A 25- $\mathrm{ft}$ ladder is leaning against a house when its base starts to slide away. By the time the base is $24 \mathrm{ft}$ from the house, the base is moving away at the rate of $14 \mathrm{ft} /$ sec
a. What is the rate of change of the height of the top of the ladder?
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?
a. The rate of change of the height of the top of the ladder is $\mathrm{ft} / \mathrm{sec}$.
(Simplify your answer.)
b. The area is changing at $\mathrm{ft}^{2} / \mathrm{sec}$
(Simplify your answer.)
c. The angle is changing at $\mathrm{rad} / \mathrm{sec}$.
(Simplify your answer.)

Step 1 ：We are given that the ladder is 25 ft long and the base is moving away from the wall at a rate of 14 ft/sec. We are asked to find the rate of change of the height of the ladder, the rate of change of the area of the triangle formed by the ladder, wall, and ground, and the rate of change of the angle between the ladder and the ground.

Step 2 ：First, we can use the Pythagorean theorem to relate the base and height of the ladder since the ladder, wall, and ground form a right triangle. The length of the ladder is the hypotenuse of the triangle, so we have the equation \(x^2 + y^2 = 25^2\).

Step 3 ：We can differentiate this equation with respect to time to find \(dy/dt\).

Step 4 ：Next, the area of the triangle is given by \(A = 1/2 * base * height = 1/2 * x * y\). We can differentiate this equation with respect to time to find \(dA/dt\).

Step 5 ：Finally, the angle \(\theta\) between the ladder and the ground can be found using the tangent function: \(\tan(\theta) = y/x\). We can differentiate this equation with respect to time to find \(d\theta/dt\).

Step 6 ：Using these equations, we find that the rate of change of the height of the top of the ladder is \(\boxed{-14 \, \text{ft/sec}}\).

Step 7 ：The area is changing at \(\boxed{168 \, \text{ft}^2/\text{sec}}\).

Step 8 ：The angle is changing at \(\boxed{-0.014 \, \text{rad/sec}}\).