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?
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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
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