A 10 foot ladder is leaning against a house. When the top of the ladder is 8 fee

Question

A 10 foot ladder is leaning against a house. When the top of the ladder is 8 feet above the ground, it is sliding down the wall at 2 feet per second. How fast is the base of the ladder sliding away from the wall? PLz Help

Explanation / Answer

length of the ladder is always constant i.e.10 feet. let vertical distance of ladder at any time be y and horizontal distance from wall of house be x x^2 + y^2 = 10^2 ------->(equation 1) is always true... differentiating w.r.t to time we get, 2x(dx/dt)+2y(dy/dt)=0---------->(equation 2) We know that sliding rate of base horizontally on the ground =(dx/dt) and sliding rate vertically against house =(dy/dt) and as y is decreasing wrt to time we have (dy/dt) negative...i.e (dy/dt)= -2 feet per second. Hence from the equation 2 obtained, sliding rate of base horizontally on the ground =(dx/dt)={[-(dy/dt)*y]/(x)} At the instant when y=8 feet, we have x=sqroot(10^2-8^2)=sqroot(36)=6...(from eqn1) i.e. x=6,y=8,(dy/dt)=-2 and hence dx/dt = (2*8)/6 =8/3=2.666 feet per second. Hence sliding rate of base horizontally on the ground = 8/3 = 2.666feet per second

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