Asked by Lynn

A 24 foot ladder is leaning against a building. Let x be the distance between the bottom of the ladder abd the building and let theta be the angle between the ladder and the ground. Suppose the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. When x=6, how fast is theta (in Radians/sec) changing?

Answers

Answered by Reiny
cosØ = x/24
-sinØ dØ/dt = (1/24)dx/dt

when x = 6, opposite side is y
y^2 + 6^2 = 24^2
y = √540 and sinØ = √540/24

dØ/dt = -1/(24sinØ)dx/dt
= (-1/24)(24/√540)(2) = -2/√540 rad/sec

simplify as needed
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