Question

A very long ladder 40 feet long, rests against a vertical wall. Suppose that the base of the ladder slides away from the wall at a rate of 5 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 24 feet from the wall? Assume that the top of the ladder remains in contact with the wall.

Answer 1

Answer:3.75 ft/s

Step-by-step explanation:

Given

Length of ladder(L)=40 feet

base of ladder moves at a rate of 5 ft/sec

Bottom of ladder from wall(x)=24 feet

let horizontal distance be x and vertical distance be y

`y=√(40^2-24^2)=32`

therefore

`x^2+y^2=L^2` from Pythagoras

differentiate

`2x* \frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0`

`x* \frac{\mathrm{d} x}{\mathrm{d} t}=-y* \frac{\mathrm{d} y}{\mathrm{d} t}`

`24* 5=-32* \frac{\mathrm{d} y}{\mathrm{d} t}`

`\frac{\mathrm{d} y}{\mathrm{d} t}=(-3)/(4)* 5=-3.75 ft/s`

negative sign indicates height is decreasing