## Related Rates

### Calculus Related Rates Problem:

How fast is the ladder’s top sliding?

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?

### 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.

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, $\dfrac{dx}{dt} = 2$ 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 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 *y* is changing, $\dfrac{dy}{dt}$, to the rate at which *x* is changing, $\dfrac{dx}{dt}$, and so we first need to write down an equation that somehow relates *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. **

\begin{align*}

\frac{d}{dt} \left( x^2 + y^2 \right) &= \frac{d}{dt} \left(100 \right) \\ \\

2x \frac{dx}{dt} + 2y \frac{dy}{dt} & = 0 \text{ }

\end{align*}

Please jump ahead in the video to 5:40 for the relevant discussion:

Are you wondering why the $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ appear? 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 $\dfrac{d}{dt}x^2 = 2x\dfrac{dx}{dt}$. Similarly, the derivative with respect to *time t* of $y^2$ is $\dfrac{d}{dt}y^2 = 2y\dfrac{dy}{dt}$.

Remember that *x* and *y* are both functions of time *t*: both positions *change* as time passes and the ladder slides down the wall. We could have captured this time-dependence explicitly by writing our relation as

$$[x(t)]^2 + [y(t)]^2 = 100$$

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

\begin{align*}

\frac{d}{dt}\left([x(t)]^2 + [y(t)]^2 \right) &= \frac{d}{dt}(100) \\ \\

2x(t) \dfrac{dx}{dt} + 2y(t) \dfrac{dy}{dt} &= 0

\end{align*}

[Recall $\dfrac{dx}{dt} = 2$ ft/s at the moment of interest, and we’re looking for $\dfrac{dy}{dt}$.]

Most people find that writing the explicit time-dependence *x(t)* and *y(t)* annoying, and so just write *x* and *y* instead. Regardless, you *must* remember that both *x* and *y* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ terms.

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

The question is asking us to find $\dfrac{dy}{dt}$ at the instant when *x* = 8 ft and $\dfrac{dx}{dt} = 2$ ft/sec. So let’s solve the preceding equation for $\dfrac{dy}{dt}$:

\begin{align*}

2x \frac{dx}{dt} + 2y \frac{dy}{dt} & = 0 \\ \\

2y \frac{dy}{dt} &= -2x \frac{dx}{dt} \\ \\

\frac{dy}{dt} &= – \frac{x}{y} \frac{dx}{dt} \text{ } [*]
\end{align*}

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:

\begin{align*}

(8)^2 + y^2 &= 100 \\

y^2 &= 100 – 64 = 36 \\

y &= 6

\end{align*}

*End subproblem.*

Substituting all of the known values into the equation marked [*] above, we have:

\begin{align*}

\frac{dy}{dt} &= – \frac{x}{y} \frac{dx}{dt} \\ \\

&= -\frac{8}{6} (2) = -\frac{8}{3} \text{ ft/s} \quad \cmark

\end{align*}

That’s the answer. The negative value indicates that the top of the ladder is sliding

*down*the wall, in the negative-

*y*direction.

**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 has already accounted for the negative sign and so to be correct you must enter a POSITIVE VALUE: $\boxed{\dfrac{8}{3}} \, \dfrac{\text{ft}}{\text{s}} \quad \checkmark$

Return to Related Rates Problems

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