Instantaneous rate of change for a volume

• Oct 20th 2011, 06:51 AM
Instantaneous rate of change for a volume
pv=c, where p=pressure, v= volume and c=constant.

When the volume is 6 cm^3, the pressure is 2 kg/cm^2 what is the instantaneous rate of change of the volume if the pressure decreases by 200g/cm^2 every minute?

So instantaneous rate of change is the same as the derivative, from what I understand. Usually to find this I would multiply d/dt into the equation, find the derivatives,isolate for the variable I am looking for and sub in the given values to solve. In this instance I just don't know where c fits in. Can anyone help me figure it out?
• Oct 20th 2011, 07:01 AM
emakarov
Re: Instantaneous rate of change for a volume
We have $\displaystyle v(t)=\frac{c}{p(t)}$, so $\displaystyle v'(t)=-\frac{c}{p^2(t)}p'(t)=-\frac{v(t)}{p(t)}p'(t)$. All three quantities are given.
• Oct 20th 2011, 07:33 AM
Re: Instantaneous rate of change for a volume
I'm not really following the third step--I don't quite understand how
-__c__ (p' (t)) = -v(t) (p'(t))
_p^2(t)________p(t)

I'm probably missing something obvious but I've been trying to wrap my brain around it and can't seem to . . .
• Oct 20th 2011, 07:38 AM
emakarov
Re: Instantaneous rate of change for a volume
$\displaystyle -\frac{c}{p^2(t)}}=-\frac{c}{p(t)}\frac{1}{p(t)}=-v(t)\frac{1}{p(t)}$ because $\displaystyle p(t)v(t)=c$.
• Oct 20th 2011, 07:51 AM
Re: Instantaneous rate of change for a volume
okay thanks, that makes sense!

Now just one more question. . . I know that I sub in 6 v(t) and 2 for p (t) but where does the .2kg come in? I thought that was the value of the average rate of change, not the instantaneous rate of change...

• Oct 20th 2011, 07:52 AM
Re: Instantaneous rate of change for a volume
sorry I meant .2kg/cm^2 (the change in pressure per minute)
• Oct 20th 2011, 08:05 AM
emakarov
Re: Instantaneous rate of change for a volume
I think that "the pressure decreases by 200g/cm^2 every minute" means that the instantaneous rate of change of pressure is 200 when measured in g / (cm^2 * min). E.g., this is the same thing as 3.33 g / (cm^2 * sec). Otherwise, it's impossible to find v'(t) because it depends on p'(t), and the average rate of change of p(t) does not determine p'(t) at some particular moment t. That is, it does not determine unless there are some additional requirements on p(t), such as that p(t) is linear, i.e., $\displaystyle p(t) = p_0 + C(t - t_0)$ for some constants $\displaystyle C$ and $\displaystyle t_0$. But in this case the instantaneous and the average rates of change of p(t) are the same.
• Oct 20th 2011, 09:08 AM