# Curriculum Derivative Error

The following problem indicates that the answer is -1/5, but I am inclined to think that it is -1/3 based on my derivative calculations. Is this an error of mine or theirs?

Note: solving only part A of the problem is necessary (not the whole thing), and that is all I provided.

Problem

A candy company needs a custom box for their truffles. The box they've chosen is the shape of a cylinder with a hemisphere of the same radius on top. The total volume of the box is $V = \frac{1}{2}\left(\frac{4\pi r^3}{3}\right) = \pi r^2(y - r)$, where $y$ is the height of the box and $r$ is the radius of the box. Originally, the candy box was designed to have a height of 6 inches and a radius of 2 inches, but the shipper suggests that the boxes be made slightly shorter. You now need to adjust the radius so that the height is reduced to 5.75 inches but the volume remains constant.

A. Find the value of $dr/dy$ at the point $r = 2, y = 6$.

My Work:

$V = 16\pi$

$V = \pi r^2(y - r)$

$16\pi = \pi r^2(y - r)$

$16 = r^2(y - r)$

Implicit Differentiation:

$\frac{dr}{dy} (16 = r^2(y - r))$

$0 = (r^2)'(y - r) + (r^2)(y - r)'$

$0 = \left(2r\frac{dr}{dy}\right)(y - r) + (r^2)\left(1 - \frac{dr}{dy}\right)$

$\frac{dr}{dy}r^2 + \frac{dr}{dy}2r^2 - \frac{dr}{dy}2ry = r^2$

$\frac{dr}{dy} (r^2 + 2r^2 - 2ry) = r^2$

$\frac{dr}{dy} = \frac{r^2}{3r^2 - 2ry} = \frac{r}{3r - 2y}$

Slope:

$\frac{dr}{dy} = \frac{2}{3(2) - 2(6)} = \frac{2}{-6} = -\frac{1}{3}$

$\frac{dr}{dy} = -1/5$

Differentiate implicitly to get $0 = 2\pi r^2 \frac{dr}{dy} + \pi \left(2ry\frac{dr}{dy} + r^2 - 3r^2 \frac{dr}{dy}\right)$

Then plug in r = 2 and y = 6 (1 point), and solve to get $\frac{dr}{dy} = −1/5$

EDIT: -1/5 Actually Correct

Was careless when I rearranged terms:

$0 = \left( 2\pi r^2 \cdot \frac{dr}{dy} \right) + \left( 2\pi r \cdot \frac{dr}{dy} \right)(y - r) + (\pi r^2)\left(1 - \frac{dr}{dy}\right)$

$0 = 2\pi r^2 \cdot \frac{dr}{dy} + 2\pi r y \cdot \frac{dr}{dy} - 2\pi r^2 \cdot \frac{dr}{dy} + \pi r^2 - \pi r^2 \cdot \frac{dr}{dy}$

$\frac{dr}{dy} \left( 2\pi r^2 + 2\pi r y - 2\pi r^2 - \pi r^2 \right) = -\pi r^2$

$\frac{dr}{dy} = \frac{-r}{2y - r} = \frac{-2}{12 - 2} = -\frac{1}{5}$

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How do you make this leap in the problem statement? $$\frac{1}{2}\frac{4\pi r^3}{3} = \pi r^2 (y-r)$$ – apnorton Jan 10 '13 at 1:49
Since the volume of the box is given to be proportional to the radius cubed, it's impossible to change the radius without changing the volume. – joriki Jan 10 '13 at 1:53
Forgot a sentence - see my post again. The box is a cylinder, and that is the volume formula they specify. I am thinking of just a regular cylinder...but maybe that is incorrect. – Biff Jan 10 '13 at 1:55
@joriki There are two variables here: $y$ (the height) and $r$ (the radius). To shorten the height to 5.75 and keep the same volume, as they propose, would mean increasing $r$; so you are correct. – Biff Jan 10 '13 at 1:58

There is a mistake in the formula $V = \frac{1}{2}\left(\frac{4\pi r^3}{3}\right) = \pi r^2(y - r)$. If the box is cylinder with a hemisphere, then $V = \frac{1}{2}\left(\frac{4\pi r^3}{3}\right) +\pi r^2(y - r)$. Then you differentiate implicitly both sides and you get $0 = 2\pi r^2 \frac{dr}{dy} + \pi \left(2ry\frac{dr}{dy} + r^2 - 3r^2 \frac{dr}{dy}\right)$ which is the answer that you wrote.

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I posted the "Answer Solution" in my original post, do you still believe the correct answer is still $-1/3$? – Biff Jan 10 '13 at 2:03
I checked with my teacher and you are indeed correct. This has taught me to verify volume equations and not expect them to be correct (should have seen that error)! – Biff Jan 11 '13 at 5:06
Actually, -1/5 is the correct answer! See my edited original post. My teacher only skimmed over what I had said. – Biff Jan 12 '13 at 22:49
@Biff I did not say that -1/3 was the correct answer, I only say that the formula of volumen has a mistake! – dwarandae Jan 12 '13 at 23:46
I misunderstood, sorry. – Biff Jan 13 '13 at 2:46

Was careless when I rearranged terms:

$0 = \left( 2\pi r^2 \cdot \frac{dr}{dy} \right) + \left( 2\pi r \cdot \frac{dr}{dy} \right)(y - r) + (\pi r^2)\left(1 - \frac{dr}{dy}\right)$

$0 = 2\pi r^2 \cdot \frac{dr}{dy} + 2\pi r y \cdot \frac{dr}{dy} - 2\pi r^2 \cdot \frac{dr}{dy} + \pi r^2 - \pi r^2 \cdot \frac{dr}{dy}$

$\frac{dr}{dy} \left( 2\pi r^2 + 2\pi r y - 2\pi r^2 - \pi r^2 \right) = -\pi r^2$

$\frac{dr}{dy} = \frac{-r}{2y - r} = \frac{-2}{12 - 2} = -\frac{1}{5}$

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