Calculus/Related Rates/Solutions

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

${\displaystyle V={\frac {4}{3}}\pi r^{3}}$
${\displaystyle {\dot {V}}=100}$
${\displaystyle r=4}$
Take the time derivative:
${\displaystyle {\dot {V}}=4\pi r^{2}{\dot {r}}}$
Solve for ${\displaystyle {\dot {r}}}$:
${\displaystyle {\dot {r}}={\frac {\dot {V}}{4\pi r^{2}}}}$
Plug in known values:

${\displaystyle {\dot {r}}={\frac {100}{4\pi 4^{2}}}=\mathbf {{\frac {25}{16\pi }}{\frac {ft}{min}}} }$
Known:

${\displaystyle V={\frac {4}{3}}\pi r^{3}}$
${\displaystyle {\dot {V}}=100}$
${\displaystyle r=4}$
Take the time derivative:
${\displaystyle {\dot {V}}=4\pi r^{2}{\dot {r}}}$
Solve for ${\displaystyle {\dot {r}}}$:
${\displaystyle {\dot {r}}={\frac {\dot {V}}{4\pi r^{2}}}}$
Plug in known values:

${\displaystyle {\dot {r}}={\frac {100}{4\pi 4^{2}}}=\mathbf {{\frac {25}{16\pi }}{\frac {ft}{min}}} }$
2. Water is pumped from a cone shaped reservoir (the vertex is pointed down) ${\displaystyle 10ft}$ in diameter and ${\displaystyle 10ft}$ deep at a constant rate of ${\displaystyle 3ft^{3}/min}$. How fast is the water level falling when the depth of the water is ${\displaystyle 6ft}$?
Known:

${\displaystyle h=2r}$
${\displaystyle V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi ({\frac {h}{2}})^{2}h={\frac {1}{12}}\pi h^{3}}$
${\displaystyle {\dot {V}}=3}$
${\displaystyle h=6}$
Take the time derivative:
${\displaystyle {\dot {V}}={\frac {1}{4}}\pi h^{2}{\dot {h}}}$
Solve for ${\displaystyle {\dot {h}}}$:
${\displaystyle {\dot {h}}={\frac {4{\dot {V}}}{\pi h^{2}}}}$
Plug in known values:

${\displaystyle {\dot {h}}={\frac {(4)(3)}{\pi 6^{2}}}=\mathbf {{\frac {1}{3\pi }}{\frac {ft}{min}}} }$
Known:

${\displaystyle h=2r}$
${\displaystyle V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi ({\frac {h}{2}})^{2}h={\frac {1}{12}}\pi h^{3}}$
${\displaystyle {\dot {V}}=3}$
${\displaystyle h=6}$
Take the time derivative:
${\displaystyle {\dot {V}}={\frac {1}{4}}\pi h^{2}{\dot {h}}}$
Solve for ${\displaystyle {\dot {h}}}$:
${\displaystyle {\dot {h}}={\frac {4{\dot {V}}}{\pi h^{2}}}}$
Plug in known values:

${\displaystyle {\dot {h}}={\frac {(4)(3)}{\pi 6^{2}}}=\mathbf {{\frac {1}{3\pi }}{\frac {ft}{min}}} }$
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 ${\displaystyle 2ft}$ in diameter. If the winch turns at a constant rate of ${\displaystyle 2rpm}$, how fast is the boat moving toward the dock?
Let ${\displaystyle R}$ be the number of revolutions made and ${\displaystyle s}$ be the distance the boat has moved toward the dock.

Known:
${\displaystyle {\frac {R}{s}}={\frac {1}{2\pi r}}}$ (each revolution adds one circumferance of distance to s)
${\displaystyle {\dot {R}}=2}$
${\displaystyle r=1}$
Solve for ${\displaystyle s}$:
${\displaystyle s=2\pi rR}$
Take the time derivative:
${\displaystyle {\dot {s}}=2\pi r{\dot {R}}}$
Plug in known values:

${\displaystyle {\dot {s}}=2\pi (1)(2)=\mathbf {4\pi {\frac {ft}{min}}} }$
Let ${\displaystyle R}$ be the number of revolutions made and ${\displaystyle s}$ be the distance the boat has moved toward the dock.

Known:
${\displaystyle {\frac {R}{s}}={\frac {1}{2\pi r}}}$ (each revolution adds one circumferance of distance to s)
${\displaystyle {\dot {R}}=2}$
${\displaystyle r=1}$
Solve for ${\displaystyle s}$:
${\displaystyle s=2\pi rR}$
Take the time derivative:
${\displaystyle {\dot {s}}=2\pi r{\dot {R}}}$
Plug in known values:

${\displaystyle {\dot {s}}=2\pi (1)(2)=\mathbf {4\pi {\frac {ft}{min}}} }$
4. At time ${\displaystyle t=0}$ a pump begins filling a cylindrical reservoir with radius 1 meter at a rate of ${\displaystyle e^{-t}}$ cubic meters per second. At what time is the liquid height increasing at 0.001 meters per second?
Known:

${\displaystyle V=\pi r^{2}h}$
${\displaystyle {\dot {V}}=e^{-t}}$
${\displaystyle r=1}$
${\displaystyle {\dot {h}}=0.001}$
Take the time derivative:
${\displaystyle {\dot {V}}=\pi r^{2}{\dot {h}}}$
Plug in the known values:
${\displaystyle e^{-t}=0.001\pi }$
Solve for t:

${\displaystyle \mathbf {t=-\ln(.001\pi )} }$
Known:

${\displaystyle V=\pi r^{2}h}$
${\displaystyle {\dot {V}}=e^{-t}}$
${\displaystyle r=1}$
${\displaystyle {\dot {h}}=0.001}$
Take the time derivative:
${\displaystyle {\dot {V}}=\pi r^{2}{\dot {h}}}$
Plug in the known values:
${\displaystyle e^{-t}=0.001\pi }$
Solve for t:

${\displaystyle \mathbf {t=-\ln(.001\pi )} }$