§3.1 Related rates (continued)

Exercises (continued)

Summary

With $\frac{dx}{dt} = 3$ and $\frac{ds}{dt} = 2$, the tip of the shadow sits at $y = x + s$. Its speed $\frac{dy}{dt}$ equals

A rope runs from a boat's bow over a pulley $5$ feet higher, with rope length $z$ and horizontal distance $x$, so $x^2 + 25 = z^2$. The rope is pulled in at $2$ feet per second. When $x = 12$ we have $z = 13$, and the boat approaches the dock at

As the skateboarder moves away from the lamppost, the rate $\frac{ds}{dt}$ of his shadow's lengthening

With $3s = 2x$ and $\frac{dx}{dt} = 3$ feet per second, compute $\frac{ds}{dt}$ in feet per second.

Sand falls at $10$ cubic feet per minute onto a conical pile with radius always equal to height. When the pile volume reaches $1000$ cubic feet, $\frac{dh}{dt} = \frac{10}{\pi h^2}$ with $h = \sqrt[3]{\frac{3000}{\pi}}$. Compute $\frac{dh}{dt}$, rounded to four decimals.

In the sailboat problem, the rope length $z$ satisfies $\frac{dz}{dt} = -2$ feet per second because the rope is being pulled in.

A pool $60$ feet long and $25$ feet wide deepens uniformly from $3$ to $15$ feet. When water stands $h$ feet deep at the deep end, its volume works out to $V = 62.5h^2$. Differentiating gives $\frac{dh}{dt} = \frac{1}{125h}\frac{dV}{dt}$, so the depth increases

For the pool with $V = 62.5h^2$ and water entering at $800$ cubic feet per minute, compute $\frac{dh}{dt}$ when the deep end reaches $h = 5$ feet.

A runner heads from second to third base at $24$ feet per second on a square diamond with $90$ foot sides. With $\tan(\theta) = \frac{x}{90}$ and $x = 30$ feet, the differentiated relation gives $\frac{d\theta}{dt} = \frac{1}{90\sec^2(\theta)} \cdot (-24)$. Compute $\frac{d\theta}{dt}$ as a decimal.