Question

A 20-foot ladder rests against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 2 ft/sec. Answer the following: a. At what rate is the top of the ladder sliding down the wall after 3 seconds if the bottom of the ladder began 10 feet from the wall? b. At what rate is the angle between the ladder and the wall changing after 2 seconds if the bottom of the ladder began 8 feet from the wall?

Answer 1

Answer:

Explanation:

Let x and y be the bottom and topmost point of the ladder where it touches the ground and the wall respectively.

x² + y² = 20²

Differentiating with respect to time

x dx/dt  = -  ydy/dt

a ) Given

dx/dt = 2 ft/s

x ( at t = 3 )

= 10 ft + 2x3

= 16 ft

y² = 20² - 16²

y = 12

Putting the values

16 x 2 = - 12 . dy/dt

dy/dt  = - 2.667 m /s downwards

b ) After 2 s ,

x =  8 + 2x2 = 12

y² = 20²-12²

y = 16

If angle between ladder and wall be θ

sin  θ = x / 20  

= Differenciating  on both sides

cos θ dθ/dt = 1/20 dx/dt

dθ/dt = 1/ (20 cos θ) x dx/dt

=(  1/20 ) x 20/12  x 2

= 1/6 ° / s  Ans