Answer to Question #337941 in Calculus for dme

Question

A ladder 5 meters long is leaning against the wall. The bottom is initially 3.5 meters away from the

wall. You are pushing it towards the wall at a rate of 0.20 m/s. How fast is the top of the ladder moving

up the wall 12 seconds after you start pushing?

Answer 1

$$x^2+y^2=5^2, y\ge 0$$

$$y=\sqrt{25-x^2}$$

$$x=x_0+v_xt$$

$$v_y=dy/dt=-\dfrac{2x}{\sqrt{25-x^2}}\cdot v_x$$

Given $x_0=3.5m, v_x=-0.20m/s.$

12 seconds after you start pushing

$$x|_{t=12}=3.5m-0.20m/s\cdot12s=1.1m$$

$$v_y|_{t=12}=-\dfrac{2(1.1m)}{\sqrt{25-(1.1)^2}}\cdot (-0.20m/s)$$

$$\approx0.09m/s$$

The top of the ladder is moving up the wall at a rate of 0.09m/s 12 seconds after you start pushing.