Calculus/Related Rates/Solutions

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

$V={\frac {4}{3}}\pi r^{3}$ ${\dot {V}}=100$ $r=4$ 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:

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

$V={\frac {4}{3}}\pi r^{3}$ ${\dot {V}}=100$ $r=4$ 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:

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

$h=2r$ $V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi ({\frac {h}{2}})^{2}h={\frac {1}{12}}\pi h^{3}$ ${\dot {V}}=3$ $h=6$ 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:

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

$h=2r$ $V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi ({\frac {h}{2}})^{2}h={\frac {1}{12}}\pi h^{3}$ ${\dot {V}}=3$ $h=6$ 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:

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

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

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

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

${\dot {s}}=2\pi (1)(2)=\mathbf {4\pi {\frac {ft}{min}}}$ 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?
Known:

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

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

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

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