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)

2 Answers
Mar 10, 2018

- 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

Mar 10, 2018

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.