Word Problem

Question

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 per second.

 

(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 θ changing between the ladder and the ground changing then?

Answer 1

1) You've got x²+y²=169, where x is the horizontal distance on the ground and y is the vertical distance on the wall. At the particular instant this problem is talking about, the horizontal distance is 12 so that you can plug it in  the equation above to get the vertical distance of 5. Since the bottom is sliding out, dx/dt = +5. 

     You can take the derivative in terms of t to find out how the rates are related. So 2x(dx/dt) + 2y(dy/dt) =0. Substitute your information to get 2*12*5 + 2*5*dy/dt =0 which leads to dy/dt= -12 ft/sec.

 

2) The area is A = 1/2 x*y. So taking the derivative in terms of t you get, dA/dt = 1/2 x(dy/dt)+1/2 y(dx/dt). Substitute to get dA/dt = 1/2 *12*-12 + 1/2 5*5 = -72+12.5=-59.5 sq in per sec.

 

3) Tan Θ = y/x, So taking the derivative you get Sec² θ dθ/dt= (x*dy/dt - y dx/dt) / x². Substituting in you get

169/144 dθ/dt = [12(-12)- 5(5)]/144

169/144 dθ/dt = -169/144

dθ/dt = -1

 

Hope this helped.