10-ft ladder is leaning against a house when its base starts to slide away. By t

Question

10-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 8 ft from the house, the base is moving away at the rate of 6 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 ground changing then? The rate of change of the height of the top of the ladder is ft/sec. (Simplify your answer.) The area is changing at ft2/sec. (Simplify your answer.) The angle is changing at rad/sec. (Simplify your answer.)

Explanation / Answer

h=sqrt(l0^2-8^2)=6

so

l^2=b^2+h^2

l is constant

diffentiate

0=2bdb/dt+2hdh/dt

dh/dt= -8*6/6=-8ft/sec

b) area =.5b*h

differentiate

da/dt = .5(bdh/dt+hdb/dt)

=.5(-8*8+6*6)

=-14

c)sin(theta) = h/l

cos(theta)*d(theta)/dt=dh/ldt

(8/10)d(theta)/dt=-8/10

d(theta)/dt = -1 rad/sec

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