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.
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.
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.
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 and you clearly know that . From the second, so and differentiate that.
But I would prefer to use the chain rule:
Now, you can argue that is approximately (Where and really are the "errors" in V and C and "~" means "approximately equal".
If they were equal, we could say that so that .
Im guessing its wrong but I dont get why.
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
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