Related Rates

We developed the general strategy we will use to solve related rates problems in our blog post 4 Steps to Solve Any Related Rates Problem - Part 1. (Link will open in a new tab.) You might find it helpful to read that post if you haven't already since we start with "How to Recognize a Related Rates Problem" and then develop the rest from there.

On this screen we have many practice problems for you to use, each with a complete solution immediately available. Practice here and get stuck in all the places students usually do so you can work through the challenges and be ready for your exam!

Related Rates: Given an Equation, Find the Rate

Practice Problem: Given an equation, find the rate

If $𝑦 = 𝑥 3 + 2 𝑥$ and $𝑑 𝑥 𝑑 𝑡 = 6$ , find $𝑑 𝑦 𝑑 𝑡$ when $𝑥 = 5 .$

View/Hide Solution

We can still use our Related Rates Problem Solving Strategy above. Instead of starting at Step 1, we simply start at Step 3:

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

$𝑦 = 𝑥 3 + 2 𝑥 𝑑 𝑑 𝑡 (𝑦) = 𝑑 𝑑 𝑡 (𝑥 3) + 𝑑 𝑑 𝑡 (2 𝑥) 𝑑 𝑦 𝑑 𝑡 = 3 𝑥 2 𝑑 𝑥 𝑑 𝑡 + 2 𝑑 𝑥 𝑑 𝑡$

4. Solve for the quantity you're after.

We want to find $𝑑 𝑦 𝑑 𝑡$ when $𝑥 = 5 .$ Recall that the problem told us that $𝑑 𝑥 𝑑 𝑡 = 6 .$ $𝑑 𝑦 𝑑 𝑡 = 3 𝑥 2 𝑑 𝑥 𝑑 𝑡 + 2 𝑑 𝑥 𝑑 𝑡 = 3 (5) 2 (6) + 2 (6) = 3 (25) (6) + 12 = 462 ✓$

[close solution]

Related Rates: Using a Simple Geometric Fact

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

the area of a circle is $𝐴 = 𝜋 𝑟 2$
the volume of a sphere is $𝑉 = 43 𝜋 𝑟 3$
the area of a rectangle is $𝐴 = 𝑙 𝑤$

and so forth.

The following problems illustrate.

As a snowball melts, its area decreases at a given rate. How fast does its radius change? (Video solution)

Practice Problem: Spherical snowball melts at constant

𝑑 𝐴 𝑑 𝑡

, radius decreases

A spherical snowball melts symmetrically such that it is always a sphere. Its surface area decreases at the rate of $𝜋$ in $2$ /min. How fast is its radius changing at the instant when $𝑟 = 2$ inches?

View/Hide Solution

[Scroll down for text-based (non-video) version of the solution.]

Let's unpack the question statement:

We're told that the snowball's area A is changing at the rate of $𝑑 𝐴 𝑑 𝑡 = - 𝜋$ in $2$ /min. (We must insert the negative sign "by hand" since we are told that the snowball is melting, and hence its area is decreasing.)
As a result, its radius is changing, at the rate $𝑑 𝑟 𝑑 𝑡$ , 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 $𝑑 𝑟 𝑑 𝑡$ at a particular instant, when $𝑟 = 2$ inches.

We of course use our 4 steps to solve this related rates problem as outlined in the Problem Solving Strategy box above:

1. Draw a picture of the physical situation.

snowball melts, its area decreases at a given rate, such that it always remains a sphere

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.

This is the hardest part of Related Rates problem for most students initially: you have to know how to develop the equation you need, how to pull that "out of thin air." By working through these problems you'll develop this skill. The key is to recognize which of the few sub-types of problem it is as described in the Problem Solving Strategy box above. In this problem, the diagram above reminds us that the snowball always remains a sphere, which is a Big Clue.

We need to develop a relationship between the rate we're given, $𝑑 𝐴 𝑑 𝑡 = - 𝜋$ in $2$ /min, and the rate we're after, $𝑑 𝑟 𝑑 𝑡$ . We thus first need to write down a relationship between the sphere's area A and its radius r. But we know that relationship since it's a simple geometric fact for a sphere that you could look up if you don't know it:
$𝐴 = 4 𝜋 𝑟 2$ That's it — that's the key relationship we need to be able to proceed with our solution.

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

𝑑 𝑑 𝑡 𝐴 = 𝑑 𝑑 𝑡 (4 𝜋 𝑟 2) 𝑑 𝐴 𝑑 𝑡 = 4 𝜋 𝑑 𝑑 𝑡 (𝑟 2) = 4 𝜋 (2 𝑟 𝑑 𝑟 𝑑 𝑡) = 8 𝜋 𝑟 𝑑 𝑟 𝑑 𝑡

4. Solve for the quantity you're after.

Solving the equation above for $𝑑 𝑟 𝑑 𝑡$ :

$𝑑 𝐴 𝑑 𝑡 = 8 𝜋 𝑟 𝑑 𝑟 𝑑 𝑡 𝑑 𝑟 𝑑 𝑡 = 1 8 𝜋 𝑟 𝑑 𝐴 𝑑 𝑡$

Now we just have to substitute values. Recall $𝑑 𝐴 𝑑 𝑡 = - 𝜋$ in $2$ /min,
and the problem asks about when $𝑟 = 2$ inches:

$𝑑 𝑟 𝑑 𝑡 = 1 8 𝜋 𝑟 𝑑 𝐴 𝑑 𝑡 = 1 8 𝜋 (2 i n) (- 𝜋 i n 2 m i n) = - 116 i n / m i n ✓$
That's the answer. The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

$Web-based homework warning icon$

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

At what rate does the radius decrease?

then the system may have already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $116 i n m i n ✓$

We can't guarantee how this problem was coded into your homework system, and so unfortunately can only advise that if you think your answer of a negative value is correct but the system is marking it as wrong, try entering it as a positive value. (Many students have thanked us for this suggestion!)

[close solution]

Snowball melts at constant rate of volume change, radius decreases

Practice Problem: Spherical snowball melts at constant

𝑑 𝑉 𝑑 𝑡

, radius decreases

A spherical snowball is melting at the rate of

2 𝜋

3

/hr. It melts symmetrically such that it is always a sphere. How fast is its radius changing at the instant

𝑟 = 10

cm?

Answer:

- 1200 c m / h r

. [If you are using a web-based homework system, see the "Caution" at the end of this solution.]

Let's unpack the question statement:

We're told that the snowball's volume V is changing at the rate of $𝑑 𝑉 𝑑 𝑡 = - 2 𝜋$ 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 $𝑑 𝑟 𝑑 𝑡$ , 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 $𝑑 𝑟 𝑑 𝑡$ at a particular instant, when $𝑟 = 10$ cm.

Now let's use our 4-step Problem Solving Strategy as outlined above:
1. Draw a picture of the physical situation.
$sphere$ 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

𝑑 𝑉 𝑑 𝑡

, and we're after the value of

𝑑 𝑟 𝑑 𝑡

. Hence we need to start with an expression that relates V to r. Since the snowball is a sphere, we know that relationship:

𝑉 = 43 𝜋 𝑟 3

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

𝑑 𝑑 𝑡 𝑉 = 𝑑 𝑑 𝑡 (43 𝜋 𝑟 3) 𝑑 𝑉 𝑑 𝑡 = 43 𝜋 𝑑 𝑑 𝑡 (𝑟 3) = 43 𝜋 (3 𝑟 2 𝑑 𝑟 𝑑 𝑡) = 4 𝜋 𝑟 2 𝑑 𝑟 𝑑 𝑡

4. Solve for the quantity you're after.
Solving the equation above for

𝑑 𝑟 𝑑 𝑡

𝑑 𝑉 𝑑 𝑡 = 4 𝜋 𝑟 2 𝑑 𝑟 𝑑 𝑡 𝑑 𝑟 𝑑 𝑡 = 1 4 𝜋 𝑟 2 𝑑 𝑉 𝑑 𝑡

Now we just have to substitute values. Recall

𝑑 𝑉 𝑑 𝑡 = - 2 𝜋

3

/hr, and the problem asks about when

𝑟 = 10

cm:

𝑑 𝑟 𝑑 𝑡 = 1 4 𝜋 𝑟 2 𝑑 𝑉 𝑑 𝑡 = 1 4 𝜋 (10) 2 (- 2 𝜋) = - 1200 c m / h r ✓

The negative value indicates that the radius is decreasing as the snowball melts, as we expect.

$Web-based homework warning icon$ Caution: IF you are using a web-based homework system and the question asks,

At what rate does the radius decrease?

then the system (probably) has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE:

1200 c m h r ✓

(Why "probably"? Because the truth is it depends on how the question got coded into the system, and problem-writers often aren't consistent here. It's annoying.)

[close solution]

Circle expands

Practice Problem: 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?

Answer:

1 4 𝜋

When a circle expands, obviously its area changes; the rate at which its area changes is

𝑑 𝐴 𝑑 𝑡

. Its radius changes as well, at the rate

𝑑 𝑟 𝑑 𝑡

. Those two quantities are related, and that recognition is the crux of this problem.

$circle$ 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

𝑑 𝐴 𝑑 𝑡

and

𝑑 𝑟 𝑑 𝑡

, we start with a relation between A and r. Fortunately, that's a particularly well-known relation:

𝐴 = 𝜋 𝑟 2

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

𝑑 𝑑 𝑡 (𝐴) = 𝑑 𝑑 𝑡 (𝜋 𝑟 2) 𝑑 𝐴 𝑑 𝑡 = 2 𝜋 𝑟 𝑑 𝑟 𝑑 𝑡

4. Solve for the quantity you're after.
The problem is asking us to find r at the instant when

𝑑 𝐴 𝑑 𝑡 = 12 𝑑 𝑟 𝑑 𝑡

, or when

(𝑑 𝐴 𝑑 𝑡) (𝑑 𝑟 𝑑 𝑡) = 12

.

Starting from the preceding equation:

𝑑 𝐴 𝑑 𝑡 = 2 𝜋 𝑟 𝑑 𝑟 𝑑 𝑡 𝑟 = 1 2 𝜋 (𝑑 𝐴 𝑑 𝑡) (𝑑 𝑟 𝑑 𝑡) = 1 2 𝜋 (12) = 1 4 𝜋 ✓

[close solution]

Rectangle, area increases as sides change

Practice Problem: Rectangle, find

𝑑 𝐴 𝑑 𝑡

given

𝑑 𝑙 𝑑 𝑡

and

𝑑 𝑤 𝑑 𝑡

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?

Answer:

𝑑 𝐴 𝑑 𝑡 = 47 c m 2 s

Let's first unpack this question:
We're given that the rectangle's length is changing at the rate

𝑑 𝑙 𝑑 𝑡 = 8

cm/s, and its width is changing at the rate

𝑑 𝑤 𝑑 𝑡 = 5

cm/s. We're looking for

𝑑 𝐴 𝑑 𝑡

at the particular instant specified in the problem.

$Rectangle labeled l, w$ 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

𝑑 𝐴 𝑑 𝑡

given values for both

𝑑 𝑙 𝑑 𝑡

and

𝑑 𝑤 𝑑 𝑡

. 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 = 𝑙 𝑤

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

𝑑 𝑤 𝑑 𝑡

and

𝑑 𝑙 𝑑 𝑡

terms:

𝑑 𝑑 𝑡 𝐴 = 𝑑 𝑑 𝑡 (𝑙 𝑤) 𝑑 𝐴 𝑑 𝑡 = 𝑑 𝑙 𝑑 𝑡 𝑤 + 𝑙 𝑑 𝑤 𝑑 𝑡

4. Solve for the quantity you're after.
We want

𝑑 𝐴 𝑑 𝑡

at the moment when

𝑙 = 3

cm and

𝑤 = 4

cm. Furthermore, recall that the problem tells us

𝑑 𝑙 𝑑 𝑡 = 8

cm/s, and its width is changing at the rate

𝑑 𝑤 𝑑 𝑡 = 5

cm/s.

Hence

𝑑 𝐴 𝑑 𝑡 = 𝑑 𝑙 𝑑 𝑡 𝑤 + 𝑙 𝑑 𝑤 𝑑 𝑡 = (8 c m s) (4 c m) + (3 c m) (5 c m s) = 32 c m 2 s + 15 c m 2 s = 47 c m 2 s ✓

[close solution]

Rectangle maintains constant area as its length increases

Practice Problem: 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?

Answer:

𝑤 = \sqrt 5008 c m

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

𝑑 𝑙 𝑑 𝑡 = 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,

𝑑 𝑤 𝑑 𝑡

, 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. $Rectangle labeled l, w$

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 = 𝑤 𝑙

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

𝑑 𝑤 𝑑 𝑡

and

𝑑 𝑙 𝑑 𝑡

terms:

𝑑 𝑑 𝑡 (500) = 𝑑 𝑑 𝑡 (𝑤 𝑙) 0 = 𝑑 𝑤 𝑑 𝑡 𝑙 + 𝑤 𝑑 𝑙 𝑑 𝑡

4. Solve for the quantity you're after.
We want to find

𝑤

at the instant

𝑑 𝑤 𝑑 𝑡 = - 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

𝑑 𝑙 𝑑 𝑡 = 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 = 𝑙 𝑤

. Hence

𝑙 = 500 𝑤

.

Then from the preceding equation:

0 = 𝑑 𝑤 𝑑 𝑡 𝑙 + 𝑤 𝑑 𝑙 𝑑 𝑡 0 = 𝑑 𝑤 𝑑 𝑡 500 𝑤 + 𝑤 𝑑 𝑙 𝑑 𝑡 𝑤 𝑑 𝑙 𝑑 𝑡 = - 500 𝑤 𝑑 𝑤 𝑑 𝑡 𝑤 2 = - 500 (𝑑 𝑤 𝑑 𝑡) (𝑑 𝑙 𝑑 𝑡)

Recall

𝑑 𝑤 𝑑 𝑡 = - 1

cm/s and

𝑑 𝑙 𝑑 𝑡 = 8

cm/s:

𝑤 2 = - (500) (- 1) 8 = 5008 𝑤 = \sqrt 5008 c m ✓

[close solution]

Square's sides grow, increasing both its perimeter and area

Practice Problem: 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 m 2 .

(a) $2.0 m s$
(b) $5.0 m 2 s$

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

𝑑 𝑥 𝑑 𝑡 = 0.5 m 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

𝑑 𝑥 𝑑 𝑡,

and we're looking for the value of

𝑑 𝑃 𝑑 𝑡,

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 = 4 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.

𝑑 𝑑 𝑡 𝑃 = 𝑑 𝑑 𝑡 (4 𝑥) 𝑑 𝑃 𝑑 𝑡 = 4 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
This step is straightforward in this problem. We have

𝑑 𝑥 𝑑 𝑡 = 0.5 m s

, so

𝑑 𝑃 𝑑 𝑡 = 4 𝑑 𝑥 𝑑 𝑡 = 4 (0.5 m s) = 2.0 m s ✓

This result makes sense conceptually: the square has four sides, each of which is growing at the rate of

0.5 m s

. Hence the perimeter grows at the rate

𝑑 𝑃 𝑑 𝑡 = 4 (0.5 m s) = 2.0 m 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

𝑑 𝑥 𝑑 𝑡 = 0.5 m s

, and we're interested in the moment when

𝐴 = 25 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

𝑑 𝑥 𝑑 𝑡,

and we're looking for the value of

𝑑 𝐴 𝑑 𝑡,

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.

𝑑 𝑑 𝑡 𝐴 = 𝑑 𝑑 𝑡 𝑥 2 𝑑 𝐴 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
We know that

𝑑 𝐴 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡

and we know that

𝑑 𝑥 𝑑 𝑡 = 0.5 m s

. We also know that we're interested in the moment when

𝐴 = 25 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 m 2 :

𝑥 2 = 25 m 2 𝑥 = 5.0 m

End subproblem.

We can now complete our calculation:

𝑑 𝐴 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡 = 2 (5.0 m) (0.5 m s) = 5.0 m 2 s ✓

[close solution]

Cube's sides grow, increasing both its volume and surface area

Practice Problem: 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

(b)

At what rate is its surface area increasing at the moment when

𝑥 = 0.6

(a) $0.024 m 3 s$
(b) $0.36 m 2 s$

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

𝑑 𝑥 𝑑 𝑡 = 0.05 m 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

𝑑 𝑥 𝑑 𝑡,

and we're looking for the value of

𝑑 𝑉 𝑑 𝑡 .

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.

𝑑 𝑑 𝑡 𝑉 = 𝑑 𝑑 𝑡 𝑥 3 𝑑 𝑉 𝑑 𝑡 = 3 𝑥 2 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
This is straightforward in this problem. We have

𝑑 𝑥 𝑑 𝑡 = 0.05 m s

and are interested in the moment when

𝑥 (𝑡) = 0.4

𝑑 𝑉 𝑑 𝑡 = 3 𝑥 2 𝑑 𝑥 𝑑 𝑡 = 3 (0.4 m) 2 (0.05 m s) = 0.024 m 3 s ✓

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

𝑑 𝑥 𝑑 𝑡 = 0.05 m s

, and we're interested in the moment when

𝑥 = 0.6

𝑑 𝑥 𝑑 𝑡,

and we're looking for the value of

𝑑 𝐴 𝑑 𝑡 .

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 = 6 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.

𝑑 𝑑 𝑡 𝐴 = 𝑑 𝑑 𝑡 (6 𝑥 2) 𝑑 𝑉 𝑑 𝑡 = 12 𝑥 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
This is straightforward in this problem. We have

𝑑 𝑥 𝑑 𝑡 = 0.05 m s

and are interested in the moment when

𝑥 (𝑡) = 0.6

𝑑 𝐴 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡 = 12 (0.6 m) (0.05 m s) = 0.36 m 2 s ✓

[close solution]

Cylinder drains: depth decreases at constant rate. Find the rate of water draining.

Practice Problem: 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 f t s

, so

𝑑 𝑦 𝑑 𝑡 = - 0.1 f t 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

(𝜋 𝑟 2)

times its height. Here, at any moment the water's height is y, and so the volume of water in the cylinder is:

𝑉 = 𝜋 𝑟 2 𝑦

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

𝑑 𝑦 𝑑 𝑡

𝑑 𝑉 𝑑 𝑡 = 𝑑 𝑑 𝑡 (𝜋 𝑟 2 𝑦) = 𝜋 𝑟 2 𝑑 𝑑 𝑡 (𝑦) = 𝜋 𝑟 2 𝑑 𝑦 𝑑 𝑡

4. Solve for the quantity you're after.
At this point we're just substituting values: the problem told us

𝑟 = 3.0 f t

and

𝑑 𝑦 𝑑 𝑡 = - 0.1 f t s

, and we're looking for

𝑑 𝑉 𝑑 𝑡

. Starting from our last expression above:

𝑑 𝑉 𝑑 𝑡 = 𝜋 𝑟 2 𝑑 𝑦 𝑑 𝑡 = 𝜋 (3.0 f t) 2 (- 0.1 f t s) = - 2.8 f t 3 s ✓

That is, water is draining from the tank at the rate

𝑑 𝑉 𝑑 𝑡 = - 2.8 f t 3 s

. The negative value indicates that the water's volume in the tank V is decreasing, which is correct.

$Web-based homework warning icon$ Caution: IF you are using a web-based homework system and the question asks,

At what rate does the volume decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE:

2.8 f t 3 s ✓

[close 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:

t a n 𝜃 = o p p o s i t e a d j a c e n t c o s 𝜃 = a d j a c e n t h y p o t e n u s e s i n 𝜃 = o p p o s i t e h y p o t e n u s e

The following problems illustrate.

Ladder slides, angle changes

Practice Problem: Ladder slides, angle changes

A 10-foot long ladder leans against a wall. The bottom of the ladder slides away from the wall at

32

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?

Answer:

- 3 16

rad/s

$Ladder slides down a wall such that the angle it makes with the ground decreases.$ 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

𝜃

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 $c o s 𝜃$ = adjacent/hypotenuse).
We need to use a trigonometric function to analyze this situation, since we need to relate the angle

𝜃

to other quantities in order to have

𝑑 𝜃 𝑑 𝑡

when we take the derivative. We're given the rate at which the ladder's bottom position changes

(𝑑 𝑥 𝑑 𝑡 = 32 f t s)

, and so it seems easiest to go with

c o s 𝜃

since that directly relates

𝜃

and x:

c o s 𝜃 = 𝑥 10

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

𝑑 𝑑 𝑡 (c o s 𝜃) = 𝑑 𝑑 𝑡 (𝑥 10) - s i n 𝜃 𝑑 𝜃 𝑑 𝑡 = 110 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
We're after

𝑑 𝜃 𝑑 𝑡

at the instant

𝑥 = 6

ft. So let's solve the preceding equation for

𝑑 𝜃 𝑑 𝑡

𝑑 𝜃 𝑑 𝑡 = - 110 1 s i n 𝜃 𝑑 𝑥 𝑑 𝑡

We're given that

𝑑 𝑥 𝑑 𝑡 = 32 f t s

. But we don't yet know what

s i n 𝜃

is when

𝑥 = 6

ft.

Begin subproblem to find $s i n 𝜃$ .
We know

s i n 𝜃 = 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:

(6) 2 + 𝑦 2 = 102 36 + 𝑦 2 = 100 𝑦 2 = 100 - 36 = 64 𝑦 = 8

Hence at this instant

s i n 𝜃 = 810 = 45

End subproblem.

Hence

𝑑 𝜃 𝑑 𝑡 = - 110 1 s i n 𝜃 𝑑 𝑥 𝑑 𝑡 = - 110 \cdot (1 4 / 5) \cdot 32 = - 110 \cdot 54 \cdot 32 = - 316 r a d / s ✓

The negative value indicates that the angle is decreasing at the ladder slides down the wall.

[close solution]

Kite flies horizontally; find the rate of change of the angle between the string and the ground

Practice Problem: 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?

Kite's string makes an angle theta with the horizontal

1. Draw a picture of the physical situation.
See the figure. We've called the length of string to the kite

ℓ .

And we've labeled the angle

𝜃

that the string makes with the ground, since the problem is asking us to find the rate at which that angle changes,

𝑑 𝜃 𝑑 𝑡

, at a particular moment — when

ℓ = 50

meters. Recall also that the kite is moving horizontally, in the x-direction, at the rate

𝑑 𝑥 𝑑 𝑡 = 2 m 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

ℓ .

B. To develop your equation, you will probably use . . . a trigonometric function (like $t a n 𝜃$ = 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

𝜃

is changing,

𝑑 𝜃 𝑑 𝑡

, to the rate at which x is changing,

𝑑 𝑥 𝑑 𝑡

, and so we first need to write down an equation that somehow relates

𝜃

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:

t a n 𝜃 = 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.

𝑑 𝑑 𝑡 t a n 𝜃 = 𝑑 𝑑 𝑡 (30 𝑥) s e c 2 𝜃 𝑑 𝜃 𝑑 𝑡 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
Let's solve the preceding equation for

𝑑 𝜃 𝑑 𝑡

s e c 2 𝜃 𝑑 𝜃 𝑑 𝑡 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡 𝑑 𝜃 𝑑 𝑡 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡 1 s e c 2 𝜃 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡 c o s 2 𝜃

Notice that

c o s 𝜃 = 𝑥 ℓ

, so

𝑑 𝜃 𝑑 𝑡 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡 (𝑥 ℓ) 2 = - 30 𝑥 2 𝑑 𝑥 𝑑 𝑡 𝑥 2 ℓ 2 = - 30 ℓ 2 𝑑 𝑥 𝑑 𝑡

We're looking for

𝑑 𝜃 𝑑 𝑡

at the instant when

ℓ = 50

m. Recall that Recall

𝑑 𝑥 𝑑 𝑡 = 2 m s

𝑑 𝜃 𝑑 𝑡 = - 30 (50 m) 2 (2 m s) = - 0.02 r a d s ✓

That's the answer. The negative value indicates that the angle is decreasing at the kite travels further out, as we expect.

$Web-based homework warning icon$ Caution: IF you are using a web-based homework system and the question asks,

At what rate does the angle decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE:

0.02 r a d s ✓

[close solution]

Rocket flies straight up; find the rate of change of the angle between the line of sight and the ground

Practice Problem: 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

𝜃

with the horizontal.

By how many radians per second is

𝜃

changing ten seconds after the launch?

Answer:

110

rad/s

$Right triangle: base 500 m, height h$

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 = 5 𝑡 2

for now.

The picture immediately suggests a relation between the angle

𝜃

and the height

ℎ

t a n 𝜃 = ℎ 500 = 5 𝑡 2 500 = 𝑡 2 100

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

𝜃

is changing with time too, and so we have to use the chain rule when we take the derivative of

t a n 𝜃

, giving a

𝑑 𝜃 𝑑 𝑡

term:

𝑑 𝑑 𝑡 t a n 𝜃 = 𝑑 𝑑 𝑡 (𝑡 2 100) s e c 2 𝜃 \cdot 𝑑 𝜃 𝑑 𝑡 = 2 𝑡 100 [𝑑 𝑑 𝜃 t a n 𝜃 = s e c 2 𝜃] 𝑑 𝜃 𝑑 𝑡 = 𝑡 50 c o s 2 𝜃 [*] [1 s e c 2 𝜃 = c o s 2 𝜃]

4. Solve for the quantity you're after.
In order to use the preceding equation, we need to know

𝜃

(so we can compute

c o s 2 𝜃

). To find

𝜃

, we return to our our figure and recall that

t a n 𝜃 = ℎ 500

. Hence if we find

ℎ

, we can then find

𝜃

.

When

𝑡 = 10

, the rocket's height is:

ℎ (10) = 5 𝑡 2 = 5 \cdot (10) 2 = 500

Thus

t a n 𝜃 = ℎ 500 = 500500 = 1

and so at

𝑡 = 10

𝜃 = 𝜋 / 4

.

We can now solve for

𝑑 𝜃 𝑑 𝑡

𝑡 = 10

seconds using the equation we found above marked [*]:

𝑑 𝜃 𝑑 𝑡 = 𝑡 50 c o s 2 𝜃 𝑑 𝜃 𝑑 𝑡 ∣ 𝑡 = 10 = 1050 c o s 2 𝜋 4 = 15 (1 \sqrt 2) 2 = 15 (12) = 110 r a d / s ✓

[close solution]

Related Rates: Using Similar Triangles

A lot of problems you encounter 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.

Note again the importance of starting with a clear figure, so that you can see the triangles!

The following problems illustrate.

Man's shadow moves along the ground as he walks away from light post (Video solution)

Practice Problem: Man walks away from light post; shadow on ground moves

A 1.8-meter tall man walks away from a 6.0-meter lamp post at the rate of 1.5 m/s. The light at the top of the post casts a shadow in front of the man. How fast is the "head" of his shadow moving along the ground?

View/Hide Solution

Answer: 2.1 m/s

[Scroll down for text (non-video) version of the solution.]

Related rates problem and solution: A man walks away from a light pole that casts his shadow. How fast is the tip of his shadow moving along the ground?

1. Draw a picture of the physical situation.

See the figure. We're calling the distance between the post and the "head" of the man's shadow $ℓ$ , and the distance between the man and the post x.

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

We are given that the man is walking away from the post at the rate $𝑑 𝑥 𝑑 𝑡 = 1.5$ m/s. We are looking for the rate at which the "head" of the man's shadow moves, which is $𝑑 ℓ 𝑑 𝑡$ . We thus need to somehow relate $ℓ$ 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 $ℓ$ and $𝑑 ℓ 𝑑 𝑡$ here because $ℓ$ is the distance from the shadow's tip to the stationary post. We're not examining the shadow's length itself (labeled $ℓ - 𝑥$ 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 post. Hence we focus on $ℓ$ and aim to compute $𝑑 ℓ 𝑑 𝑡$ .

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

Two similar triangles: the hypotenuse of each goes from the light to the head of the man's shadow.

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 $(ℓ - 𝑥 ℓ)$ is equal to the ratio of their heights $(1.8 m 6.0 m)$ :
$ℓ - 𝑥 ℓ = 1.8 m 6.0 m = 0.30 ℓ - 𝑥 = 0.30 ℓ ℓ - 0.30 ℓ = 𝑥 (1 - 0.30) ℓ = 𝑥 0.70 ℓ = 𝑥$

3. Take the derivative with respect to time of both sides of your equation.

$𝑑 𝑑 𝑡 (0.70 ℓ) = 𝑑 𝑑 𝑡 (𝑥) 0.70 𝑑 ℓ 𝑑 𝑡 = 𝑑 𝑥 𝑑 𝑡$

4. Solve for the quantity you're after.

We're looking for $𝑑 ℓ 𝑑 𝑡$ :

$0.70 𝑑 ℓ 𝑑 𝑡 = 𝑑 𝑥 𝑑 𝑡 𝑑 ℓ 𝑑 𝑡 = 1 0.70 𝑑 𝑥 𝑑 𝑡 = 1 0.70 (1.5 m s) = 2.1 m s ✓$

[close solution]

Regarding the preceding problem, students often ask what changes if the person walks toward instead of away from the light pole. The simple answer is that $𝑑 𝑥 𝑑 𝑡$ gains a negative sign that you must insert by hand, which then propagates through the solution so that $𝑑 ℓ 𝑑 𝑡$ emerges as a positive rather than negative value as it did in that problem. But it's probably worth tracing that through so you can see for yourself:

Man's shadow moves along the ground as he walks toward light pole

Practice Problem: Man walks toward light pole; 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?

Answer:

- 2.1

m/s

$Light pole casts a shadow behind a man walking toward it$

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

ℓ

, 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

𝑑 𝑥 𝑑 𝑡 = - 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

𝑑 ℓ 𝑑 𝑡

. We thus need to somehow relate

ℓ

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

ℓ

and

𝑑 ℓ 𝑑 𝑡

here because

ℓ

is the distance from the shadow's tip to the stationary pole. We're not examining the shadow's length itself (labeled

ℓ - 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

ℓ

and aim to compute

𝑑 ℓ 𝑑 𝑡

.

B. To develop your equation, you will probably use . . . similar triangles.
$Two similar triangles: the hypotenuse of each goes from the light to the head of the man's shadow.$
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

(ℓ - 𝑥 ℓ)

is equal to the ratio of their heights

(1.8 m 6.0 m)

ℓ - 𝑥 ℓ = 1.8 m 6.0 m = 0.30 ℓ - 𝑥 = 0.30 ℓ ℓ - 0.30 ℓ = 𝑥 0.70 ℓ = 𝑥

3. Take the derivative with respect to time of both sides of your equation.

𝑑 𝑑 𝑡 (0.70 ℓ) = 𝑑 𝑑 𝑡 (𝑥) 0.70 𝑑 ℓ 𝑑 𝑡 = 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
We're looking for

𝑑 ℓ 𝑑 𝑡

0.70 𝑑 ℓ 𝑑 𝑡 = 𝑑 𝑥 𝑑 𝑡 𝑑 ℓ 𝑑 𝑡 = 1 0.70 𝑑 𝑥 𝑑 𝑡 = 1 0.70 (- 1.5 m s) = - 2.1 m s ✓

[close solution]

Man's shadow on the wall changes length as he walks

Practice Problem: 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?

Spotlight on ground creates man of shadow on the wall. Man's distance from spotlight is x; shadow's height is y.

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

𝑑 𝑥 𝑑 𝑡 = 1.5 m s .

We're calling the height of his shadow on the wall y. The problem is asking us to find

𝑑 𝑦 𝑑 𝑡

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

𝑑 𝑥 𝑑 𝑡

and asks us to find the rate

𝑑 𝑦 𝑑 𝑡 .

We thus need to first relate y to x. The figure above suggests that we can use similar triangles to do so. Triangle 1: spotlight, man's feet and head. Triangle 2: spotlight, shadow's bottom and shadow's top

Triangle 1: spotlight, man's feet and head. Triangle 2: spotlight, shadow's bottom and shadow's top

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

𝑦 1.8

is equal to the ratio of their bases

20 𝑥 .

That is,

𝑦 1.8 = 20 𝑥 𝑦 = (1.8) (20) 𝑥 = 36 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.

𝑑 𝑑 𝑡 𝑦 = 𝑑 𝑑 𝑡 (36 𝑥) 𝑑 𝑦 𝑑 𝑡 = - 36 𝑥 2 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
We're given that

𝑑 𝑥 𝑑 𝑡 = 1.5 m s

. The problem asks us to find

𝑑 𝑦 𝑑 𝑡

at the moment the man is 5.0 meters from the wall. Hence at the moment of interest the value of x is:

𝑥 = 20 m - 5.0 m = 15.0 m

Finally, then:

𝑑 𝑦 𝑑 𝑡 = - 36 𝑥 2 𝑑 𝑥 𝑑 𝑡 = - 36 (15.0) 2 (1.5 m s) = - 0.24 m s ✓

That is, the man's shadow is changing at the rate

𝑑 𝑦 𝑑 𝑡 = - 0.24 m s

. The negative value indicates that the shadow's height y is decreasing, as we expect.

$Web-based homework warning icon$ Caution: IF you are using a web-based homework system and the question asks,

At what rate does the shadow's height decrease?

then the system has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE:

0.24 m s ✓

[close solution]

Water drains from a cone at constant rate; water's level falls at what rate?

How fast is the water level falling as water drains from the cone?

Water drains from a cone at 15 cubic-cm each second.

An inverted cone is 20 cm tall, has an opening radius of 8 cm, and was initially full of water. The water drains from the cone at the constant rate of 15 cm $3$ each second. The water's surface level falls as a result. At what rate is the water level falling when the water is halfway down the cone?
(Note: The volume of a cone is $13 𝜋 𝑟 2 ℎ$ . You may leave $𝜋$ in your answer; do not use a calculator to find a decimal answer.)

View/Hide Solution

Let's unpack the question statement:

We're told that volume of water in the cone V is changing at the rate of $𝑑 𝑉 𝑑 𝑡 = - 15$ cm $3$ /s. (We must insert the negative sign "by hand" since we are told that the water is draining out, and so its volume is decreasing.)
As a result, the water's height in the cone h is changing at the rate $𝑑 ℎ 𝑑 𝑡$ , which is the quantity we're after.
The inverted cone has a radius of 8 cm at its top, and a full height of 20 cm.
The problem is asking us about $𝑑 ℎ 𝑑 𝑡$ at a particular instant, when the water is halfway down the cone, and so when $ℎ = 10$ cm. We'll use this value toward the end of our solution.

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 height of the water changes as time passes, so we're going to keep that height as a variable, h.

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

This is the hardest part of Related Rates problem for most students initially: you have to know how to develop the equation you need, how to pull that "out of thin air." By working through these problems you'll develop this skill. The key is to recognize which of the few sub-types of problem it is, as listed in our Problem Solving Strategy box at the top of the page.

Here we need to develop a relationship between the rate we're given, $𝑑 𝑉 𝑑 𝑡 = - 15$ cm $3$ /s, and the rate we're after, $𝑑 ℎ 𝑑 𝑡$ . We thus first need to write down a relationship between the water's volume V and its height-in-the-cone h. But we know that relationship since it was given in the problem statement:
$𝑉 = 13 𝜋 𝑟 2 ℎ$
Notice that this relation expresses the water's volume as the function of two variables, r and h. We can only take the derivative with respect to one variable, so we need to eliminate one of those two. Since the question asks us to find the rate at which the water is falling when its at a particular height, let's keep h and eliminate r as a variable using similar triangles.

Begin subproblem to eliminate r as a variable.

The water's volume and the full cone form similar triangles.

The figure is the same as in Step 1, but with the rest of the cone removed for clarity. Note that there are two triangles, a small one inside a larger one. Because these are similar triangles, the ratio of the base of the small triangle to that of the big triangle $(𝑟 8)$ must equal the ratio of the height of the small triangle to that of the big triangle $(ℎ 20)$ :

𝑟 8 = ℎ 20

Hence

𝑟 = 820 ℎ = 25 ℎ

End subproblem.

Now let's substitute the expression we just found for r into our relation for V:

𝑉 = 13 𝜋 𝑟 2 ℎ = 13 𝜋 (25 ℎ) 2 ℎ = 13 𝜋 425 ℎ 3 = 475 𝜋 ℎ 3

That's it. That's the key relation we need to be able to proceed with the rest of the solution.

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

$𝑑 𝑑 𝑡 𝑉 = 𝑑 𝑑 𝑡 (475 𝜋 ℎ 3) 𝑑 𝑉 𝑑 𝑡 = 475 𝜋 𝑑 𝑑 𝑡 (ℎ 3) = 475 𝜋 (3 ℎ 2 𝑑 ℎ 𝑑 𝑡) = 425 𝜋 ℎ 2 𝑑 ℎ 𝑑 𝑡$

4. Solve for the quantity you're after.

We have $𝑑 𝑉 𝑑 𝑡 = - 15 c m 3 s$ , and want to find $𝑑 ℎ 𝑑 𝑡$ at the instant when h = 10 cm.
Starting from our preceding expression, let's first solve for $𝑑 ℎ 𝑑 𝑡$ and then substitute the values we're given:

$𝑑 𝑉 𝑑 𝑡 = 425 𝜋 ℎ 2 𝑑 ℎ 𝑑 𝑡 𝑑 ℎ 𝑑 𝑡 = 25 4 𝜋 ℎ 2 𝑑 𝑉 𝑑 𝑡 = 25 4 𝜋 (10) 2 (- 15) = 25 4 𝜋 (100) (- 15) = - 15 16 𝜋 c m / s ✓$
That's the answer. The negative value indicates that the water's height h is decreasing, as we expect.

$Web-based homework warning icon$

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

At what rate does the water's height decrease?

then the system has possibly already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $15 16 𝜋 c m s ✓$

This has everything to do with how the coder entered this problem into your homework system and nothing to do with the math or your understanding!

[close solution]

The following problem is essentially the same as the preceding one, except that water flows into instead of out of the cone. We include it here so you can see how the sign of $𝑑 𝑉 𝑑 𝑡$ propogates through to change the sign of $𝑑 ℎ 𝑑 𝑡,$ something students often ask about.

Water fills a cone at constant rate; water's level rises at what rate?

Practice Problem: Water fills a cone

Water is poured at a uniform rate of

15 c m 3 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

13 𝜋 𝑟 2 ℎ

. You may leave

𝜋

in your answer; do not use a calculator to find a decimal answer.)

Answer:

5 3 𝜋

cm/s

$Water fills cone to height h and has surface radius r$ 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

𝑑 𝑉 𝑑 𝑡 = 15 c m 3 s

. We want to find the rate at which the water is rising,

𝑑 ℎ 𝑑 𝑡

, 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 = 13 𝜋 𝑟 2 ℎ

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

(𝑟 6)

must be equal to the ratio of their heights

(ℎ 16)

𝑟 6 = ℎ 16 𝑟 = 616 ℎ = 38 ℎ

Then substituting this expression into our relation for

𝑉

𝑉 = 13 𝜋 (38 ℎ) 2 ℎ = 364 𝜋 ℎ 3

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

𝑑 𝑉 𝑑 𝑡 = 𝑑 𝑑 𝑡 (364 𝜋 ℎ 3) = 364 𝜋 (3 ℎ 2 𝑑 ℎ 𝑑 𝑡) = 964 𝜋 ℎ 2 𝑑 ℎ 𝑑 𝑡

4. Solve for the quantity you're after.
We have

𝑑 𝑉 𝑑 𝑡 = 15 c m 3 s

, and want to find

𝑑 ℎ 𝑑 𝑡

, at the instant when

ℎ = 8

cm.

𝑑 ℎ 𝑑 𝑡 = 64 9 𝜋 ℎ 2 𝑑 𝑉 𝑑 𝑡 = 64 9 𝜋 (8) 2 (15) = 64 3 𝜋 (64) (5) = 5 3 𝜋 c m / s ✓

[close solution]

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.

Note again the importance of starting with a clear figure, so that you can see the right triangle!

The following problems illustrate.

Ladder slides down a wall; how fast does the top move down (video solution)

How fast is the ladder's top sliding down?

A 10-ft ladder is leaning against a house on flat ground. The house is to the left of the ladder. The base of the ladder starts to slide away from the house. When the base has slid to 8 ft from the house, it is moving horizontally at the rate of 2 ft/sec. How fast is the ladder's top sliding down the wall when the base is 8 ft from the house?

View/Hide Solution

We have a video for this solution; a full written-out solution is below.

1. Draw a picture of the physical situation.

Ladder leaning against a wall slides away. How fast is the ladder's top sliding at a particular instant?

Let x be the horizontal distance, in feet, from the wall to the bottom of the ladder.

Let y be the distance, in feet, from the ground to the top of the ladder.

The problem tells us that at the moment of interest, when x = 8 ft, $𝑑 𝑥 𝑑 𝑡 = 2$ ft/sec. We'll use these values only 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.
In this situation, both x and y change as the ladder slides, so we will leave both quantities as variables.

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
This is the hardest part of Related Rates problem for most students initially: you have to know how to develop the equation you need, how to pull that "out of thin air." By working through these problems you'll develop this skill. The key is to recognize which of the few sub-types of problem it isas described in the Problem Solving Strategy box at the top of this page. 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 y is changing, $𝑑 𝑦 𝑑 𝑡$ , to the rate at which x is changing, $𝑑 𝑥 𝑑 𝑡$ , and so we first need to write down an equation that somehow relates x and y.

While x and y change as the ladder slides, the hypotenuse of the right triangle shown is always equal to the ladder's length, 10 ft. Hence the Pythagorean theorem applies:
$𝑥 2 + 𝑦 2 = (10) 2 = 100$
That's it. That's the key relation we need to be able to proceed with the rest of the solution.

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

$𝑑 𝑑 𝑡 (𝑥 2 + 𝑦 2) = 𝑑 𝑑 𝑡 (100) 2 𝑥 𝑑 𝑥 𝑑 𝑡 + 2 𝑦 𝑑 𝑦 𝑑 𝑡 = 0$

4. Solve for the quantity you're after.

The question is asking us to find $𝑑 𝑦 𝑑 𝑡$ at the instant when x = 8 ft and $𝑑 𝑥 𝑑 𝑡 = 2$ ft/sec. So let's solve the preceding equation for $𝑑 𝑦 𝑑 𝑡$ :
$2 𝑥 𝑑 𝑥 𝑑 𝑡 + 2 𝑦 𝑑 𝑦 𝑑 𝑡 = 02 𝑦 𝑑 𝑦 𝑑 𝑡 = - 2 𝑥 𝑑 𝑥 𝑑 𝑡 𝑑 𝑦 𝑑 𝑡 = - 𝑥 𝑦 𝑑 𝑥 𝑑 𝑡 [*]$
To complete the calculation, we need to know the value of y at the instant when x = 8.

Begin subproblem to find the value of y at the instant when x = 8 ft.

We can find this value by using the Pythagorean theorem:
$(8) 2 + 𝑦 2 = 100 𝑦 2 = 100 - 64 = 36 𝑦 = 6$
End subproblem.

Substituting all of the known values into the equation marked [*] above, we have:
$𝑑 𝑦 𝑑 𝑡 = - 𝑥 𝑦 𝑑 𝑥 𝑑 𝑡 = - 86 (2) = - 83 f t / s ✓$
That's the answer. The negative value indicates that the top of the ladder is sliding down the wall, in the negative-y direction.

$Web-based homework warning icon$

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

At what rate does the ladder slide down the wall?

then the system may have already accounted for the negative sign and so to be correct you may need to enter a POSITIVE VALUE: $83 f t s ✓$

This is annoying we agree, and has everything to do with whoever entered the problem into the system. Making matters worse, since different problems were probably entered by different people, within a given problem set you may find inconsistencies here; the best we can do is help you be aware.

[close solution]

Rectange maintains constant area as its sides change; find the rate its diagonal changes

Practice Problem: 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.)

$Rectangle with diagonal d$ 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

𝑑 𝑤 𝑑 𝑡

and

𝑑 𝑙 𝑑 𝑡

terms:

𝑑 𝑑 𝑡 𝑑 2 = 𝑑 𝑑 𝑡 (𝑤 2 + 𝑙 2) 2 𝑑 𝑑 𝑑 𝑑 𝑡 = 2 𝑤 𝑑 𝑤 𝑑 𝑡 + 2 𝑙 𝑑 𝑙 𝑑 𝑡 𝑑 𝑑 𝑑 𝑑 𝑡 = 𝑤 𝑑 𝑤 𝑑 𝑡 + 𝑙 𝑑 𝑙 𝑑 𝑡 [*]

4. Solve for the quantity you're after.
We want to find

𝑑 𝑑 𝑑 𝑡

at the instant when

𝑤 = 10

cm. The problem statement tells us that

𝑑 𝑙 𝑑 𝑡 = 8

cm/s. In order to use the preceding equation (marked [*]), we need to determine

𝑑

and

𝑙

at this instant, and also

𝑑 𝑤 𝑑 𝑡

at this instant.

First, we can find

𝑙

by noting that the rectangle has constant area,

𝐴 = 500 = 𝑙 𝑤

. Hence when

𝑤 = 10

cm:

𝑙 = 𝐴 𝑤 = 50010 = 50 c m

Second, we can find

𝑑

at this instant using the Pythagorean theorem:

𝑑 2 = 𝑤 2 + 𝑙 2 = (10) 2 + (50) 2 = 100 + 2500 = 2600 𝑑 = \sqrt 2600 = 10 \sqrt 26

Next, we can find

𝑑 𝑤 𝑑 𝑡

at this instant by taking the derivative of

𝐴 = 500 = 𝑙 𝑤

𝑑 𝑑 𝑡 (500) = 𝑑 𝑑 𝑡 (𝑙 𝑤) 0 = 𝑙 𝑑 𝑤 𝑑 𝑡 + 𝑤 𝑑 𝑙 𝑑 𝑡 𝑙 𝑑 𝑤 𝑑 𝑡 = - 𝑤 𝑑 𝑙 𝑑 𝑡

Hence

𝑑 𝑤 𝑑 𝑡 = - (𝑤 𝑙) (𝑑 𝑙 𝑑 𝑡) = - (1050) (8) = - 85 c m / s

Then substituting values into the equation marked [*] above:

𝑑 𝑑 𝑑 𝑑 𝑡 = 𝑤 𝑑 𝑤 𝑑 𝑡 + 𝑙 𝑑 𝑙 𝑑 𝑡 (10 \sqrt 26) 𝑑 𝑑 𝑑 𝑡 = (10) (- 85) + (50) (8) = - 16 + 400 = 384 𝑑 𝑑 𝑑 𝑡 = 384 10 \sqrt 26 = 38.4 \sqrt 26 c m / s ✓

[close solution]

Kite flies horizontally; find the rate at which the length of the string is changing

Practice Problem: 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?

A kite travels horizontally 30 meters above the ground

1. Draw a picture of the physical situation.
See the figure. We've called the length of string to the kite

ℓ .

The problem is asking for the rate at which that length changes (= the rate at which the string is unspooling),

𝑑 ℓ 𝑑 𝑡

, at a particular moment — when

ℓ = 50

meters. Recall also that the kite is moving horizontally, in the x-direction, at the rate

𝑑 𝑥 𝑑 𝑡 = 2 m s

ℓ .

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

ℓ

is changing,

𝑑 ℓ 𝑑 𝑡

, to the rate at which x is changing,

𝑑 𝑥 𝑑 𝑡

. Hence we must first write down an equation that somehow relates

ℓ

and x. The Pythagorean theorem gives us that relation:

ℓ 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.

𝑑 𝑑 𝑡 ℓ 2 = 𝑑 𝑑 𝑡 𝑥 2 + 𝑑 𝑑 𝑡 302 0 2 ℓ 𝑑 ℓ 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡 ℓ 𝑑 ℓ 𝑑 𝑡 = 𝑥 𝑑 𝑥 𝑑 𝑡

4. Solve for the quantity you're after.
Let's solve the preceding equation for

𝑑 ℓ 𝑑 𝑡

𝑑 ℓ 𝑑 𝑡 = 𝑥 ℓ 𝑑 𝑥 𝑑 𝑡

To complete the solution, we need to know the value of x at the instant when

ℓ = 50

Begin subproblem to find x at the moment of interest.
Recall from the Pythagorean theorem that at every moment

ℓ 2 = 𝑥 2 + (30) 2

Hence when

ℓ = 50

𝑥 2 = ℓ 2 - 302 = (50) 2 - 302 = 1600 𝑥 = 40

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:

𝑑 ℓ 𝑑 𝑡 = 𝑥 ℓ 𝑑 𝑥 𝑑 𝑡

We have

𝑥 = 40

ℓ = 50

m, and

𝑑 𝑥 𝑑 𝑡 = 2 m s

𝑑 ℓ 𝑑 𝑡 = 𝑥 ℓ 𝑑 𝑥 𝑑 𝑡 = 40 m 50 m (2 m s) = 1.6 m s ✓

[close solution]

Two joggers run toward an intersection; find the rate at which the distance between them changes

Practice Problem: 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:

- 395

mph

$One jogger headed north, another west$ 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

𝑑 𝑑 𝑑 𝑡

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.

Pythagorean triangle

B. To develop your equation, you will probably use . . . the Pythagorean theorem.
We can relate

𝑑

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.

𝑑 𝑑 𝑡 (𝑑 2) = 𝑑 𝑑 𝑡 (𝑥 2 + 𝑦 2) 2 𝑑 𝑑 𝑑 𝑑 𝑡 = 2 𝑥 𝑑 𝑥 𝑑 𝑡 + 2 𝑦 𝑑 𝑦 𝑑 𝑡 𝑑 𝑑 𝑑 𝑑 𝑡 = 𝑥 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡 [*]

4. Solve for the quantity you're after.
We want to find

𝑑 𝑑 𝑑 𝑡

at noon, when

𝑥 = 3

miles and

𝑦 = 4

miles. We also know that

𝑑 𝑥 𝑑 𝑡 = - 5

mph, and

𝑑 𝑦 𝑑 𝑡 = - 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

𝑑 𝑑 𝑑 𝑡

, we also need to know

𝑑

; this we can find by using the Pythagorean theorem at this instant:

𝑑 2 = 𝑥 2 + 𝑦 2 = 32 + 42 = 25 𝑑 = 5

Then substituting values into the equation marked [*] above:

𝑑 𝑑 𝑑 𝑑 𝑡 = 𝑥 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡 (5) 𝑑 𝑑 𝑑 𝑡 = (3) (- 5) + (4) (- 6) = - 15 - 24 = - 39 𝑑 𝑑 𝑑 𝑡 = - 395 m p h ✓

The negative value indicates that the distance between the joggers is decreasing as they jog toward each other.

[close solution]

Two ships travel at right angles; find the rate at which the distance between them changes

Practice Problem: 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?

Ship Red and ship Blue's initial positions, and their positions later at time t

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

ℓ .

We're looking for

𝑑 ℓ 𝑑 𝑡

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

𝑑 𝑥 𝑑 𝑡

and

𝑑 𝑦 𝑑 𝑡,

and asked to find the value for

𝑑 ℓ 𝑑 𝑡

at a particular moment. We thus first need to relate

ℓ

to x and y.

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have

ℓ 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.

𝑑 𝑑 𝑡 [ℓ 2] = 𝑑 𝑑 𝑡 [(100 - 𝑥) 2] + 𝑑 𝑑 𝑡 [𝑦 2] 2 ℓ 𝑑 ℓ 𝑑 𝑡 = 2 (100 - 𝑥) (- 1) 𝑑 𝑥 𝑑 𝑡 + 2 𝑦 𝑑 𝑦 𝑑 𝑡 ℓ 𝑑 ℓ 𝑑 𝑡 = - (100 - 𝑥) 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡

4. Solve for the quantity you're after.
Let's solve the preceding equation for

𝑑 ℓ 𝑑 𝑡 :

ℓ 𝑑 ℓ 𝑑 𝑡 = - (100 - 𝑥) 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡 𝑑 ℓ 𝑑 𝑡 = - (100 - 𝑥) 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡 ℓ (*)

We know

𝑑 𝑥 𝑑 𝑡 = 30 k m h r

and

𝑑 𝑦 𝑑 𝑡 = 20 k m h r .

To complete the problem, we need to know the values of x,y, and

ℓ

at the moment of interest.

Begin subproblem to find x, y, and $ℓ$ .
We're interested in the particular moment 12:00-noon, which is t = 3 hours after our initial picture at 9:00 AM. Hence

𝑥 = (30 k m h r) (3 h r) = 90 k m 𝑦 = (20 k m h r) (3 h r) = 60 k m

And using the Pythagorean theorem we can find

ℓ

at this moment:

ℓ 2 = (100 - 𝑥) 2 + 𝑦 2 = (100 - 90) 2 + (60) 2 = 3700 ℓ = \sqrt 3700 = 60.83 k m

End subproblem.

We finally substitute all of these values into our equation marked (*):

𝑑 ℓ 𝑑 𝑡 = - (100 - 𝑥) 𝑑 𝑥 𝑑 𝑡 + 𝑦 𝑑 𝑦 𝑑 𝑡 ℓ = - (100 - 90) (30) + (60) (20) 60.83 = - 300 + 1200 60.83 = 14.8 k m h r ✓

[close solution]

Rocket flies straight up; find the rate at which the distance between the rocket and an observer changes

Practice Problem: Rocket, distance changes

A rocket is launched straight up, and its altitude is

𝑦 = 5 𝑡 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?

Rocket travels vertically to height y, at a distance 500 m from the observer.

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

ℓ .

The question is asking us to find

𝑑 ℓ 𝑑 𝑡

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

𝑑 ℓ 𝑑 𝑡,

and we know (actually: we can easily determine) the rate at which the rocket's y- position is changing,

𝑑 𝑦 𝑑 𝑡

. We thus first need to related

ℓ

to y.

The figure above suggests how to do so: the Pythagorean theorem. Specifically, at every moment we have

ℓ 2 = (500) 2 + 𝑦 2

𝑑 𝑑 𝑡 (ℓ 2) = 𝑑 𝑑 𝑡 (500) 2 0 + 𝑑 𝑑 𝑡 (𝑦 2) 2 ℓ 𝑑 ℓ 𝑑 𝑡 = 2 𝑦 𝑑 𝑦 𝑑 𝑡 ℓ 𝑑 ℓ 𝑑 𝑡 = 𝑦 𝑑 𝑦 𝑑 𝑡

4. Solve for the quantity you're after.
Let's solve the preceding equation for

𝑑 ℓ 𝑑 𝑡 :

ℓ 𝑑 ℓ 𝑑 𝑡 = 𝑦 𝑑 𝑦 𝑑 𝑡 𝑑 ℓ 𝑑 𝑡 = 𝑦 ℓ 𝑑 𝑦 𝑑 𝑡 (*)

To complete our solution, we need to know the values of y,

ℓ,

and

𝑑 ℓ 𝑑 𝑡

6 seconds after launch.

Begin subproblems to find y, $ℓ,$ and $𝑑 ℓ 𝑑 𝑡$ .
A. Find y.
The problem states that

𝑦 = 5 𝑡 2

meters after t seconds. Hence 6 seconds after launch

𝑦 = 5 (6) 2 = 180 m

B. Find $ℓ .$
By the Pythagorean theorem, at this instant

ℓ 2 = (500) 2 + (180) 2 = 282400 ℓ = \sqrt 282400 = 531 m

C. Find $𝑑 𝑦 𝑑 𝑡 .$
Since

𝑦 = 5 𝑡 2,

𝑑 𝑦 𝑑 𝑡 = 10 𝑡

Hence at

𝑡 = 6

seconds,

𝑑 𝑦 𝑑 𝑡 ∣ 6 s = 10 (6) = 60 m s

End subproblems.

We can now substitute these values into our equation marked (*) above:

𝑑 ℓ 𝑑 𝑡 = 𝑦 ℓ 𝑑 𝑦 𝑑 𝑡 = 180531 (60) = 20.3 m s ✓

[close solution]

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