A ladder 10ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2ft/s, how fast is the angle between the top of the ladder and the wall changing when the angle is #pi/4# rad?

1 Answer
Jim H
Apr 6, 2018

#sqrt2/5# #rad#/#s#

Explanation:

Let #x# be the distance between the base of the wall and the bottom of the ladder. And let #theta# be the angle between the top of the ladder and the wall,

Then #x/10 = sin theta#, so #x = 10sintheta#.

Differentiating with respect to time #t# gets us

#dx/dt = 10 cos theta (d theta)/dt#

We were told that #dx/dt = 2# #ft#/#s# and we seek #(d theta)/dt# when #theta = pi/4#

#2 = 10 ( cos(pi/4)) (d theta)/dt#

#2 = 10 (sqrt2/2) (d theta)/dt = 5sqrt2 (d theta)/dt#

#(d theta)/dt = 2/(5sqrt2) = sqrt2/5# #rad#/#s#

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