Calculus/Related Rates/Solutions

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1. A spherical balloon is inflated at a rate of 100 ft^3/min. Assuming the rate of inflation remains constant, how fast is the radius of the balloon increasing at the instant the radius is 4 ft?

V=\frac{4}{3}\pi r^{3}
Take the time derivative:
\dot{V}=4\pi r^{2}\dot{r}
Solve for \dot{r}:
\dot{r}=\frac{\dot{V}}{4\pi r^{2}}
Plug in known values:
\dot{r}=\frac{100}{4\pi4^{2}}=\mathbf{\frac{25}{16\pi} \frac{ft}{min}}

2. Water is pumped from a cone shaped reservoir (the vertex is pointed down) 10 ft in diameter and 10 ft deep at a constant rate of 3 ft^3/min. How fast is the water level falling when the depth of the water is 6 ft?

V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}
Take the time derivative:
\dot{V}=\frac{1}{4}\pi h^{2}\dot{h}
Solve for \dot{h}:
\dot{h}=\frac{4\dot{V}}{\pi h^{2}}
Plug in known values:
\dot{h}=\frac{(4)(3)}{\pi6^{2}}=\mathbf{\frac{1}{3\pi} \frac{ft}{min}}

3. A boat is pulled into a dock via a rope with one end attached to the bow of a boat and the other wound around a winch that is 2ft in diameter. If the winch turns at a constant rate of 2rpm, how fast is the boat moving toward the dock?

Let R be the number of revolutions made and s be the distance the boat has moved toward the dock.
\frac{R}{s}=\frac{1}{2\pi r} (each revolution adds one circumferance of distance to s)
Solve for s:
s=2\pi rR
Take the time derivative:
\dot{s}=2\pi r\dot{R}
Plug in known values:

4. At time t=0 a pump begins filling a cylindrical reservoir with radius 1 meter at a rate of e^{-t} cubic meters per second. At what time is the liquid height increasing at 0.001 meters per second?

V=\pi r^{2}h
Take the time derivative:
\dot{V}=\pi r^{2}\dot{h}
Plug in the known values:
Solve for t: