The top of a ladder slides down a vertical wall at a rate of 0.125 m/s. at the moment when the bottom of the ladder is 5 m from the wall, it slides away from the wall at a rate of 0.3 m/s. how long is the ladder?

Vertically (y-axis), horizontally(x-axis)

dy/dt = -0.125 m/s (-ve since y decreasing )

dx/dt = +0.3 m/s (+ve, x increases)

and, x = 5 m

length of the ladder = k (a constant)

k^2 = x^2 + y^2

differentiating it wrt t,

0 = 2x dx/dt + 2y dy/dt

0 = 2(5)(0.3) + 2(y)(-0.125)

y = 12

which means, when the bottom of the ladder is 5m from the wall, the top of the ladder is 12m from the bottom.

thus, k^2 = 5^2 + 12^2

length of the ladder, k = 13 m

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