Elementary Calculus Problem involving Derivatives

[kierlain]

I am currently working on old tests to prepare for my final in Elementary Calculus. I came across this problem and have no idea what to do. Any help would be greatly appreciated.

A new Japanese restaurant is pricing a koi pond. A 4 foot deep, 8 foot radius circular pond looks nice, but requires over 6000 gallons of water. The owner only wants a 5000 gallon pond, but assumes that will be too small. You suggest reducing the radius by half a foot, but the owner scoffs that 5999.5 gallons is no better. You explain there are nearly 7.5 gallons in each cubic foot, but the owner still thinks you are only saving 3.75 gallons.

Since the volume of the pond is given by 4πr2, use the power rule to explain why you are saving about 200 times as much as the owner's last estimate.

 

[MarkFL]

Since we are instructed to use differentiation, we could use a linear approximation.

\(\displaystyle \frac{\Delta V}{\Delta r}\approx \d{V}{r}=8\pi r\)

\(\displaystyle \Delta V\approx8\pi r\Delta r\)

Now, with $r=8,\,\Delta r=\dfrac{1}{2}$, we find the approximation for the change in volume is given by:

\(\displaystyle \Delta V\approx8\pi(8)\left(\frac{1}{2}\right)\)

Since the owner believes the change in volume is only 0.5 cubic feet, we can then show that the actual change is more by a factor $f$, given by:

\(\displaystyle f=\frac{8\pi(8)\left(\dfrac{1}{2}\right)}{\dfrac{1}{2}}=64\pi\approx201.061929829747\)

 

[Prove It]

Since we are instructed to use differentiation, we could use a linear approximation.

\(\displaystyle \frac{\Delta V}{\Delta r}\approx \d{V}{r}=8\pi r\)

\(\displaystyle \Delta V\approx8\pi r\Delta r\)

Now, with $r=8,\,\Delta r=\dfrac{1}{2}$, we find the approximation for the change in volume is given by:

\(\displaystyle \Delta V\approx8\pi(8)\left(\frac{1}{2}\right)\)

Since the owner believes the change in volume is only 0.5 cubic feet, we can then show that the actual change is more by a factor $f$, given by:

\(\displaystyle f=\frac{8\pi(8)\left(\dfrac{1}{2}\right)}{\dfrac{1}{2}}=64\pi\approx201.061929829747\)

Actually since the radius is being reduced, $\displaystyle \begin{align*} \Delta r = -\frac{1}{2} \end{align*}$. It doesn"t make much difference to the actual calculation though...