A 10-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 6 ft from the house, the base is moving away at the rate of $$ 16 \mathrm{ft} / \mathrm{sec} $$. a. What is the rate of change of the height of the top of the ladder 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?

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We start with the fact that the ladder is 10 ft long, so at any time the horizontal distance x and the vertical height y satisfy

x² + y² = 10² = 100.

a. To find the rate at which the top of the ladder is descending (dy/dt) when x = 6 ft (and y = ?), differentiate the equation with respect to time t:

2x (dx/dt) + 2y (dy/dt) = 0  
  => x (dx/dt) + y (dy/dt) = 0  
  => y (dy/dt) = - x (dx/dt)  
  => dy/dt = - (x (dx/dt)) / y.

When x = 6 ft, we find y by:

6² + y² = 100  
  36 + y² = 100  
  y² = 64  
  y = 8 ft   (choosing the positive value for height).

Given that dx/dt = 16 ft/sec, we have

dy/dt = - (6 ft × 16 ft/sec) / 8 ft  
  dy/dt = - 96/8 = -12 ft/sec.

Thus, the top of the ladder is sliding down at 12 ft/sec (the negative sign indicates it is decreasing).

b. The area A of the right triangle formed by the ladder, wall, and ground is

A = ½ x y.

Differentiate with respect to time:

dA/dt = ½ (x (dy/dt) + y (dx/dt)).

Substitute x = 6 ft, y = 8 ft, dx/dt = 16 ft/sec, and dy/dt = -12 ft/sec:

dA/dt = ½ (6 × (-12) + 8 × 16)  
  = ½ (-72 + 128)  
  = ½ (56)  
  = 28 ft²/sec.

So the area is increasing at 28 ft²/sec.

c. Let θ be the angle between the ladder and the ground. The cosine of the angle is given by

cosθ = adjacent/hypotenuse = x/10.

Differentiate both sides with respect to time:

-dθ/dt · sinθ = (1/10) (dx/dt)
  => dθ/dt = - (dx/dt) / (10 sinθ).

When x = 6 ft and y = 8 ft, we have

sinθ = y/10 = 8/10 = 0.8.

Thus,

dθ/dt = - (16 ft/sec) / (10 × 0.8)  
  = -16 / 8  
  = -2 rad/sec.

Alternatively, using the sine relationship:
  sinθ = y/10  => d/dt (sinθ) = (1/10) (dy/dt) 
and 
  cosθ dθ/dt = (1/10)(dy/dt)  
with cosθ = 6/10 = 0.6. Then,
  dθ/dt = (dy/dt)/(10 cosθ) = (-12)/(10×0.6) = -12/6 = -2 rad/sec.
In either case, the angle between the ladder and the ground is decreasing at 2 rad/sec.

Summary of answers:
a. The height of the top is decreasing at 12 ft/sec.
b. The area is increasing at 28 ft²/sec.
c. The angle between the ladder and the ground is decreasing at 2 rad/sec.
a. The top of the ladder is sliding down at 12 ft/sec. b. The area of the triangle is increasing at 28 ft²/sec. c. The angle between the ladder and the ground is decreasing at 2 rad/sec.

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