Answer to Question #348050 in Calculus for Cecel

Question

A ladder 20 meters long leans against a wall. If the bottom of the ladder is being pushed towards the wall at the rate of 20 m/min, how fast is the top of the ladder moving up the wall when the top of the ladder is 6 meters from the ground?

Answer 1

Let $AB$  be the ladder, where $AB=20 m.$ Let at time $t$  minutes, the end $A$  of the ladder be $x$  metres from the wall and the end $B$  be $y$  metres from the ground.

Since, $AOB$  is a right angled triangle, by Pythagoras theorem.

$$x^2+y^2=20^2$$

Differentiate both sides wth respect to $t$

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

Solve for $\dfrac{dy}{dt}$

$$\dfrac{dy}{dt}=-\dfrac{x}{y}(\dfrac{dy}{dt})$$

Given $\dfrac{dx}{dt}=-20m/min, y=6m.$

Substitute

$$x=\sqrt{20^2-6^2}=2\sqrt{91}$$

$$\dfrac{dy}{dt}=-\dfrac{2\sqrt{91}}{6}(-20m/min)=\dfrac{20\sqrt{91}}{3}m/min$$

The top of the ladder is sliding up the wall, at the rate of  $\dfrac{20\sqrt{91}}{3}m/min.$