Derivative Error Confirmation

**Biff** · January 9th 2013, 02:29 PM

Hello,

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 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:

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

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

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

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

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

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

Slope:

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

**BobP** · January 10th 2013, 12:51 AM

Your differentiation looks to be correct, but where does that formula come from ?

$V=\frac{1}{2}\left(\frac{4\pi r^{3}}{3} \right)=\pi r^{2}(y-r).$

The middle term is the volume of a hemishere, but what is that term on the right, and why are they equal.

**Biff** · January 10th 2013, 05:51 AM

Hi Bob,

I was wondering that as well and just discovered that there must have been a typo; the formula should instead be:

$V = \frac{1}{2} \left(\frac{4\pi r^3}{3}\right) + \pi r^2 (y - r)$

Differentiating this out again still yields -1/3, so I believe that my school is incorrect?

$V = \frac{1}{2}\left( \frac{4\pi r^3}{3} \right) + (\pi r^2)(y - r)$

$\frac{dr}{dy} \left( V = \frac{1}{2}\left( \frac{4\pi r^3}{3} \right) + (\pi r^2)(y - r) \right)$

$0 = \left( \frac{4\pi r^3}{6} \right)' + (\pi r^2)'(y - r) + (\pi r^2)(y - r)'$

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

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

$\frac{dr}{dy} = \left( 2 \pi r y - 2\pi r^2 - \pi r^2 \right) = -\pi r^2 \\ \\ \frac{dr}{dy} = \frac{-\pi r^2}{2\pi r y - 3\pi r^2} = \frac{-r}{2y - 3r} \\ \\ \frac{dr}{dy} = \frac{-r}{2y - 3r} = \frac{-2}{2(6) - 3(2)} = -\frac{1}{3}$

**Biff** · January 10th 2013, 07:11 AM

Hello,

I contacted my teacher with the work that I did and she said that it was indeed an error. The slope is -1/3.

**BobP** · January 10th 2013, 01:39 PM

Well, I'm glad that you got that sorted, but I still can't see where that $\pi r^{2}(y-r)$ comes from.

**Biff** · January 10th 2013, 04:06 PM

Hi Bob,

Here is a diagram that I came up with:

Derivative Error Confirmation-cylindrical_hemisphere.png

You see that $y$ is the total height, so they did $(y - r)$ for the cylinder volume to get the proper height.

**BobP** · January 11th 2013, 12:08 AM

O.K. that explains that, but now look at your calculation in post 3, specifically in going from five lines from the bottom to four lines from the bottom. What happened to the first term ?
I get (-1/5) for the final answer.

**Biff** · January 11th 2013, 05:46 AM

Ah, you are right. ;-)

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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