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. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall?
(b) 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 changing when the base of the ladder is 7 feet from the wall.

see below

Step-by-step explanation:

We know that we have a right triangle with x^2 + y^2 = 25^2

x^2 + y^2 =625

Taking the derivative with respect to time on each side

d/dt( x^2 + y^2) =d/dt 625

The derivative of x^2 with respect to t is 2 x  times dx/dt and

the derivative of y^2 with respect to t is 2 y  times dy/dt and

and the derivative of a constant is zero

2 x  dx/dt  + 2 y dy/dt = 0

We are trying to find dy/dt or the rate it is sliding down the wall

Subtracting 2y dy/dt

2 x  dx/dt  = -  2 y dy/dt

Divide each side by  -2y

2x/- 2y * dx/dt = dy/dt

-x/y * dx/dt = dy/dt

We know that dx/dt = 2

If the base is 7   x^2 + y^2 =625  7^2 + y^2 = 625  so y = sqrt(625 - 49) =24

-7/24 * 2 = dy/dt

-7/12 ft/sec= dy/dt  when x=7

If the base is 15   x^2 + y^2 =625  15^2 + y^2 = 625  so y = sqrt(625 - 225) =20

-15/20 * 2 = dy/dt

-3/2ft/sec= dy/dt  when x=15

If the base is 24   x^2 + y^2 =625  24^2 + y^2 = 625  so y = sqrt(625 - 576) =7

-24/7 * 2 = dy/dt

-48/7 ft/sec= dy/dt  when x=24

Now we need to find the rate at which the area is changing

A = 1/2 xy

Taking the derivative of each side

dA/dt =d/dt ( 1/2 xy)

Using the product rule of derivatives

        = 1/2 ( x dy/dt + y dx/dt)

 Using the information for 7 ft from the wall

x = 7, dy/dt = -7/12, y = 24 and dx/dt =2

      =  1/2 ( 7 * -7/12 + 24*2)

      = 1/2 ( -49/12+ 48)

  = 527/24 ft^2/ sec

  a) -7/12, -3/2, -48/7 . . . feet per second

  b) 21 23/24 square feet per second

Step-by-step explanation:

a) Let x represent the distance of the base from the wall. Then the Pythagorean theorem relates r, x, and the length of the ladder:

  r^2 +x^2 = 25^2

  2r·r' +2x·x' = 0 . . . . . derivative with respect to time

  r' = -x·x'/r

  r' = -x·x'/√(25^2 -x^2)

For x = 7, r' = -7(2)/√(625 -49) = -14/24 = -7/12 ft/s

For x = 15, r' = -15(2)/(√(625 -225) = -30/20 = -3/2 ft/s

For x = 24, r' = -24(2)/√(625 -576) = -48/7 ft/s

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b) The area is ...

  A = 1/2rx

Then the rate of change of area is ...

  A' = (1/2)(r'x +rx') . . . . differentiate with respect to time

At x=7, this is ...

  A' = (1/2)(-7/12×7 +24×2) = 21 23/24 . . . ft^2/s

can you complete your sentence? i could !