Question

Need help with Differentiation (Rates of Change) - The volume, V cm^3, of a cone of height h is [(pi)(h^3)]/(12). If h increases at a constant rate of 0.2 cm s^-1 and the initial height is 2 cm, express V in terms of t and find the rate of change of V at time t.

Answer 1

You're given\[V=\frac{1}{12}\pi h^3,\]\[\frac{dh}{dt}=\frac{1}{5}\]and are asked to find\[\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}\]where\[\frac{dV}{dt}=\frac{1}{4}\pi h^2.\]Then\[\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}=\frac{1}{4}\pi h^2\cdot\frac{1}{5}=\frac{1}{20}\pi h^2.\]

Answer 2

I meant to write "... where \[\frac{dV}{dh}=\frac{1}{4}\pi h^2..."\]

Answer 3

Do you know how to express V in terms of t?

Answer 4

Hey! Thanks A LOT for the help! Your explanation does make sense, but somehow the answer doesn't match up with the one given in my textbook..

Answer 5

Another approach is to start by expressing \(h\) in terms of \(t\). Using the second equation across wrote
\(\frac{dh}{dt}=\frac{1}{5} \implies h(t)=\frac{1}{5}t+c\), but we have h(0)=2cm which gives c=0. Thus \(h(t)=\frac{1}{5}t+2\).

Answer 6

Now substitute this into \[V=\frac{\pi h^3}{12}=\frac{\pi(\frac{1}{5}t+2)^3}{12}=..\]

Answer 7

I meant \(c=2\)...

Answer 8

Wait, I don't get the whole 'c' part..we haven't done that in our course..Isn't there any other method to go about this frustrating question? Thanks, btw!

Answer 9

Do you know anti-derivatives?

Answer 10

Nope. I guess we're supposed to do it next time. So are you saying this question can't be answered without knowing them, then?

Answer 11

It's a simple differential equation xd\[\frac{dh}{dt}=\frac{1}{5},\]\[dh=\frac{1}{5}dt,\]\[\int dh=\int\frac{1}{5}dt,\]\[h=\frac{1}{5}t+c.\]c is a constant of integration.

Answer 12

Oh I see..integration..that's why I can't wrap my head around this question. We're yet to start it, but I don't know what this question is doing in a supposedly 'basic' differentiation chapter..sorry, my bad

Answer 13

The part about finding the rate of change of V at a time t can be found without the need of anti-derivatives. But, I don't think you can express V in terms of t unless you use anti-derivatives or integration as across did above.

Answer 14

Good luck!

Answer 15

Thanks for all your help, Mr. Math and across!