Question

The bottom of a 17-foot ladder is on the ground and the top rests against a vertical wall. If the bottom is pushed toward the wall at the rate of 3 feet per second, how fast is the top moving up the wall when the bottom of the ladder is 15 feet from the base of the wall?

Answer 1

Answer:

Approximately $5.6\; {\rm ft \cdot s^{-1}}$

Step-by-step explanation:

Let $t$ denote time. Let $f(t)$ denote the distance between the bottom of the ladder and the base of the wall at time $t\!$. Similarly, let $g(t)$ denote the height of the top of the ladder at time $t$.

This question is an example of a related rates problem. Let $t_{0}$ denote the time at this instant.

• Rate of change in the horizontal distance $f$ at this instant is given: $f^{\prime}(t_{0}) = (-3)\; {\rm ft \cdot s^{-1}}$. This rate of change is negative since the ladder is moving towards the wall, such that the value of $f\!$ is being reduced.
• Value of $f$ at this instant is given: $f(t_{0}) = 15\; {\rm ft}$.
• Rate of change in height $g$ at this instant, $g^{\prime}(t_{0})$, needs to be found.
• Refer to the diagram attached. The value $g(t)\!$ is not directly given as a function of $t$. However, since the length of the ladder (hypotenuse) is given, the value of $g(t)$ can be expressed in terms of $f(t)$ using the Pythagorean Theorem:
• $\displaystyle g(t) = \sqrt{17^{2} - (f(t))^{2}}$.

To solve this related rates problem and find $g^{\prime}(t_{0})$, start by differentiating the expression for $g(t)$ (with respect to $t$). The goal is to express $g(t_{0})$ in terms of the known quantities: $f(t_{0}) = 15\; {\rm ft}$ and $f^{\prime}(t_{0}) = (-3)\; {\rm ft \cdot s^{-1}}$.

$\begin{aligned} g^{\prime}(t) &= \displaystyle \frac{d}{d t}\left[\sqrt{17^{2} - (f(t))^{2}}\right] \\ &= \displaystyle \frac{d}{d t}\left[\left(17^{2} - (f(t))^{2}\right)^{1/2}\right] \\ &= \frac{1}{2}\, \left(17^{2} - (f(t))^{2}\right)^{-1/2}\, \frac{d}{d t}\left[17^{2}-(f(t))^{2}\right]\\ &= \frac{1}{2}\, \left(17^{2} - (f(t))^{2}\right)^{-1/2}\, (-1)\, \frac{d}{d t}\left[(f(t))^{2}\right]\end{aligned}$.

To differentiate $(f(t))^{2}$ with respect to $t$ implicitly, apply the chain rule of differentiation:

$\begin{aligned} \frac{d}{dt}\left[(f(t))^{2}\right] &= 2\, (f(t))\, \frac{d}{dt}\left[f(t)\right] = 2\, f(t)\, f^{\prime}(t)\end{aligned}$.

Hence, the expression for $g^{\prime}(t)$ becomes:

$\begin{aligned} g^{\prime}(t) &= \displaystyle \frac{d}{d t}\left[\sqrt{17^{2} - (f(t))^{2}}\right] \\ &= \cdots \\ &= \frac{1}{2}\, \left(17^{2} - (f(t))^{2}\right)^{-1/2}\, (-1)\, \frac{d}{d t}\left[(f(t))^{2}\right] \\ &= \frac{1}{2}\, \left(17^{2} - (f(t))^{2}\right)^{-1/2}\, (-1)\, \left(2\, f(t)\, f^{\prime}(t)\right) \\ &= \frac{-\, f(t)\, f^{\prime}(t)}{\sqrt{17^{2} - (f(t))^{2}}}\end{aligned}$.

Substitute in $t = t_{0}$ into $g^{\prime}(t)$ and evaluate to find the rate of change in $g$ at this instant, $g^{\prime}(t_{0})$:

$\begin{aligned} g^{\prime}(t_{0}) &= \frac{-\, f(t_{0})\, f^{\prime}(t_{0})}{\sqrt{17^{2} - (f(t_{0}))^{2}}} \\ &= \frac{-(15\; {\rm ft})\, ((-3)\; {\rm ft\cdot s^{-1}})}{\sqrt{(17\; {\rm ft})^{2} - (15\; {\rm ft})^{2}}} \\ &= \frac{45}{8}\; {\rm ft\cdot s^{-1}} \\ &\approx 5.6\; {\rm ft\cdot s^{-1}}\end{aligned}$.

In other words, the top of the ladder would be moving upward at approximately $5.6\; {\rm ft \cdot s^{-1}}$ at this instance.