<s>[INST] 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:
$\frac{dr}{dy} = \frac{2}{3(2) - 2(6)} = \frac{2}{-6} = -\frac{1}{3}$
Answer Solution
$\frac{dr}{dy} = -1/5$
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:
$0 = \left( 2\pi r^2 \cdot \frac{dr}{dy} \right) + \left( 2\pi r \cdot \frac{dr}{dy} \right)(y - r) + (\pi r^2)\left(1 - \frac{dr}{dy}\right)$
$0 = 2\pi r^2 \cdot \frac{dr}{dy} + 2\pi r y \cdot \frac{dr}{dy} - 2\pi r^2 \cdot \frac{dr}{dy} + \pi r^2 - \pi r^2 \cdot \frac{dr}{dy}$
$\frac{dr}{dy} \left( 2\pi r^2 + 2\pi r y - 2\pi r^2 - \pi r^2 \right) = -\pi r^2$
$\frac{dr}{dy} = \frac{-r}{2y - r} = \frac{-2}{12 - 2} = -\frac{1}{5}$
 [/INST] 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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