \\[12px] \sin{\theta} &= \frac{\text{opposite}}{\text{hypotenuse}} \end{align*} The following problems demonstrate.

\begin{align*} \tan{\theta} &= \frac{\text{opposite}}{\text{adjacent}} \\[12px] \cos{\theta} &= \frac{\text{adjacent}}{\text{hypotenuse}} \\[12px] \sin{\theta} &= \frac{\text{opposite}}{\text{hypotenuse}} \end{align*} The following problems demonstrate.

Related Rates: Ladder slides, angle changes

A 10-foot long ladder leans against a wall. The bottom of the ladder slides away from the wall at $\frac{3}{2}$ ft/s. How fast is the angle between the ladder and the ground changing at the moment when the ladder is 6 feet from the wall?

Show/Hide Solution

Answer: $\dfrac{-3}{16}$ rad/s

**1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the ladder’s position along the ground*x*, and its position along the wall *y*. These both change as the situation unfolds, so we’ve left them as variables. (The ladder’s length of 10 feet, by contrast, is a constant.) We’ve also labeled the angle $\theta$ that the ladder makes with the ground, since the problem is asking us to find its rate of change at a particular instant.

**2. Write an equation that relates the quantities of interest.**

*B. To develop your equation, you will probably use . . . a trigonometric function (like $\cos{\theta}$ = adjacent/hypotenuse). *

We need to use a trigonometric function to analyze this situation, since we need to relate the angle $\theta$ to other quantities in order to have $\dfrac{d \theta}{dt}$ when we take the derivative. We’re given the rate at which the ladder’s bottom position changes $\left(\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}\right)$, and so it seems easiest to go with $\cos \theta$ since that directly relates $\theta$ and*x*:

$$\cos{\theta} = \frac{x}{10}$$

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*} \frac{d}{dt}( \cos{\theta}) &= \frac{d}{dt} \left( \frac{x}{10} \right) \\ \\ -\sin{\theta}\, \frac{d\theta}{dt} &= \frac{1}{10}\frac{dx}{dt} \end{align*}

**4. Solve for the quantity you’re after.**

We’re after $\dfrac{d\theta}{dt}$ at the instant $x = 6$ ft. So let’s solve the preceding equation for $\dfrac{d\theta}{dt}$:

$$\frac{d\theta}{dt} = -\frac{1}{10} \frac{1}{\sin{\theta}} \frac{dx}{dt}$$ We’re given that $\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}$. But we don’t yet know what $\sin \theta$ is when $x = 6$ ft.

*Begin subproblem to find $\sin \theta$.*

We know $$\sin \theta = \frac{y}{10}$$ So we need to know the value of $y$ at the instant when $x = 6$ ft. We can find this by using the Pythagorean theorem:

\begin{align*} (6)^2 + y^2 &= 10^2 \\[8px] 36 + y^2 &= 100 \\[8px] y^2 &= 100 – 36 = 64 \\[8px] y &= 8 \end{align*} Hence at this instant $$\sin\theta = \dfrac{8}{10} = \dfrac{4}{5}$$*End subproblem.*

Hence \begin{align*} \frac{d\theta}{dt} & = -\frac{1}{10} \, \frac{1}{\sin{\theta}}\, \frac{dx}{dt}\\[12px] &= -\frac{1}{10}\cdot \left(\frac{1}{4/5} \right)\cdot \dfrac{3}{2}\\[12px] &= – \frac{1}{10}\cdot \frac{5}{4}\cdot \dfrac{3}{2}\\[12px] &= – \frac{3}{16} \text{ rad/s} \quad \cmark \end{align*} The negative value indicates that the angle is decreasing at the ladder slides down the wall.

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It’s all free: our only aim is to support your learning via our community-supported and ad-free site.

See the figure. We’ve labeled the ladder’s position along the ground

We need to use a trigonometric function to analyze this situation, since we need to relate the angle $\theta$ to other quantities in order to have $\dfrac{d \theta}{dt}$ when we take the derivative. We’re given the rate at which the ladder’s bottom position changes $\left(\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}\right)$, and so it seems easiest to go with $\cos \theta$ since that directly relates $\theta$ and

$$\cos{\theta} = \frac{x}{10}$$

\begin{align*} \frac{d}{dt}( \cos{\theta}) &= \frac{d}{dt} \left( \frac{x}{10} \right) \\ \\ -\sin{\theta}\, \frac{d\theta}{dt} &= \frac{1}{10}\frac{dx}{dt} \end{align*}

We’re after $\dfrac{d\theta}{dt}$ at the instant $x = 6$ ft. So let’s solve the preceding equation for $\dfrac{d\theta}{dt}$:

$$\frac{d\theta}{dt} = -\frac{1}{10} \frac{1}{\sin{\theta}} \frac{dx}{dt}$$ We’re given that $\dfrac{dx}{dt} = \dfrac{3}{2} \, \tfrac{\text{ft}}{\text{s}}$. But we don’t yet know what $\sin \theta$ is when $x = 6$ ft.

We know $$\sin \theta = \frac{y}{10}$$ So we need to know the value of $y$ at the instant when $x = 6$ ft. We can find this by using the Pythagorean theorem:

\begin{align*} (6)^2 + y^2 &= 10^2 \\[8px] 36 + y^2 &= 100 \\[8px] y^2 &= 100 – 36 = 64 \\[8px] y &= 8 \end{align*} Hence at this instant $$\sin\theta = \dfrac{8}{10} = \dfrac{4}{5}$$

Hence \begin{align*} \frac{d\theta}{dt} & = -\frac{1}{10} \, \frac{1}{\sin{\theta}}\, \frac{dx}{dt}\\[12px] &= -\frac{1}{10}\cdot \left(\frac{1}{4/5} \right)\cdot \dfrac{3}{2}\\[12px] &= – \frac{1}{10}\cdot \frac{5}{4}\cdot \dfrac{3}{2}\\[12px] &= – \frac{3}{16} \text{ rad/s} \quad \cmark \end{align*} The negative value indicates that the angle is decreasing at the ladder slides down the wall.

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Related Rates: Kite, angle changes

A kite flies 30 meters above ground. It travels horizontally at 2 meters per second. At what rate is the angle between the string and the ground changing when the length of string to the kite is 50 meters? **1. Draw a picture of the physical situation.**

See the figure. We’ve called the length of string to the kite $\ell.$ And we’ve labeled the angle $\theta$ that the string makes with the ground, since the problem is asking us to find the rate at which that angle changes, $\dfrac{d\theta}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the*x*-direction, at the rate $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}} $. We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

*B. To develop your equation, you will probably use . . . a trigonometric function (like $\tan{\theta}$ = opposite/adjacent). *

In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which $\theta$ is changing, $\dfrac{d\theta}{dt}$, to the rate at which*x* is changing, $\dfrac{dx}{dt}$, and so we first need to write down an equation that somehow relates $\theta$ and *x*. Such a relation *must* be trigonometric.

Specifically, we notice that*x* is the side of the triangle that is adjacent to the angle. Furthermore, the opposite leg of the triangle remains constant throughout the problem, since the kite’s height is always 30 m. Hence at every moment:
$$\tan{\theta} = \frac{30}{x}$$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \begin{align*}
\frac{d}{dt} \tan{\theta} &= \frac{d}{dt} \left( \frac{30}{x} \right) \\[8px]
\sec^2{\theta}\, \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}
\end{align*}

**4. Solve for the quantity you’re after.**

Let’s solve the preceding equation for $\dfrac{d\theta}{dt}$: \[ \begin{align*} \sec^2{\theta}\, \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\\[8px] \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\frac{1}{\sec^2{\theta}} \\[8px] &= -\frac{30}{x^2}\frac{dx}{dt}\cos^2 {\theta} \end{align*} \] Notice that $\cos{\theta} = \dfrac{x}{\ell}$, so \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\left( \dfrac{x}{\ell}\right) ^2 \\[8px] &= -\frac{30}{\cancel{x^2}}\frac{dx}{dt}\frac{\cancel{x^2}}{\ell^2} \\[8px] &= -\frac{30}{\ell^2}\frac{dx}{dt} \end{align*} \] We’re looking for $\dfrac{d\theta}{dt}$ at the instant when $\ell =50$ m. Recall that Recall $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{(50 \text{ m})^2 } \left( 2 \, \tfrac{\text{m}}{\text{s}}\right) \\[8px] &= -0.02 \, \tfrac{\text{rad}}{\text{s}} \quad \cmark \end{align*} \] That’s the answer. The negative value indicates that the angle is decreasing at the kite travels further out, as we expect.

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See the figure. We’ve called the length of string to the kite $\ell.$ And we’ve labeled the angle $\theta$ that the string makes with the ground, since the problem is asking us to find the rate at which that angle changes, $\dfrac{d\theta}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the

The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which $\theta$ is changing, $\dfrac{d\theta}{dt}$, to the rate at which

Specifically, we notice that

Open to read why the d(theta)/dt and dx/dt are there.

Are you wondering why the $\dfrac{d\theta}{dt}$ and $\dfrac{dx}{dt}$ appear? The answer is the Chain Rule. While the derivative of $\tan \theta$ with respect to angle $\theta$ is
$$\dfrac{d}{d\theta}\tan \theta = \sec^2 \theta,$$
the derivative of $\tan \theta$ with respect to *time t* is
$$\dfrac{d}{dt}\tan \theta = \sec^2 \theta \,\dfrac{d\theta}{dt}.$$
Similarly, while the derivative of $\dfrac{1}{x}$ with respect to *x* is
$$\dfrac{d}{dx}\left(\dfrac{1}{x} \right) = -\dfrac{1}{x^2},$$
the derivative of $\dfrac{1}{x}$ with respect to *time t* is
$$\dfrac{d}{dt}\left(\dfrac{1}{x} \right) = -\dfrac{1}{x^2}\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that $\theta$ and*x* are both functions of time *t*: the angle *changes* as time passes and the kite’s *x*-position *changes* as the kite travels horizontally. We could have captured this time-dependence explicitly by writing our relation as
$$ \tan \theta(t) = \dfrac{30}{x(t)}$$
to remind ourselves that both $\theta$ and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}\tan \theta(t) &= \frac{d}{dt}\left( \dfrac{30}{x(t)}\right) \\ \\
\left(\sec^2 \theta(t)\right) \dfrac{d\theta(t)}{dt}&= -\frac{30}{[x(t)]^2} \dfrac{dx(t)}{dt}
\end{align*}

Remember that $\theta$ and

[Recall $\dfrac{dx(t)}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{d\theta(t)}{dt}$ at the moment when $\ell = 50$ m.]
Most people find that writing the explicit time-dependence $\theta(t)$ and *x(t)* annoying, and so just write $\theta$ and *x* instead. Regardless, you *must* remember that both $\theta$ and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{d\theta}{dt}$ and $\dfrac{dx}{dt}$ terms.

[collapse]

Let’s solve the preceding equation for $\dfrac{d\theta}{dt}$: \[ \begin{align*} \sec^2{\theta}\, \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\\[8px] \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\frac{1}{\sec^2{\theta}} \\[8px] &= -\frac{30}{x^2}\frac{dx}{dt}\cos^2 {\theta} \end{align*} \] Notice that $\cos{\theta} = \dfrac{x}{\ell}$, so \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{x^2}\frac{dx}{dt}\left( \dfrac{x}{\ell}\right) ^2 \\[8px] &= -\frac{30}{\cancel{x^2}}\frac{dx}{dt}\frac{\cancel{x^2}}{\ell^2} \\[8px] &= -\frac{30}{\ell^2}\frac{dx}{dt} \end{align*} \] We’re looking for $\dfrac{d\theta}{dt}$ at the instant when $\ell =50$ m. Recall that Recall $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \frac{d\theta}{dt} &= -\frac{30}{(50 \text{ m})^2 } \left( 2 \, \tfrac{\text{m}}{\text{s}}\right) \\[8px] &= -0.02 \, \tfrac{\text{rad}}{\text{s}} \quad \cmark \end{align*} \] That’s the answer. The negative value indicates that the angle is decreasing at the kite travels further out, as we expect.

At what rate does the angle *decrease*?

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Related Rates: Rocket, angle changes

A rocket is launched straight up, and its altitude is $h = 5 t^2$ meters after $t$ seconds. You are on the ground 500 meters from the launch site. The line of sight from you to the rocket makes an angle $\theta$ with the horizontal.

By how many radians per second is $\theta$ changing ten seconds after the launch?

By how many radians per second is $\theta$ changing ten seconds after the launch?

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Answer: $\dfrac{1}{10}$ rad/s

**1. Draw a picture of the physical situation.**

See the figure.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

In this case, the height changes as the rocket travels, so keep calling it $h = 5t^2$ for now.

The picture immediately suggests a relation between the angle $\theta$ and the height $h$.

$$\tan{\theta} = \frac{h}{500} = \frac{5 t^2}{500} = \frac{t^2}{100} $$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

Remember that $\theta$ is changing with time too, and so we have to use the chain rule when we take the derivative of $\tan{\theta}$, giving a $\dfrac{d\theta}{dt}$ term:

\begin{align*} \frac{d}{dt}\tan\theta &= \frac{d}{dt} \left(\frac{t^2}{100} \right)\\ \\ \sec^2{\theta} \cdot \frac{d\theta}{dt} &= \frac{2t}{100} && [\frac{d}{d\theta} \tan{\theta} = \sec^2{\theta}] \\ \\ \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \text{ } [*]&& [\frac{1}{\sec^2{\theta}} = \cos^2{\theta}] \\ \end{align*}**4. Solve for the quantity you’re after.**

In order to use the preceding equation, we need to know $\theta$ (so we can compute $\cos^2 \theta$). To find $\theta$, we return to our our figure and recall that $\tan{\theta} = \dfrac{h}{500}$. Hence if we find $h$, we can then find $\theta$.

When $t =10$, the rocket’s height is:

$$h(10) = 5 t^2 = 5 \cdot (10)^2 = 500$$ Thus $$\tan{\theta} = \frac{h}{500} = \frac{500}{500} = 1$$ and so at $t = 10$, $\theta = \pi/4$.

We can now solve for $\dfrac{d\theta}{dt}$ at $t = 10$ seconds using the equation we found above marked [*]:

\begin{align*} \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \\ \\ \left. \frac{d\theta}{dt} \right|_{t=10} &= \frac{10}{50} \cos^2{\frac{\pi}{4}} \\ \\ &= \frac{1}{5} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{5} \left(\frac{1}{2}\right) \\ \\ &= \frac{1}{10} \text{ rad/s} \quad \cmark \end{align*}

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See the figure.

In this case, the height changes as the rocket travels, so keep calling it $h = 5t^2$ for now.

The picture immediately suggests a relation between the angle $\theta$ and the height $h$.

$$\tan{\theta} = \frac{h}{500} = \frac{5 t^2}{500} = \frac{t^2}{100} $$

Remember that $\theta$ is changing with time too, and so we have to use the chain rule when we take the derivative of $\tan{\theta}$, giving a $\dfrac{d\theta}{dt}$ term:

\begin{align*} \frac{d}{dt}\tan\theta &= \frac{d}{dt} \left(\frac{t^2}{100} \right)\\ \\ \sec^2{\theta} \cdot \frac{d\theta}{dt} &= \frac{2t}{100} && [\frac{d}{d\theta} \tan{\theta} = \sec^2{\theta}] \\ \\ \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \text{ } [*]&& [\frac{1}{\sec^2{\theta}} = \cos^2{\theta}] \\ \end{align*}

In order to use the preceding equation, we need to know $\theta$ (so we can compute $\cos^2 \theta$). To find $\theta$, we return to our our figure and recall that $\tan{\theta} = \dfrac{h}{500}$. Hence if we find $h$, we can then find $\theta$.

When $t =10$, the rocket’s height is:

$$h(10) = 5 t^2 = 5 \cdot (10)^2 = 500$$ Thus $$\tan{\theta} = \frac{h}{500} = \frac{500}{500} = 1$$ and so at $t = 10$, $\theta = \pi/4$.

We can now solve for $\dfrac{d\theta}{dt}$ at $t = 10$ seconds using the equation we found above marked [*]:

\begin{align*} \frac{d\theta}{dt} &= \frac{t}{50} \cos^2{\theta} \\ \\ \left. \frac{d\theta}{dt} \right|_{t=10} &= \frac{10}{50} \cos^2{\frac{\pi}{4}} \\ \\ &= \frac{1}{5} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{5} \left(\frac{1}{2}\right) \\ \\ &= \frac{1}{10} \text{ rad/s} \quad \cmark \end{align*}

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Related Rates: Using Similar Triangles

A lot of problems you encounter will require that, when you look at your figure, you see a smaller triangle embedded in a larger triangle. Because the smaller triangle and the larger triangle have identical angles, they are *similar triangles*, and hence the ratios of the corresponding sides are equal.

The following problems illustrate.

The following problems illustrate.

Related Rates: Man walks, shadow on ground moves

A 1.8-meter tall man walks toward a 6.0-meter light pole at the rate of 1.5 m/s. The light at the top of the pole casts a shadow behind the man. How fast is the "head" of his shadow moving along the ground?

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Answer: $-2.1$ m/s

**1. Draw a picture of the physical situation.** See the figure. We’re calling the distance between the pole and the “head” of the man’s shadow $\ell$, and the distance between the man and the pole *x*.

**2. Write an equation that relates the quantities of interest.**

We are given that the man is walking toward the pole at the rate $\dfrac{dx}{dt} = -1.5$ m/s. (We’ve made the value negative because the distance*x* is decreasing as the man walks toward the pole.) We are looking for the rate at which the “head” of the man’s shadow moves, which is $\dfrac{d \ell}{dt}$. We thus need to somehow relate $\ell$ to *x*, so we can then develop the relationship between their time-derivatives.

There’s a subtlety to this problem that typically goes unaddressed: We’re focusing on $\ell$ and $\dfrac{d \ell}{dt}$ here because $\ell$ is the distance from the shadow’s tip to the*stationary pole*. We’re not examining the shadow’s length itself (labeled $\ell – x$ in the left figure below) because that length is relative to the man’s feet, which are also moving. So we’d find a different answer if we calculated the rate at which that gray shadow is changing. This problem asks us to find the rate the shadow’s head as it moves along the (stationary) ground, so it’s best to make our measurements from a point that isn’t also moving—namely, from the pole. Hence we focus on $\ell$ and aim to compute $\dfrac{d \ell}{dt}$.

*B. To develop your equation, you will probably use . . . similar triangles.*

In the figure above we’ve separated out the two triangles. Notice that the angles are identical in the two triangles, and hence they are similar. The ratio of their respective components are thus equal as well. Hence the ratio of their bases $\left(\dfrac{\ell – x}{\ell} \right)$ is equal to the ratio of their heights $\left( \dfrac{1.8\, \text{m}}{6.0\, \text{m}}\right)$: \begin{align*} \dfrac{\ell – x}{\ell} &= \frac{1.8 \, \text{m}}{6.0 \, \text{m}} \\[12px] &= 0.30 \\[12px] \ell – x &= 0.30 \ell \\[12px] \ell – 0.30 \ell &= x \\[12px] 0.70 \ell &= x \end{align*}**3. Take the derivative with respect to time of both sides of your equation. **

\begin{align*} \dfrac{d}{dt}(0.70 \ell) &= \dfrac{d}{dt}(x) \\[12px] 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \end{align*}**4. Solve for the quantity you’re after.**

We’re looking for $\dfrac{d \ell}{dt}$:

\begin{align*} 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \\[12px] \dfrac{d \ell}{dt} &= \frac{1}{0.70} \dfrac{dx}{dt} \\[12px] &= \frac{1}{0.70} \left( -1.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[12px] &= -2.1\, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*}

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We are given that the man is walking toward the pole at the rate $\dfrac{dx}{dt} = -1.5$ m/s. (We’ve made the value negative because the distance

There’s a subtlety to this problem that typically goes unaddressed: We’re focusing on $\ell$ and $\dfrac{d \ell}{dt}$ here because $\ell$ is the distance from the shadow’s tip to the

In the figure above we’ve separated out the two triangles. Notice that the angles are identical in the two triangles, and hence they are similar. The ratio of their respective components are thus equal as well. Hence the ratio of their bases $\left(\dfrac{\ell – x}{\ell} \right)$ is equal to the ratio of their heights $\left( \dfrac{1.8\, \text{m}}{6.0\, \text{m}}\right)$: \begin{align*} \dfrac{\ell – x}{\ell} &= \frac{1.8 \, \text{m}}{6.0 \, \text{m}} \\[12px] &= 0.30 \\[12px] \ell – x &= 0.30 \ell \\[12px] \ell – 0.30 \ell &= x \\[12px] 0.70 \ell &= x \end{align*}

\begin{align*} \dfrac{d}{dt}(0.70 \ell) &= \dfrac{d}{dt}(x) \\[12px] 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \end{align*}

We’re looking for $\dfrac{d \ell}{dt}$:

\begin{align*} 0.70 \dfrac{d \ell}{dt} &= \dfrac{dx}{dt} \\[12px] \dfrac{d \ell}{dt} &= \frac{1}{0.70} \dfrac{dx}{dt} \\[12px] &= \frac{1}{0.70} \left( -1.5 \, \tfrac{\text{m}}{\text{s}}\right) \\[12px] &= -2.1\, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*}

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Related Rates: Man walks, shadow on wall changes length

A 1.8-meter tall man walks toward a wall at 1.5 m/s. There is a spotlight on the ground behind him, 20 meters away from the wall, that projects his shadow onto the wall. At what rate is the length of his shadow changing when he is 5.0 meters from the wall? **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the distance from the spotlight to the man*x.* The man is thus moving at the rate $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}.$ We’re calling the height of his shadow on the wall *y.* The problem is asking us to find $\dfrac{dy}{dt}$ at a particular moment, when the man is 5.0 meters from the wall. We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The man’s position changes as he walks, so we’re calling his distance from the spotlight*x.*

*B. To develop your equation, you will probably use . . . similar triangles. *

The problem gives us the rate $\dfrac{dx}{dt}$ and asks us to find the rate $\dfrac{dy}{dt}.$ We thus need to first relate*y* to *x.* The figure above suggests that we can use similar triangles to do so. Here we’ve separated out the two triangles. Notice that their angles are identical, and hence the triangles are similar. The ratio of their respective components are thus equal: the ratio of their heights $\dfrac{y}{1.8}$ is equal to the ratio of their bases $\dfrac{20}{x}.$ That is,
\[ \begin{align*}
\frac{y}{1.8} &= \frac{20}{x} \\[8px]
y &= \frac{(1.8)(20)}{x} \\[8px]
&= \frac{36}{x}
\end{align*} \]
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\[ \begin{align*} \frac{d}{dt}y &= \frac{d}{dt}\left( \frac{36}{x}\right) \\[8px] \frac{dy}{dt} &= \frac{-36}{x^2}\, \dfrac{dx}{dt} \end{align*} \] **4. Solve for the quantity you’re after.**

We’re given that $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$. The problem asks us to find $\dfrac{dy}{dt}$ at the moment the man is 5.0 feet*from the wall.* Hence at the moment of interest the value of *x* is:
\[ \begin{align*}
x &= 20 \text{ m} – 5.0 \text{ m} \\[8px]
&= 15.0 \text{ m}
\end{align*} \]
Finally, then:
\[ \begin{align*}
\frac{dy}{dt} &= \frac{-36}{x^2}\, \dfrac{dx}{dt} \\[8px]
&= \frac{-36}{(15.0)^2} \left(1.5 \, \tfrac{\text{m}}{\text{s}} \right) \\[8px]
&= -0.24 \, \tfrac{\text{m}}{\text{s}} \quad \cmark
\end{align*} \]
That is, the man’s shadow is changing at the rate $\dfrac{dy}{dt} = -0.24\, \tfrac{\text{m}}{\text{s}}$. The negative value indicates that the shadow’s height *y* is decreasing, as we expect.

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Show/Hide Solution

See the figure. We’ve labeled the distance from the spotlight to the man

The man’s position changes as he walks, so we’re calling his distance from the spotlight

The problem gives us the rate $\dfrac{dx}{dt}$ and asks us to find the rate $\dfrac{dy}{dt}.$ We thus need to first relate

\[ \begin{align*} \frac{d}{dt}y &= \frac{d}{dt}\left( \frac{36}{x}\right) \\[8px] \frac{dy}{dt} &= \frac{-36}{x^2}\, \dfrac{dx}{dt} \end{align*} \]

Open to read why the dx/dt is there.

Are you wondering why $\dfrac{dx}{dt}$ appears? The answer is the Chain Rule.
While the derivative of $\dfrac{1}{x}$ with respect to *x* is
$$\dfrac{d}{dx}\left(\dfrac{1}{x} \right) = \dfrac{-1}{x^2},$$
the derivative of $\dfrac{1}{x}$ with respect to *time t* is
$$\dfrac{d}{dt}\left(\dfrac{1}{x} \right) = \dfrac{-1}{x^2}\dfrac{dx}{dt}.$$
(Recall that that rate is $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$ in this problem.)

Remember that*x* is a function of time *t*: the man’s position *x* *changes* as time passes, which in turn causes the length of his shadow to change. We could have captured this time-dependence explicitly by writing our relation as
$$ y(t) = \dfrac{36}{x(t)}$$
to remind ourselves that both *y* and *x* are functions of time *t*. Then when we take the derivative,
\begin{align*}
\frac{d}{dt}y(t) &= \frac{d}{dt}\dfrac{36}{x(t)} \\[8px]
\frac{dy}{dt}&= \dfrac{-36}{\left[x(t) \right]^2} \, \dfrac{dx}{dt}
\end{align*}

Remember that

[Recall $\dfrac{dx(t)}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$, and we’re looking for $\dfrac{dy(t)}{dt}$ at the moment when $x(t) = 5.0$ m.]
Most people find that writing the explicit time-dependence $y(t)$ and *x(t)* annoying, and so just write *y* and *x* instead. Regardless, you *must* remember that both *y* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ term.

[collapse]

We’re given that $\dfrac{dx}{dt} = 1.5 \, \tfrac{\text{m}}{\text{s}}$. The problem asks us to find $\dfrac{dy}{dt}$ at the moment the man is 5.0 feet

At what rate does the shadow’s height *decrease*?

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Related Rates: Water fills a cone

Water is poured at a uniform rate of $15 \, \tfrac{\text{cm}^3}{\text{s}}$ into a cup whose inside is shaped like a cone. The radius of the opening is 6 cm, and the height of the cup is 16 cm. How fast is the water level rising when the water is halfway up? (Note: The volume of a cone is $\dfrac{1}{3}\pi r^{2}h$. You may leave $\pi$ in your answer; do not use a calculator to find a decimal answer.) **Answer:** $\dfrac{5}{3\pi}$ cm/s

**1. Draw a picture of the physical situation.**

See the top figure.

**2. Write an equation that relates the quantities of interest.**

We are given that the volume of water in the cup is increasing at the rate $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$. We want to find the rate at which the water is rising, $\dfrac{dh}{dt}$, at the instant when the water is halfway up the cup, so when $h = 8$ cm.

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.* The height of the water changes as time passes, as does the water’s surface area, so we’re going to keep both of those as a variables, *h* and *r*.

We have a relation between the volume of water in the cup at any moment and those variables:

$$V = \frac{1}{3} \pi r^2h $$*B. To develop your equation, you will probably use . . . similar triangles.*

We don’t want the radius $r$ to appear as a separate variable, so we need to eliminate it. We can do so by using the similar triangles shown in the lower figure. Because the small triangle is similar to the larger triangle it’s embedded in, the ratio of their bases $\left(\dfrac{r}{6} \right)$ must be equal to the ratio of their heights $\left(\dfrac{h}{16} \right)$

\begin{align*} \frac{r}{6} &= \frac{h}{16} \\[12px] r &= \frac{6}{16}h\\[12px] &=\frac{3}{8} h \end{align*} Then substituting this expression into our relation for $V$:

$$V = \frac{1}{3} \pi \left(\frac{3}{8} h \right)^2h = \frac{3}{64} \pi h^3$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*} \frac{dV}{dt} &= \frac{d}{dt}\left(\frac{3}{64} \pi h^3 \right) \\ \\ &= \frac{3}{64} \pi \left(3h^2 \frac{dh}{dt} \right) \\ \\ &= \frac{9}{64} \pi h^2 \frac{dh}{dt} \end{align*}**4. Solve for the quantity you’re after.**

We have $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$, and want to find $\dfrac{dh}{dt}$, at the instant when $h = 8$ cm.

\begin{align*} \frac{dh}{dt} &= \frac{64}{9\pi h^2} \frac{dV}{dt} \\ \\ &= \frac{64}{9\pi (8)^2} (15) \\ \\ &= \frac{64}{3\pi (64)} (5) \\ \\ &= \frac{5}{3\pi} \text{ cm/s} \quad \cmark \end{align*}

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See the top figure.

We are given that the volume of water in the cup is increasing at the rate $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$. We want to find the rate at which the water is rising, $\dfrac{dh}{dt}$, at the instant when the water is halfway up the cup, so when $h = 8$ cm.

We have a relation between the volume of water in the cup at any moment and those variables:

$$V = \frac{1}{3} \pi r^2h $$

We don’t want the radius $r$ to appear as a separate variable, so we need to eliminate it. We can do so by using the similar triangles shown in the lower figure. Because the small triangle is similar to the larger triangle it’s embedded in, the ratio of their bases $\left(\dfrac{r}{6} \right)$ must be equal to the ratio of their heights $\left(\dfrac{h}{16} \right)$

\begin{align*} \frac{r}{6} &= \frac{h}{16} \\[12px] r &= \frac{6}{16}h\\[12px] &=\frac{3}{8} h \end{align*} Then substituting this expression into our relation for $V$:

$$V = \frac{1}{3} \pi \left(\frac{3}{8} h \right)^2h = \frac{3}{64} \pi h^3$$

\begin{align*} \frac{dV}{dt} &= \frac{d}{dt}\left(\frac{3}{64} \pi h^3 \right) \\ \\ &= \frac{3}{64} \pi \left(3h^2 \frac{dh}{dt} \right) \\ \\ &= \frac{9}{64} \pi h^2 \frac{dh}{dt} \end{align*}

We have $\dfrac{dV}{dt} = 15 \, \tfrac{\text{cm}^3}{\text{s}}$, and want to find $\dfrac{dh}{dt}$, at the instant when $h = 8$ cm.

\begin{align*} \frac{dh}{dt} &= \frac{64}{9\pi h^2} \frac{dV}{dt} \\ \\ &= \frac{64}{9\pi (8)^2} (15) \\ \\ &= \frac{64}{3\pi (64)} (5) \\ \\ &= \frac{5}{3\pi} \text{ cm/s} \quad \cmark \end{align*}

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Related Rates: Using the Pythagorean theorem

And probably more than any other approach, you will find yourself invoking the Pythagorean theorem often. The typical clue to use this approach will be that you've drawn a right triangle, and are asked something about a distance that happens to equal the hypotenuse.

The following problems illustrate.

The following problems illustrate.

Related Rates: Morphing rectangle

A rectangle has constant area 500 square centimeters. Its length is increasing at the rate of 8 centimeters per second.
At what rate is the diagonal of the rectangle changing at the instant when its width is 10 centimeters? (Do not use a calculator. You may leave a square root in your answer.) **1. Draw a picture of the physical situation.** See the figure.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet. * Both the width and the length of the rectangle are changing, so let’s leave them both as variables.

*B. To develop your equation, you will probably use . . .the Pythagorean theorem.* We need to relate the rectangle’s diagonal to its width and length:

$$d^2 = w^2 + l^2$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

Both $w$ and $l$ are changing with time, and so when we take the derivative we must include $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}d^2 &= \frac{d}{dt}(w^2 + l^2) \\[12px] 2d \frac{dd}{dt} &= 2w \frac{dw}{dt} + 2l \frac{dl}{dt} \\[12px] d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \text{ [*]} \end{align*}**4. Solve for the quantity you’re after.**

We want to find $\dfrac{dd}{dt}$ at the instant when $w = 10$ cm. The problem statement tells us that $\dfrac{dl}{dt} = 8$ cm/s. In order to use the preceding equation (marked [*]), we need to determine $d$ and $l$ at this instant, and also $\dfrac{dw}{dt}$ at this instant.

First, we can find $l$ by noting that the rectangle has constant area, $A = 500 = lw$. Hence when $w = 10$ cm:

\begin{align*} l &= \frac{A}{w} \\[8px] &= \frac{500}{10} = 50 \text{ cm} \end{align*} Second, we can find $d$ at this instant using the Pythagorean theorem:

\begin{align*} d^2 &= w^2 + l^2 \\[12px] &= (10)^2 + (50)^2 = 100 + 2500 = 2600 \\[12px] d &= \sqrt{2600} = 10\sqrt{26} \end{align*} Next, we can find $\dfrac{dw}{dt}$ at this instant by taking the derivative of $A = 500 = lw$ :

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(lw) \\[12px] 0 &= l \frac{dw}{dt} + w\frac{dl}{dt} \\[12px] l \frac{dw}{dt} &= – w\frac{dl}{dt} \end{align*} Hence \begin{align*} \dfrac{dw}{dt} &= – \left( \frac{w}{l} \right) \left( \frac{dl}{dt} \right) \\[12px] &= – \left( \frac{10}{50} \right) \left( 8\right) = -\dfrac{8}{5} \text{ cm/s} \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \\[12px] (10\sqrt{26}) \frac{dd}{dt} & = (10) \left(-\dfrac{8}{5}\right) + (50) (8) \\[12px] &= -16 + 400 = 384 \\[12px] \frac{dd}{dt} &= \frac{384}{10\sqrt{26}} \\[12px] &= \frac{38.4}{\sqrt{26}} \text{ cm/s } \quad \cmark \end{align*}

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$$d^2 = w^2 + l^2$$

Both $w$ and $l$ are changing with time, and so when we take the derivative we must include $\dfrac{dw}{dt}$ and $\dfrac{dl}{dt}$ terms:

\begin{align*} \frac{d}{dt}d^2 &= \frac{d}{dt}(w^2 + l^2) \\[12px] 2d \frac{dd}{dt} &= 2w \frac{dw}{dt} + 2l \frac{dl}{dt} \\[12px] d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \text{ [*]} \end{align*}

We want to find $\dfrac{dd}{dt}$ at the instant when $w = 10$ cm. The problem statement tells us that $\dfrac{dl}{dt} = 8$ cm/s. In order to use the preceding equation (marked [*]), we need to determine $d$ and $l$ at this instant, and also $\dfrac{dw}{dt}$ at this instant.

First, we can find $l$ by noting that the rectangle has constant area, $A = 500 = lw$. Hence when $w = 10$ cm:

\begin{align*} l &= \frac{A}{w} \\[8px] &= \frac{500}{10} = 50 \text{ cm} \end{align*} Second, we can find $d$ at this instant using the Pythagorean theorem:

\begin{align*} d^2 &= w^2 + l^2 \\[12px] &= (10)^2 + (50)^2 = 100 + 2500 = 2600 \\[12px] d &= \sqrt{2600} = 10\sqrt{26} \end{align*} Next, we can find $\dfrac{dw}{dt}$ at this instant by taking the derivative of $A = 500 = lw$ :

\begin{align*} \frac{d}{dt}(500) &= \frac{d}{dt}(lw) \\[12px] 0 &= l \frac{dw}{dt} + w\frac{dl}{dt} \\[12px] l \frac{dw}{dt} &= – w\frac{dl}{dt} \end{align*} Hence \begin{align*} \dfrac{dw}{dt} &= – \left( \frac{w}{l} \right) \left( \frac{dl}{dt} \right) \\[12px] &= – \left( \frac{10}{50} \right) \left( 8\right) = -\dfrac{8}{5} \text{ cm/s} \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= w \frac{dw}{dt} + l \frac{dl}{dt} \\[12px] (10\sqrt{26}) \frac{dd}{dt} & = (10) \left(-\dfrac{8}{5}\right) + (50) (8) \\[12px] &= -16 + 400 = 384 \\[12px] \frac{dd}{dt} &= \frac{384}{10\sqrt{26}} \\[12px] &= \frac{38.4}{\sqrt{26}} \text{ cm/s } \quad \cmark \end{align*}

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Related Rates: Kite, length changes

A kite flies 30 meters above ground. It travels horizontally at 2 meters per second. How fast is the string unspooling at the moment the distance to the kite is 50 meters? **1. Draw a picture of the physical situation.**

See the figure. We’ve called the length of string to the kite $\ell.$ The problem is asking for the rate at which that length changes (= the rate at which the string is unspooling), $\dfrac{d\ell}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the*x*-direction, at the rate $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}} $. We’ll use these values at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

*B. To develop your equation, you will probably use . . . the Pythagorean theorem. *

In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which the string’s length $\ell$ is changing, $\dfrac{d\ell}{dt}$, to the rate at which*x* is changing, $\dfrac{dx}{dt}$. Hence we must first write down an equation that somehow relates $\ell$ and *x*. The Pythagorean theorem gives us that relation:
$$\ell^2 = x^2 + (30)^2 $$
That’s it. That’s the key relationship that will allow us to complete the solution.

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}\ell^2 &= \dfrac{d}{dt}x^2 + \cancelto{0}{\dfrac{d}{dt}30^2} \\[8px]
2\ell \dfrac{d\ell}{dt} &= 2x \dfrac{dx}{dt} \\[8px]
\ell \dfrac{d\ell}{dt} &= x \dfrac{dx}{dt}
\end{align*} \] **4. Solve for the quantity you’re after.**

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}$: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ To complete the solution, we need to know the value of*x* at the instant when $\ell = 50$ m.

*Begin subproblem to find* x *at the moment of interest.*

Recall from the Pythagorean theorem that at every moment $$\ell^2 = x^2 + (30)^2 $$ Hence when $\ell = 50$ m, \[ \begin{align*} x^2 &= \ell^2 – 30^2 \\[8px] &= (50)^2 – 30^2 \\[8px] &= 1600\\[8px] x &= 40 \end{align*} \] As a faster approach, you might have noticed that this is a 30-___-50 right triangle, and so as a “3-4-5 right triangle” the missing length has to be 40.

*End subproblem.*

We can now substitute values into our preceding equation: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ We have $x = 40$ m, $\ell = 50$ m, and $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{x}{\ell} \dfrac{dx}{dt} \\[8px] &= \frac{40 \text{ m}}{50 \text{ m}}(2 \, \tfrac{\text{m}}{\text{s}}) \\[8px] &= 1.6 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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See the figure. We’ve called the length of string to the kite $\ell.$ The problem is asking for the rate at which that length changes (= the rate at which the string is unspooling), $\dfrac{d\ell}{dt}$, at a particular moment — when $\ell = 50$ meters. Recall also that the kite is moving horizontally, in the

The string’s length changes as the situation progresses, so we’re calling that the variable $\ell.$

In this problem, the diagram above immediately suggests that we’re dealing with a right triangle. Furthermore, we need to related the rate at which the string’s length $\ell$ is changing, $\dfrac{d\ell}{dt}$, to the rate at which

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}$: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ To complete the solution, we need to know the value of

Recall from the Pythagorean theorem that at every moment $$\ell^2 = x^2 + (30)^2 $$ Hence when $\ell = 50$ m, \[ \begin{align*} x^2 &= \ell^2 – 30^2 \\[8px] &= (50)^2 – 30^2 \\[8px] &= 1600\\[8px] x &= 40 \end{align*} \] As a faster approach, you might have noticed that this is a 30-___-50 right triangle, and so as a “3-4-5 right triangle” the missing length has to be 40.

We can now substitute values into our preceding equation: $$\dfrac{d\ell}{dt} = \frac{x}{\ell} \dfrac{dx}{dt}$$ We have $x = 40$ m, $\ell = 50$ m, and $\dfrac{dx}{dt} = 2 \, \tfrac{\text{m}}{\text{s}}$: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{x}{\ell} \dfrac{dx}{dt} \\[8px] &= \frac{40 \text{ m}}{50 \text{ m}}(2 \, \tfrac{\text{m}}{\text{s}}) \\[8px] &= 1.6 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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Related Rates: Jogging toward an intersection

Two students, Ann and Frederick, are jogging along straight roads that intersect at right angles. Ann is heading West and approaching the intersection at a speed of 5 mph. Frederick is heading North and is approaching the intersection at a speed of 6 mph. At noon, as both Ann and Frederick are approaching the intersection, Ann is 3 miles from the intersection and Frederick is 4 miles from the intersection.
At what rate is the distance between Ann and Frederick decreasing at noon? **Answer:** $-\dfrac{39}{5}$ mph

**1. Draw a picture of the physical situation.**

See the figure. We’ll call Ann’s distance from the intersection $x$, and Frederick’s distance from the intersection $y$.

Let’s call the distance between them at any instant $d$, as indicated in the lower figure. The question is asking us to find $\dfrac{dd}{dt}$ at noon.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.* In this scenario, both of the jogger’s positions change over time, so we’re leaving each of their positions as a variable.

*B. To develop your equation, you will probably use . . . the Pythagorean theorem.*

We can relate $d$ to $x$ and $y$ using the Pythagorean theorem:

$$d^2 = x^2 + y^2$$**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*} \frac{d}{dt}(d^2) &= \frac{d}{dt}\left(x^2 + y^2 \right) \\[12px] 2d \frac{dd}{dt} &= 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\[12px] d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \text{ [*]} \end{align*}**4. Solve for the quantity you’re after.**

We want to find $\dfrac{dd}{dt}$ at noon, when $x = 3$ miles and $y = 4$ miles. We also know that $\dfrac{dx}{dt} = -5$ mph, and $\dfrac{dy}{dt} = -6$ mph. (Note that we must insert both negative signs “by hand” since both $x$ and $y$ are*decreasing* as the joggers head toward the intersection.)

To solve the preceding equation for $\dfrac{dd}{dt}$, we also need to know $d$; this we can find by using the Pythagorean theorem at this instant:

\begin{align*} d^2 &= x^2+y^2 \\ &= 3^2 + 4^2 = 25 \\ d &= 5 \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \\[12px] (5) \frac{dd}{dt} &= (3)(-5) + (4)(-6) = -15-24 = -39 \\[12px] \frac{dd}{dt} & = -\dfrac{39}{5} \text{ mph} \quad \cmark \end{align*} The negative value indicates that the distance between the joggers is decreasing as they jog toward each other.

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See the figure. We’ll call Ann’s distance from the intersection $x$, and Frederick’s distance from the intersection $y$.

Let’s call the distance between them at any instant $d$, as indicated in the lower figure. The question is asking us to find $\dfrac{dd}{dt}$ at noon.

We can relate $d$ to $x$ and $y$ using the Pythagorean theorem:

$$d^2 = x^2 + y^2$$

\begin{align*} \frac{d}{dt}(d^2) &= \frac{d}{dt}\left(x^2 + y^2 \right) \\[12px] 2d \frac{dd}{dt} &= 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\[12px] d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \text{ [*]} \end{align*}

We want to find $\dfrac{dd}{dt}$ at noon, when $x = 3$ miles and $y = 4$ miles. We also know that $\dfrac{dx}{dt} = -5$ mph, and $\dfrac{dy}{dt} = -6$ mph. (Note that we must insert both negative signs “by hand” since both $x$ and $y$ are

To solve the preceding equation for $\dfrac{dd}{dt}$, we also need to know $d$; this we can find by using the Pythagorean theorem at this instant:

\begin{align*} d^2 &= x^2+y^2 \\ &= 3^2 + 4^2 = 25 \\ d &= 5 \end{align*} Then substituting values into the equation marked [*] above:

\begin{align*} d\frac{dd}{dt} &= x \frac{dx}{dt} + y \frac{dy}{dt} \\[12px] (5) \frac{dd}{dt} &= (3)(-5) + (4)(-6) = -15-24 = -39 \\[12px] \frac{dd}{dt} & = -\dfrac{39}{5} \text{ mph} \quad \cmark \end{align*} The negative value indicates that the distance between the joggers is decreasing as they jog toward each other.

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Related Rates: Ships travel

Ship Blue travels north, in the *y-*direction, at 20 kilometers per hour. Ship Red travels east, in the *x-*direction, at 30 kilometers per hour. At 9:00 AM, ship Blue is 100 kilometers east of Ship Red. At what rate is the distance between the ships increasing at noon? **1. Draw a picture of the physical situation.**

See the figures. On the left, we’ve shown the situation at 9:00 AM, with ship Blue 100 km east of ship Red. On the right, we’ve shown the ships at some arbitrary time*t* later: ship Red has moved to the east a distance *x,* and ship Blue has moved to the north a distance *y.* We’ve labeled the distance between the ships $\ell.$ We’re looking for $\dfrac{d\ell}{dt}$ at particular moment, 12:00-noon, a fact we’ll use at the *end* of the solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

Both ship Red and ship Blue’s positions change, so we’ve called the distances they travel*x* and *y* respectively.

*B. To develop your equation, you will probably use . . . the Pythagorean theorem.*

In this problem, we’re given values for $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt},$ and asked to find the value for $\dfrac{d\ell}{dt}$ at a particular moment. We thus first need to relate $\ell$ to*x* and *y.*

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (100 -x)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}\left[\ell^2 \right] &= \dfrac{d}{dt} \left[(100 -x)^2 \right] + \dfrac{d}{dt} \left[ y^2\right] \\[8px]
2 \ell \dfrac{d\ell}{dt} &= 2(100 – x)(-1) \dfrac{dx}{dt} + 2 y \dfrac{dy}{dt} \\[8px]
\ell \dfrac{d\ell}{dt} &= -(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}
\end{align*} \]

**4. Solve for the quantity you’re after.**

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= -(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} &(*) \end{align*} \] We know $\dfrac{dx}{dt} = 30 \, \tfrac{\text{km}}{\text{hr}}$ and $\dfrac{dy}{dt} = 20 \, \tfrac{\text{km}}{\text{hr}}.$ To complete the problem, we need to know the values of*x,* *y,* and $\ell$ at the moment of interest.

*Begin subproblem to find x, y, and $\ell$.*

We’re interested in the particular moment 12:00-noon, which is*t* = 3 hours after our initial picture at 9:00 AM. Hence
\[ \begin{align*}
x &= \left(30 \, \tfrac{\text{km}}{\text{hr}} \right) (3 \text{ hr}) = 90 \text{ km} \\[8px]
y &= \left(20 \, \tfrac{\text{km}}{\text{hr}} \right) (3 \text{ hr}) = 60 \text{ km}
\end{align*} \]
And using the Pythagorean theorem we can find $\ell$ at this moment:
\[ \begin{align*}
\ell^2 &= (100 -x)^2 + y^2 \\[8px]
&= (100 – 90)^2 + (60)^2 \\[8px]
&= 3700 \\[8px]
\ell &= \sqrt{3700} \\[8px]
&= 60.83 \text{ km}
\end{align*} \] *End subproblem.*

We finally substitute all of these values into our equation marked (*): \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} \\[8px] &= \frac{-(100 – 90)\, (30) + (60)\,(20)}{60.83} \\[8px] &= \frac{-300 + 1200}{60.83} \\[8px] &= 14.8 \, \tfrac{\text{km}}{\text{hr}} \quad \cmark \end{align*} \]

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See the figures. On the left, we’ve shown the situation at 9:00 AM, with ship Blue 100 km east of ship Red. On the right, we’ve shown the ships at some arbitrary time

Both ship Red and ship Blue’s positions change, so we’ve called the distances they travel

In this problem, we’re given values for $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt},$ and asked to find the value for $\dfrac{d\ell}{dt}$ at a particular moment. We thus first need to relate $\ell$ to

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (100 -x)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= -(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} &(*) \end{align*} \] We know $\dfrac{dx}{dt} = 30 \, \tfrac{\text{km}}{\text{hr}}$ and $\dfrac{dy}{dt} = 20 \, \tfrac{\text{km}}{\text{hr}}.$ To complete the problem, we need to know the values of

We’re interested in the particular moment 12:00-noon, which is

We finally substitute all of these values into our equation marked (*): \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{-(100 – x) \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{\ell} \\[8px] &= \frac{-(100 – 90)\, (30) + (60)\,(20)}{60.83} \\[8px] &= \frac{-300 + 1200}{60.83} \\[8px] &= 14.8 \, \tfrac{\text{km}}{\text{hr}} \quad \cmark \end{align*} \]

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[hide solution]

Related Rates: Rocket, distance changes

A rocket is launched straight up, and its altitude is $y=5t^2$ meters after *t* seconds. You are on the ground 500 meters from the launch site. At what rate is the distance from you to the rocket changing 6 seconds after the launch? **1. Draw a picture of the physical situation.**

See the figure. We’ve labeled the rocket’s vertical position at any moment after launch*y,* and the distance from you to the rocket $\ell.$ The question is asking us to find $\dfrac{d\ell}{dt}$ at a particular moment, 6 seconds after launch—a fact we’ll use at the *end* of our solution.

**2. Write an equation that relates the quantities of interest.**

*A. Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.*

The rocket’s vertical position changes as the time passes, so we’re calling that the variable*y.*

*B. To develop your equation, you will probably use . . . the Pythagorean theorem. *

We’re looking for $\dfrac{d\ell}{dt},$ and we know (actually: we can easily determine) the rate at which the rocket’s*y-* position is changing, $\dfrac{dy}{dt}$. We thus first need to related $\ell$ to *y.*

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (500)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. ** \[ \begin{align*}
\dfrac{d}{dt}\left(\ell^2 \right) &= \cancelto{0}{\dfrac{d}{dt}(500)^2} + \dfrac{d}{dt}\left(y^2 \right) \\[8px]
2 \ell \dfrac{d\ell}{dt} &= 2 y \dfrac{dy}{dt} \\[8px]
\ell \dfrac{d\ell}{dt} &= y \dfrac{dy}{dt} \\[8px]
\end{align*} \]

**4. Solve for the quantity you’re after.**

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} &(*) \end{align*} \] To complete our solution, we need to know the values of*y,* $\ell,$ and $\dfrac{d\ell}{dt}$ 6 seconds after launch.

*Begin subproblems to find y, $\ell,$ and $\dfrac{d\ell}{dt}$.*

*A. Find y.*

The problem states that $y=5t^2$ meters after*t* seconds. Hence 6 seconds after launch
$$y=5\,(6)^2 = 180 \text{ m}$$ *B. Find $\ell.$*

By the Pythagorean theorem, at this instant \[ \begin{align*} \ell^2 &= (500)^2 + (180)^2 \\[8px] &= 282400 \\[8px] \ell &= \sqrt{282400} \\[8px] &= 531 \text{ m} \end{align*} \]*C. Find $\dfrac{dy}{dt}.$*

Since $y=5t^2,$ $$\dfrac{dy}{dt} = 10t $$ Hence at $t = 6$ seconds, $$\dfrac{dy}{dt} \big|_{6\text{ s}} = 10(6) = 60 \, \tfrac{\text{m}}{\text{s}}$$*End subproblems.*

We can now substitute these values into our equation marked (*) above: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} \\[8px] &= \frac{180}{531} (60) \\[8px] &= 20.3 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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Show/Hide Solution

See the figure. We’ve labeled the rocket’s vertical position at any moment after launch

The rocket’s vertical position changes as the time passes, so we’re calling that the variable

We’re looking for $\dfrac{d\ell}{dt},$ and we know (actually: we can easily determine) the rate at which the rocket’s

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have $$\ell^2 = (500)^2 + y^2$$ That’s it. That’s the key relationship that will allow us to complete the solution. (Note, by the way, how critical drawing the figure correctly is to being able to solve this problem.)

Let’s solve the preceding equation for $\dfrac{d\ell}{dt}:$ \[ \begin{align*} \ell \dfrac{d\ell}{dt} &= y \dfrac{dy}{dt} \\[8px] \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} &(*) \end{align*} \] To complete our solution, we need to know the values of

The problem states that $y=5t^2$ meters after

By the Pythagorean theorem, at this instant \[ \begin{align*} \ell^2 &= (500)^2 + (180)^2 \\[8px] &= 282400 \\[8px] \ell &= \sqrt{282400} \\[8px] &= 531 \text{ m} \end{align*} \]

Since $y=5t^2,$ $$\dfrac{dy}{dt} = 10t $$ Hence at $t = 6$ seconds, $$\dfrac{dy}{dt} \big|_{6\text{ s}} = 10(6) = 60 \, \tfrac{\text{m}}{\text{s}}$$

We can now substitute these values into our equation marked (*) above: \[ \begin{align*} \dfrac{d\ell}{dt} &= \frac{y}{\ell}\dfrac{dy}{dt} \\[8px] &= \frac{180}{531} (60) \\[8px] &= 20.3 \, \tfrac{\text{m}}{\text{s}} \quad \cmark \end{align*} \]

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