Find and express a ratio that compares the rate at which the volume of the sphere changes with respect to the rate at which the surface area of the sphere changes (the ratio will be a function of $r$). I have found what the change in the volume with respect to time ($4\pi(r)^2*dr/dt$) and the change in the surface area with respect to time ($8\pi(r)*dr/dt$). I am confused on how I should write this ratio and if I should write it with respect to $r$ or $t$. Any help is appreciated!

  • $\begingroup$ I don't think that there should be a time variable. You have a formula for volume $V(r)$ and a formula for surface area $A(r)$. You should be able to find the radius as a formula of surface area $r(A)$ then you can substitute and find a formula for volume in terms of surface area $V(r(A))$. Then simply differentiate $\frac{dV}{dA}$ for your solution. $\endgroup$ – ShawSa Jan 15 at 21:49
  • $\begingroup$ By the chain rule, $\frac{dV}{dA}=\frac{dV}{dt} \frac{dt}{dA}$. Just plug in what you have for $\frac{dV}{dt}$ and $\frac{dt}{dA}=\frac{1}{\frac{dA}{dt}}$. $\endgroup$ – greelious Jan 15 at 21:51

You already have your rates. The ratio of two things is just one thing divided by the other thing. So take your rate for the change in volume and divide it by the change in area.

If you write the division in the format shown below, $$ \frac{\text{rate of change of volume}}{\text{rate of change of area}}, $$ then you may notice some opportunities to cancel factors in the numerator and denominator. Note that $dr/dt$ is just another number when you’re comparing the two rates at the same time. (If you were looking at $dr/dt$ at two different times, you should not assume it stayed the same, but we can be reasonably sure we were supposed to compare two things at the same time; the question makes no sense otherwise.)

  • $\begingroup$ Should I use the derivatives with respect to time that I found? $\endgroup$ – math lover Jan 15 at 23:07
  • $\begingroup$ At first, yes, because they're part of what it means to have a "rate of change" of volume or area. But if you put each formula in its proper place in the ratio, you should find that you have one factor of $dr/dt$ in the numerator and one in the denominator. They'll end up canceling each other--but better not take my word for that, write them out and see that they do that. $\endgroup$ – David K Jan 16 at 1:22
  • $\begingroup$ So, I am allowed to have a ratio that is just a fraction? $\endgroup$ – math lover Jan 16 at 1:46
  • $\begingroup$ Things like "gear ratio" usually get written with a colon, like $3:1$. But a mathematical ratio can just be a number, like $\frac54.$ If you're dealing with someone who really wants to see a colon then the ratio $\frac54$ could be written $1.25:1$. $\endgroup$ – David K Jan 16 at 1:52
  • $\begingroup$ Thank you! Is the proof dv/da=dv/dt * dt/da a valid proof? $\endgroup$ – math lover Jan 16 at 2:01

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