For each of the following, it may help to find some equation that relates the information in the problem, then use implicit differentiation to find the related rates.

1. A 13-foot ladder is leaning against a house when its base starts to slide away. By the time the base is 12 feet from the house, the base is moving at the rate of 6 feet per second. What is the rate the ladder sliding down the wall at this time?

\( \begin{align}
x^2 + y^2 &= 13^2 \\
2x \frac{dx}{dt} + 2y \frac{dy}{dt} &= 0 \\
2 \cdot 12 \cdot 6 + 2 \cdot 5 \frac{dy}{dt} &= 0 \\
\frac{dy}{dt} &= -14.4 \,\mathrm{ft/sec}
\end{align} \)

2. Assume that the infected area of an injury is circular and the radius of the injury is growing at a rate of 0.5 mm per hour. At what rate is the area of the injury increasing when the radius is 6mm?

\( \begin{align}
A &= \pi r^2 \\
\frac{dA}{dt} &= \pi \cdot 2r \frac{dr}{dt} \\
\frac{dA}{dt} &= \pi \cdot 2 \cdot 6 \cdot 0.5 \\
\frac{dA}{dt} &= 6 \pi \,\mathrm{mm/hr}
\end{align} \)

3. An airplane is flying away from an airport at an altitude of 4 miles. At a specific time, the radar at the airport detects the diagonal distance between the plane and the airport is changing at a rate of 240 miles per hour and the horizontal distance is 40 miles. If the plane flies at a constant altitude, what is the horizontal velocity of the airplane at this time?

\( \begin{align}
x^2 + 4^2 &= d^2 \\
2x \frac{dx}{dt} + 0 &= 2d \frac{dd}{dt} \\
2 \cdot 40 \frac{dx}{dt} &= 2 \sqrt{1616} \cdot 240 \\
\frac{dx}{dt} &= 6 \sqrt{1616} = 24 \sqrt{101} \,\mathrm{mi/hr}
\end{align} \)

4. A conical tank is twice as wide as it is high. Water is draining from the tank at the rate of 6 cubic feet per minute. How fast is the depth of water in the tank declining when it is 4 feet deep?

Twice as wide as high implies \( d = 2h \rightarrow 2r = 2h \rightarrow r = h \)

\( \begin{align}
V = \frac{1}{3} \pi r^2 h &= \frac{1}{3} \pi h^3 \\
\frac{dV}{dt} &= \pi h^2 \frac{dh}{dt} \\
-6 &= \pi 4^2 \frac{dh}{dt} \\
-\frac{6}{16 \pi} &= \frac{dh}{dt}
\end{align} \)

Answer is 6/(16 pi) feet/minute because question asks for speed (always positive) not rate.

5. A balloon leaves the ground 500 feet from an observer and rises vertically at a rate of 140 feet per minute. At what rate is the angle of inclination from the observer increasing the instant the balloon is 500 feet off the ground?

\( \begin{align}
\tan{\theta} &= \frac{y}{500} \\
\sec^2(\theta) \frac{d\theta}{dt} &= \frac{1}{500} \cdot \frac{dy}{dt} \\
\sec^2\left(\frac{\pi}{4}\right) \frac{d\theta}{dt} &= \frac{1}{500}\cdot 140 \\
\frac{d\theta}{dt} &= \frac{7}{50}\mathrm{radians/min}
\end{align} \)

6. The volume of a cube is increasing at a rate of 0.1 cubic cm per second when the side length is 10 cm. What is rate of the surface area at this same time?

\( \begin{align}
V &= s^3 \\
\frac{dV}{dt} &= 3s^2 \frac{ds}{dt} \\
0.1 &= 3 \cdot 10^2 \frac{ds}{dt} \\
\frac{1}{3000} &= \frac{ds}{dt} \\
SA &= 6 s^2 \\
\frac{dSA}{dt} &= 12 s \frac{ds}{dt} \\
\frac{dSA}{dt} &= 12 \cdot 10 \cdot \frac{1}{3000} = \frac{120}{3000} = \frac{1}{25} \,\mathrm{cm^2/sec}
\end{align} \)