## Related Rates

### Calculus Related Rates Problem:

At what rate does the angle change as a ladder slides away from a house?

A 10-ft ladder leans 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 at 2 ft/s. At what rate is the angle between the ladder and the ground changing when the base is 8 ft from the house?

### Calculus Solution

To solve this problem, we will use our standard 4-step Related Rates Problem Solving Strategy.

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

See the figure. We’ve labeled the angle $\theta$ that the ladder 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 $x = 8$. Recall also that $\dfrac{dx}{dt} = 2$ ft/s. We’ll use these values at the *end* of our solution.

**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). *

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; we’ve listed each on our Related Rates 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 $\theta$ 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 hypotenuse of the triangle remains constant throughout the problem, since the ladder’s length is always 10 ft. Hence at every moment:

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

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} \cos{\theta} &= \frac{d}{dt} \left( \frac{x}{10} \right) \\ \\

&= \frac{1}{10} \frac{d}{dt}(x) \\ \\

-\sin{\theta}\, \frac{d\theta}{dt} &= \frac{1}{10}\frac{dx}{dt}

\end{align*}

While the derivative of $\cos \theta$ with respect to angle $\theta$ is

$$\dfrac{d}{d\theta}\cos \theta = -\sin \theta,$$

the derivative of $\cos \theta$ with respect to *time t* is

$$\dfrac{d}{dt}\cos \theta = -\sin \theta \,\dfrac{d\theta}{dt}.$$

Similarly, while the derivative of *x* with respect to *x* is

$$\dfrac{d}{dx}x = 1,$$

the derivative of *x* with respect to *time t* is

$$\dfrac{d}{dt}x = \dfrac{dx}{dt}.$$

(Recall that that rate is $\dfrac{dx}{dt} = 2$ ft/s in this problem.)

Remember that $\theta$ and *x* are both functions of time *t*: the angle *changes* as time passes and the ladder’s *x*-position *changes* as the ladder slides down the wall. We could have captured this time-dependence explicitly by writing our relation as

$$ \cos \theta(t) = \dfrac{x(t)}{10}$$

to remind ourselves that both $\theta$ and *x* are functions of time *t*. Then when we take the derivative,

\begin{align*}

\frac{d}{dt}\cos \theta(t) &= \frac{d}{dt}\left( \dfrac{x(t)}{10}\right) \\ \\

\left(-\sin \theta(t)\right) \dfrac{d\theta(t)}{dt}&= \dfrac{1}{10}\dfrac{dx(t)}{dt}

\end{align*}

[Recall $\dfrac{dx(t)}{dt} = 2$ ft/s, and we’re looking for $\dfrac{d\theta(t)}{dt}$ at the moment when *x* = 8 ft.]

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.

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

Let’s solve the preceding equation for $\dfrac{d\theta}{dt}$:

\begin{align*}

-\sin{\theta}\, \frac{d\theta}{dt} &= \frac{1}{10}\frac{dx}{dt} \\ \\

\frac{d\theta}{dt} &= -\frac{1}{\sin{\theta}}\frac{1}{10}\frac{dx}{dt}

\end{align*}

To complete the solution, we need to know the value of $\sin \theta$ at the moment when x = 8 ft.

*Begin subproblem to find* $\sin \theta$ *at the moment of interest.*

You can use any of these three approaches:

**Approach #1**:

Looking back at the figure, we see that

$$ \sin \theta = \dfrac{y}{10}$$

Next, recognize that at this instant the triangle is a “3-4-5 right triangle,” with the actual proportions 6-8-10. Hence *y* = 6 ft at this instant, and so

$$\sin\theta = \dfrac{y}{10} = \dfrac{6}{10} = \dfrac{3}{5}$$

**Approach #2**:

Looking back at the original figure, we see that

$$ \sin \theta = \dfrac{y}{10}$$

So we need to know the value of *y* when *x* = 8 ft. The Pythagorean theorem as applied to the triangle lets us solve for *y* at this instant:

\begin{align*}

x^2 + y^2 &= 10^2 \\[4px]
8^2 + y^2 &= 10^2 \\[4px]
64 + y^2 &= 100 \\[4px]
y^2 &= 100 – 64 = 36 \\[4px]
y &= 6

\end{align*}

Hence at this instant

$$\sin\theta = \dfrac{y}{10} = \dfrac{6}{10} = \dfrac{3}{5}$$

**Approach #3**.

Recall the trig identity $\sin^2 \theta + \cos^2 \theta = 1$:

\begin{align*}

\sin^2 \theta + \cos^2 \theta &= 1 \\[4px]
\sin^2 \theta + \left(\frac{8}{10} \right)^2 &= 1 \\[4px]
\sin^2 \theta + \left(\frac{4}{5} \right)^2 &= 1 \\[4px]
\sin^2 \theta + \frac{16}{25} &= 1 \\[4px]
\sin^2 \theta &= 1 – \frac{16}{25} = \frac{9}{25} \\[4px]
\sin \theta = \frac{3}{5}

\end{align*}

*End subproblem.*

We can now substitute values into our preceding equation:

$$\frac{d\theta}{dt} = -\frac{1}{\sin{\theta}}\frac{1}{10}\frac{dx}{dt}$$

We have $\sin \theta = \dfrac{3}{5}$ and $\dfrac{dx}{dt} =2$:

\begin{align*}

\frac{d\theta}{dt} & = -\frac{1}{\sin{\theta}}\frac{1}{10}\frac{dx}{dt} \\ \\

&= \, -\frac{1}{3/5}\frac{1}{10}(2)\\ \\

&=\, – \frac{5}{3}\frac{1}{10}(2)\\ \\

&= \, – \frac{1}{3} \text{ rad/s} \quad \cmark

\end{align*}

That’s the answer. The negative value indicates that the angle is decreasing at the ladder slides down the wall, as we expect.

**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: $\boxed{\dfrac{1}{3}} \, \dfrac{\text{rad}}{\text{s}} \quad \checkmark$

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you need.

COULD IT WORK WITH TANGENT

[Note that you may need to reload your webpage for this reply to format correctly.]I imagine you mean can you start with

\[\tan \theta = \frac{y}{x}\]

instead of using $\cos \theta = x/10$ as we did above?

Assuming that right: great question, and we love that you’re thinking this through!

Writing $\tan \theta = y/x$ in Step 2 would be a correct representation of the situation given the figure we drew in Step 1. So up to that point, yes.

However, you’d run into a problem in the following steps: When you take the derivative in Step 3, because both

xandyare changing as the ladder slides down the wall, you would have both $\dfrac{dx}{dt}$and$\dfrac{dy}{dt}$ terms. And then when you come to Step 4, we don’t know the value of that latter quantity, so you’d be stuck. (Note also that the derivative would be much harder to take in Step 3, since bothxandyare variables and so you’d have to use the quotient rule. By contrast, when using $\cos \theta$ the 10 in the denominator on the right is a constant and so the derivative is quite simple to write down.)So in short: as a practical matter, no, you can’t use $\tan \theta$ to arrive at the solution here. (At least not nearly as easily. You actually find $\dfrac{dy}{dt}$ as its own problem – see “How fast is the ladder’s top sliding?” But then you’re solving an entirely different problem as another sub-part of this problem when it’s not necessary.)

But we want to emphasize that starting your solution attempt with $\tan \theta = y/x$ is perfectly reasonable. It’s only through experience (and heading down a

lotof ultimately-unproductive paths) while we were learning this stuff ourselves that we now know more quickly which the best approach to take is. So: keep practicing!!I hope that answers your question, and please let us know if not! : )