Related Rates

Following the first midterm, we are turning to focus on the use of derivatives - how do properties of the derivative constrain the behavior of the function?

Example: Air is being pumped from a ballloon at a rate of 100 cubic centimeters a second. If the balloon is 50cm, how quickly is the radius changing? How much is the surface area stretching?

Solution: Differentiate $V=4/3\pi r^3$ wtih respect to time Once you know the rate of change in radius, can you find the surface area change by differentiating $S=4\pi r^2$?

A snowball is melting so that it looses volume at a rate proportional to its surface area (say, the change in volume is -0.1* surface area for specificity). If the snowball starts out 10cm across, how long does it take until half the volume has melted away?

Solution: Write everything we know in terms of radius $r$: $V=4/3\pi r^3$, $S=4\pi r^2$ and $dV/dt=0.1 S$ Calculate these $$\frac{dV}{dt}=-0.1 S = \frac{d}{dt}(\frac{4}{3}\pi r^3)$$ $$=4\pi r^2\frac{dr}{dt}$$

Thus, $dr/dt = -0.1$! This tells us the radius of the snowball is shrinking at a \emph{constant} rate! If the original radius was 10cm; what’s the radius when it’s half that volume? $$V_0=4/3 \pi (10)^3\hspace{1cm}V_{new}=\frac{1}{2}V_0$$ Find $r_{new}$, then can easily figure out how long it takes to go from $r_0$ to this new radius: divide their difference by $0.1$!

Example: A man walks along a straight line across a stage at 4ft/sec. A spotlight located 20ft from stage center is rotating to keep up with him. How fast is the spotlight rotating when the man is 15ft from stage center?

Solution: Given: Draw a triangle: $x$ is distance from stage center, 20 is other leg to spotlight. Know $dx/dt=4$, want $d\theta/dt$ when $x=15$. Use trig to get the relationship $x/20=\tan\theta$.

Example: as the sun sets, your (6ft tall person) shadow on the ground grows. Your friend tries to walk to keep up with your shadow’s head. How long is your shadow when your friend can no longer keep up (15ft/sec)?

Example: What shape does water make when it runs out of a faucet? Solution: the volume passing through any cross section over a fixed interval of time is constant (all the water has to go somewhere): so we get $area*vel=K$. What is $vel$? We want to know $radius(h)$ so need velocity as a function of height: $v=\sqrt{v_o^2+2gh}$

$$K=\pi r_0^2 v_0 \Delta t = \pi r(h)^2 v(h) \Delta t$$ $$K=\pi r