SOLUTION: Q5 A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/ sec. a. How

Algebra ->  Pythagorean-theorem -> SOLUTION: Q5 A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/ sec. a. How       Log On


   

Question 1033516: Q5 A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/ sec.
a. How fast is the top of the ladder sliding down the wall then?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
c. At what rate is the angle between the ladder and the ground changing then?
Answer by addingup(3677) About Me  (Show Source):
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Let x be the distance between the base of the wall and the base of the ladder. Let y be the height the top of the ladder is off the ground. The given is
that dx/dt = +5 at the moment x = 12. The length of the ladder, 13 feet, is a constant, since its length doesn't change.
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(a) How fast is the top of the ladder sliding down the wall then?
x^2+y^2 = 13^2
2x%28dx%2Fdt%29%2B2y%28dy%2Fdt%29+=+0
Solve for dy/dt in terms of x,y, and dx/dt:
use your calculator and you get y = 5. Therefore:
dy/dx = -x*dx/y*dt = (-12/5)*5 = -12 Note that the minus sign indicates the ladder is sliding down. So, the top of the ladder is sliding down at 12 ft/second
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b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing.
Here the unknown is the rate of change of the area, dA/dt
dA/dt = 1/2 (x*dy/dt+y*dx/dt)
Substitute with x = 12 and dx/dt = 5 and y=5 dy/dt = -12:
dA/dt = 1//2(12*(-12)5*5)= -144+25/2 = -119/2 = 59.5 ft^2/sec
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c. At what rate is the angle between the ladder and the ground changing then?
ormula that relates the angle θ with the sides x and y of a right triangle:
Answer
1 rad/sec