# Estimate the percentage error

• Jan 6th 2011, 01:45 PM
bcahmel
Estimate the percentage error
The circumference of a sphere is measured at C = 100 cm. Estimate the maximum percentage error in V if the error in C is at most 3 cm.

so C=2(pi)r^2
Dc=4(pi)r (dr) do I use implicit differentiation?

I don't understand what I do from here. Or if I even did that right. Any help would be greatly appreciated.
• Jan 6th 2011, 02:18 PM
HallsofIvy
Quote:

Originally Posted by bcahmel
The circumference of a sphere is measured at C = 100 cm. Estimate the maximum percentage error in V if the error in C is at most 3 cm.

so C=2(pi)r^2
Dc=4(pi)r (dr) do I use implicit differentiation?

I don't understand what I do from here. Or if I even did that right. Any help would be greatly appreciated.

The problem as for a percentage error in V given a percentage error in C. What you need is $\displaystyle \frac{dV}{dC}$. You can do that by using the chain rule: $\displaystyle \frac{dV}{dC}= \frac{\frac{dV}{dr}}{\frac{dC}{dr}}$
• Jan 9th 2011, 10:52 AM
bcahmel
so I still don't get this. What is dV?
Can I use linearization? So C=2(pi)r^2 + 3 cm. I know the actual value is 100. so delta x, which is 3 divided by 100 would give .03 error. Why can't I do this? Im guessing its wrong but I dont get why.
• Jan 9th 2011, 12:16 PM
mr fantastic
Quote:

Originally Posted by bcahmel
so I still don't get this. What is dV?
Can I use linearization? So C=2(pi)r^2 + 3 cm. I know the actual value is 100. so delta x, which is 3 divided by 100 would give .03 error. Why can't I do this? Im guessing its wrong but I dont get why.

You have been shown one way of getting dV/dC. Another way (and the way I'd guess you're probbaly expected to use) is to first get an expression for the volume of a sphere in terms of its circumference. Then you can get dV/dC.

Then I suggest you refer to your class notes or textbook for the appropriate formula to get the percentage error (possibly under the heading linear approximation). Your reference material will have examples.
• Jan 9th 2011, 03:09 PM
HallsofIvy
Quote:

Originally Posted by bcahmel
so I still don't get this. What is dV?
Can I use linearization? So C=2(pi)r^2 + 3 cm. I know the actual value is 100. so delta x, which is 3 divided by 100 would give .03 error. Why can't I do this?

It's wrong because it gives you the error in C, not V! You haven't said anything about the volume!

Strictly speaking, dV and dC are "infinitesmal" changes in V and C but they can be approximated by small changes in V and C- in other woreds the error in measurement.

You could, as mr. fantastic suggests, actually find V as a function of C. You know, I hope, that $\displaystyle V= \frac{4}{3}\pi r^3$ and you clearly know that $\displaystyle C= 4\pi r^2$. From the second, $\displaystyle r= \sqrt{\frac{C}{4\pi}}$ so $\displaystyle V= \frac{4}{3}r^3= \frac{4}{3}\pi\left(\sqrt{\frac{C}{4\pi}}\right)^3 = \frac{4}{3}\pi\left(C^{3/2}{8\pi^{3/2}}\right)$ and differentiate that.

But I would prefer to use the chain rule:
$\displaystyle \frac{dV}{dC} = \frac{dV}{dr} \frac{dr}{dC} = \frac{\frac{dV}{dr}}{\frac{dC}{dr}}$

Now, you can argue that $\displaystyle \frac{\Delta V}{\Delta C}$ is approximately $\displaystyle \frac{dV}{dC}$ (Where $\displaystyle \Delta V$ and $\displaystyle \Delta C$ really are the "errors" in V and C and "~" means "approximately equal".

If they were equal, we could say that $\displaystyle \frac{dV}{dC}= \frac{\Delta V}{\Delta C}$ so that $\displaystyle \Delta V= \left(\frac{dV}{dC}\right)\Delta C$.

Quote:

Im guessing its wrong but I dont get why.
• Jan 10th 2011, 04:09 PM
bcahmel
ok it's starting to make sense. It's funny though, I talked to my calc teacher and he said I didn't even need to use calculus. He basically said that
C=2pir=100,so r=50/pi
and C=2pir=103, so r=103/2pi

Then he said to plug these r values into the V=4/3pir formula, and then subtract their differences. Then basically find a percent from there. I don't think he was too sure though....

and HallofIvy, you're right I'm trying to find the percentage area in V. Guess I was slow that day, I didn't connect V with volume somehow...

so if I try to find the percent error the chain rule way it's V/C basically. So dV/dC w/ the chain rule becomes (dV/dR)/(dC/dR)...where dv=4pir^2, dC=4pir, and what is dR? wow, this is bad, I never imagined this problem would give me this much trouble. But really, thanks for all the help so far! You're all great