Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.8 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?

(That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall)

Answer 1

`- 2/15` radians per second

Explanation:

Denoting the distance in feet between the wall and the base of the ladder by `x` and the angle in radians between the ladder and the ground by `y`, it is noted

`cos(y) = x / 10`

which implies

`y = arccos(x/10)`

Denoting time in seconds by t, it is further noted that

`dy / dt = dy/dx dx/dt` (chain rule)

Noting (using standard table of derivatives for convenience)

`dy/dx = - 1 / (sqrt(1 - (0.1 x)^2)) (0.1)` (also by chain rule)

that is

`dy/dx = - 0.1/(sqrt(1 - 0.01 x^2))`

It is noted from the question that in this particular system

`dx/dt = 0.8` feet per second

So (denoting the derivative as a function of `x`)

`dy/dt (x) = dy/dx dx/dt = - 0.08/(sqrt(1 - 0.01 x^2))`

So

`dy/dt (8) = dy/dx dx/dt = - 0.08/(sqrt(1 - 0.01 (64)))`

`= - 0.08/(sqrt(1 - 0.64)) = - 0.08/(sqrt(0.36))`

`- 0.08/0.6 = - 8/60 = - 2/15` radians per second

Answer 2

Please see below.

Explanation:

Variables:
`x` = distance between the bottom of the wall and the bottom of the ladder

`theta` = the angle between the ladder and the ground.

Rates of change
`dx/dt = 0.8` `ft`/`sec`

Find `(d theta)/dt` when `x = 8` `ft`

Equation relating variables
`cos theta = x/10`, so we'll use

`x = 10costheta`

Equation relating the rates of change

`d/dt(x) = d/dt(10costheta)`

`dx/dt = -10sin theta (d theta)/ dt`

Finish
We know `dx/dt = 0.8`.
To find `(d theta)/dt` we need `sin theta` when `x = 8`.

But when `x = 8` we know `cos theta = 8/10`, so we can find `sin theta = 6/10 = 3/5`

`dx/dt = -10sin theta (d theta)/ dt`

`0.8 = -10 (3/5) (d theta)/ dt`

So `(d theta)/dt = -0.8/6 = -8/60 = -2/15 ~~ - 0.133` `rad`/`s`

The angle is decreasing at a rate of `2/5` (or approximately `0.133`) radians per second.