A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the highest point of the ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of

2 m/sec

$x^2+y^2=25 \Rightarrow 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$

but $\frac{d x}{d t}$ = 1.5 m/sec

∴ $2(x)(1.5)+2(y) \frac{d y}{d t}=0 \Rightarrow \frac{d y}{d t}=-\frac{1.5 x}{y}$

when x = 4, y = 3

∴ $\frac{d y}{d t}=-\frac{1.5 \times 4}{3}$ = -2 m/sec

∴ Height of the wall is decreasing at the rate of 2 m/sec.