A 25-ft ladder is leaning against a house when its base starts to slide away By

Question

A 25-ft ladder is leaning against a house when its base starts to slide away By the time the base is 24 ft from the house, the base is moving away at the rate of 14 ft/sec. What is the rate of change of the height of the top of the ladder? At what rate is the area of the triangle formed by the ladder, wall and ground changing then? At what rate is the angle between the ladder and the pound changing then? The rate of change of the height of the top of the ladder is -48 ft/sec. (Simplify your answer ) The area is changing at -527 ft^2/sec. (Simplify your answer) The angle is changing at rad/sec. (Simplify your answer.)

Explanation / Answer

Let the height be 'h', base be 'b',
ladder's length be 'l', then by pythagoras theorem, we have,
(h^2) + (b^2) =(l^2)   given b= 24 ft ; l=25 ft

h^2 = 25^2 -24^2 = 49

h = 7
as the base is continuously increasing, so, we can use use differentiation with respect to time 't',
so, 2h*(dh/dt) + 2b*(db/dt) =0 , zero because the length of the length of the ladder remains unchanged,
so, find the rate of change of height = dh/dt

h dh/dt = -b db/dt

7 dh/dt = -24 * 14                                  given db/dt = 14

so dh/dt = -48

area of the triangle =A= (1/2)*base * height =(1/2)*bh
rate of change of area of the triangle=dA/dt = (1/2)*[b*(dh/dt) +h*(db/dt)]
                                                              = (1/2)*[24*(-48) + 7*14] = -527

B = angle between ladder and ground
cos(B) = b/25
-sin(B)dB/dt = (1/25)db/dt
dB/dt = (-1/25)(db/dt)/sin(B)
dB/dt = (-1/25)(14)/(7/25)   = 2 RAD/SEC

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