# Math Help - Estimate the percentage error

1. ## 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.

2. 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 $\frac{dV}{dC}$. You can do that by using the chain rule: $\frac{dV}{dC}= \frac{\frac{dV}{dr}}{\frac{dC}{dr}}$

3. 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.

4. 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.

5. 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 $V= \frac{4}{3}\pi r^3$ and you clearly know that $C= 4\pi r^2$. From the second, $r= \sqrt{\frac{C}{4\pi}}$ so $V= \frac{4}{3}r^3= \frac{4}{3}\pi\left(\sqrt{\frac{C}{4\pi}}\right)^3 = \frac{4}{3}\pi\left(C^{3/2}{8\pi^{3/2}}\right)$ and differentiate that.

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

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

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

Im guessing its wrong but I dont get why.

6. 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