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A ladder 13m long leans against a wall. The foot of the ladder ispulled along the ground away from the wall, at the rate of 1.5m/sec. How fastis the angle `theta`betweenthe ladder and the ground is changing when the foot of the ladder is 12m awayfrom the wall. |
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Answer» `tan theta = y/x` `sin theta = y/13` differentiating wrt t `cos theta* (d theta)/dt = ((dy)/dt)/13` `12/13*(d theta)/dt = (dy/dt)/13` `(d theta)/dt = (dy/dt)/12` eqn (1) now, `x^2 + y^2 = 13^2` `2xdy/dt + 2ydy/dt = 0` `2*12*1.5 = -2*5*dy/dt` `dy/dt = 24*1.5/-10= 36/-10` putting in eqn (1) `(d theta)/dt = -36/(10*12) = -0.3`rad/sec |
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