I've got this problem:

The stem of a particular mushroom has a cylindrical shape. A stem with $2$ cm of height and $r$ centimeters of radius which has a volume $V=2 \pi r^2$. Use differentials to determine the approximated increase of the stem's volume where its radius grows from $0.4$cm to $0.5$cm.

So I got the $dr$ from the difference of $0.4 - 0.5$ which means its result was $0.1$. Right after that, $Dv=4 \pi r dr$. I calculated this without the value of $r$ and I think you can't do it that way, the result was $1.256r \ \text{cm}^3$ but I think there is something left.

  • $\begingroup$ When you write "phi", do you mean "pi" ($\pi$)? Phi is the name of a different letter, which looks like $\phi$ or $\varphi$. $\endgroup$ – hmakholm left over Monica Oct 21 '11 at 21:36
  • $\begingroup$ It is not a differential equation, you were probably asked to use differentials. You could use your calculator to find, in this case, the exact change in volume. Your answer is much larger than the actual change, which is kind of an indication your answer may be wrong. The procedure, as far as I can guess from what is written, is right. You got $4\pi r\,dr$. Ask yourself now, what is the appropriate $r$? What is the appropriate $dr$? $\endgroup$ – André Nicolas Oct 21 '11 at 21:37
  • $\begingroup$ I edited the differential equation to differentials, sorry, my bad but I still can't figure it out. $\endgroup$ – Theil Oct 21 '11 at 21:59
  • $\begingroup$ I do not have the value of $r$ . should I clear $r$ ? or should I simply express it that way? $\endgroup$ – Theil Oct 21 '11 at 22:08
  • $\begingroup$ You do have the value of $r$. Rather, you have two values of $r$, as you are told the radius grows from .4 to .5. The starting value of $r$ is the one you want to use when talking about the "augment" (or, increase) in volume. $\endgroup$ – Gerry Myerson Oct 21 '11 at 22:26

Given that the volume is $V=2\pi r^2$, if $r$ increases to $r+dr$ you have the new volume $V'=2\pi (r + dr)^2$. You are asked for the increase in volume, which is $Dv=2\pi (r + dr)^2 - 2\pi r^2$. (I am using your notation-I would suggest if the volume is $V$, the change in volume should be $DV$ or $dV$). The method of differentials says that $(dr)^2$ is negligible compared with $dr$, so you are correct in finding $Dv=4 \pi r\ dr$. Although $r$ varies in the problem, we have agreed to ignore the variation except where it is indicated specifically, so it would be consistent to use anything in the range $0.4 - 0.5$. If you care about the difference in the answers you get, you need to use a more accurate method.

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