Calculus question!

Question

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changed when the base of the ladder is 11 feet from the wall. Round your answer to two decimal places.

Answer 1

√Let x = distance of the base of the ladder from the wall at time t

 

     y = distance of the top of the ladder from the bottom of the wall at              time t

 

        = √ [(25)² - x²] = √[625-x²]

 

     A = area of triangle at time t  = xy = x√[625-x²]

 

Given:  dx/dt = 2

 

Find:  dA/dt when x = 11

 

Solution:  A = x(625-x²)^½

 

             dA/dt = (dx/dt)(625-x²)^½ + (½)x(625-x²)^-1/2(-2x)(dx/dt)

 

                      = (dx/dt)[√(625-x²) - x²/√(625-x²)]

 

                      = (dx/dt)[(625-2x²)/√(625-x²)]

 

            When x = 11, dA/dt = 2[(625-242)/√(625-121)]

 

                                         = 766/ √504 ≈ 34.1 ft²/sec