§3.1 Related rates (continued)

Related Rates Problems

Sand forms a right circular cone whose radius $r$ and height $h$ both depend on time. Using $V = \frac{1}{3}\pi r^2 h$, the rates $\frac{dV}{dt}$, $\frac{dr}{dt}$, and $\frac{dh}{dt}$ are related by

Sand falls at $10$ cubic feet per minute so that the pile height is always half the radius. Writing $h = \frac{r}{2}$ gives $V = \frac{\pi}{6}r^3$ and $\frac{dV}{dt} = \frac{\pi}{2}r^2\frac{dr}{dt}$. Compute $\frac{dh}{dt}$ when $r = 4$ feet, rounded to three decimals.

Sketching the conical tank at several water levels suggests which quantities change and which stay fixed. The correct inventory is

A conical tank has depth $8$ feet and top radius $6$ feet. If water stands at depth $h$ with surface radius $r$, similar triangles give

With $r = \frac{3}{4}h$ and $V = \frac{1}{3}\pi r^2 h$, the water volume becomes a function of depth alone, namely

Once $V = \frac{3}{16}\pi h^3$ expresses the water volume through depth alone, differentiating with respect to $t$ gives

Water enters the tank at $4$ cubic feet per minute, so $\frac{dh}{dt} = \frac{64}{9\pi h^2}$. The water level rises most rapidly

For the tank with $\frac{dV}{dt} = 4$ cubic feet per minute, compute $\frac{dh}{dt}$ when $h = 3$ feet, rounded to four decimals.

A television camera sits $4000$ feet from a launch pad tracking an ascending rocket. Which quantities change, and which stay constant?

With $\tan(\theta) = \frac{h}{4000}$, so $h = 4000\tan(\theta)$, differentiating with respect to $t$ gives

With $\frac{dh}{dt} = 600$ feet per second at $h = 3000$ feet, the triangle gives $\cos(\theta) = \frac{4}{5}$. Using $\frac{dh}{dt} = 4000\sec^2(\theta)\frac{d\theta}{dt}$, compute $\frac{d\theta}{dt}$ as a decimal.

With $h^2 + 4000^2 = z^2$ relating rocket height to camera distance, differentiating with respect to $t$ gives

A camera $4000$ feet from a rocket's launch pad tracks the rocket, with $h^2 + 4000^2 = z^2$. If the rocket rises at $600$ feet per second at $h = 3000$, compute $\frac{dz}{dt}$ at that instant.

If the rocket keeps rising at $600$ feet per second past $3000$ feet, the distance rate $\frac{dz}{dt} = \frac{h}{z} \cdot 600$ when the elevation reaches $4000$ feet is

A skateboarder rides away from a lamppost under a light $15$ feet high, and the rider is $6$ feet tall. Letting $x$ be his distance from the post and $s$ his shadow length, which quantities change?

With rider height $6$, post height $15$, distance $x$, and shadow $s$, similar triangles give $\frac{s}{6} = \frac{s+x}{15}$, which simplifies to

A skateboarder rides at $\frac{dx}{dt} = 3$ feet per second under a lamppost with $3s = 2x$. His shadow length grows at a constant $2$ feet per second.