Question 1033516
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(dx/dt)+2y(dy/dt) = 0}}}
Solve for dy/dt in terms of x,y, and dx/dt:
{{{(dy/dt) = (-x/y)(dy/dt)
The problem tells us that right now x = 12, dx/dt = 5. Find y by setting x to 12:
{{{x^2+y^2 = 13^2}}} 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