Related Rates III

Problem 165: A 10ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2ft per sec, how fast is the bottom of the ladder moving away from the wall when the bottom of the ladder is 5ft from the wall?

Solution: The ladder and the wall make a right triangle. We are given the length of one side $y = 5$ ft and the length of the hypotenuse $z = 10$ ft. That is, we find the other side

(1) $\begin{equation*} z^2 = x^2 + y^2 \quad \Rightarrow \quad x = \sqrt{10^2 - 5^2} =\sqrt{75}. \end{equation*}$

We are asked to find $dy/dt$ . Thus

(2) $\begin{align*} z^2 & = x^2 + y^2 \\ 2z \frac{dz}{dt} & = 2x \frac{dx}{dt} + 2 y \frac{dy}{dt} \\ 0 & = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \\ \frac{dy}{dt} & = \frac{2x}{2y} \frac{dx}{dt} \\ & = \frac{x}{y} \frac{dx}{dt} \\ & = \frac{\sqrt{75}}{5} \cdot 2 \\ & = 2\sqrt{3} \ \text{ft}/\text{sec}.\end{align*}$

Note that $dz/dt = 0$ because the $z$ direction does not change, and $dx/dt$ is negative since the ladder is sliding down. That is, the bottom of the ladder is moving at a rate of $2\sqrt{3} \ \text{ft}/\text{sec}$ .

Leave a Reply Cancel reply