Posted by **ron** on Sunday, November 4, 2012 at 1:37pm.

A 13 foot ladder leaning against a wall makes an angle of degree radians with the ground. The base of the ladder is pulled away from the wall at a rate of 2ft/sec. How fast is degree changing when the base of the ladder is 5 feet way.

- calculus -
**Steve**, Sunday, November 4, 2012 at 1:41pm
if the base is x from the wall,

cosθ = x/13

-sinθ dθ/dt = 1/13 dx/dt

when x=5, sinθ = 12/13

-12/13 dθ/dt = 1/13 (2)

dθ/dt = -1/6

note the "-" sign: the angle decreases as the ladder slips down.

- calculus -
**Damon**, Sunday, November 4, 2012 at 1:50pm
5, 12, 13 right triangle

call angle ladder to ground T

cos T = x/13 where x is base of ladder to base of wall

-sin T dT/dt = (1/13) dx/dt

at x = 5, t = 0, dx/dt = 5

cos T = 5/13 so T = 67.4 degrees or 1.18 radians

sin T = sin 67.4 = 12/13 = .923

-.923 dT/dt = (1/13)(2)

so

dT/dt = -.167 radians/second

times 180/pi = -9.55 degrees/second

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