Question

A 13-foot ladder leans against the side of a building, forming an angle with the ground. Given that the foot of the ladder is being pulled away from the building at the rate of 0.1 feet per second, what is the rate of change of when the top of the ladder is 12 feet above the ground

Answer 1

Step-by-step explanation:

We have $\mathrm{dx} / \mathrm{dt}=0.1 \mathrm{ft} / \mathrm{sec},$ and we want

dy/dt.

$x$ and $y$ are related by the Pythagorean Theorem $x^{2}+y^{2}=(13 f t)^{2}$

Differentiate both sides of this equation with respect to $t$ to get $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$

$\frac{d y}{d t}=-\frac{x}{y} \frac{d x}{d t}$

When $x=12 \mathrm{ft},$  

we have,  

$y=\sqrt{(13 \mathrm{ft})^{2}-(12 \mathrm{ft})^{2}}=5 \mathrm{ft}$

therefore

$\frac{d y}{d t}=-\frac{12f t}{5 f t} \cdot \frac{1 f}{\sec }=-\frac{12}{5} \frac{f t}{\sec }$