Related rates problem with Ohm's law V=IR

[kreil]

Question:
Ohm's law for electrical circuits states that V=IR, where V is voltage, I is current in amperes, and R is the resistance in ohms. Suppose that V is increasing at the rate of 1 volt/sec while I is decreasing at the rate of 1/3 amp/sec. Let t denote time in seconds.
a)what is the value of dV/dt?
b)what is the value of dI/dt?
c)what equation relates dR/dt to dV/dt and dI/dt?
d)Find the rate at which R is changing when V=12 volts and I=2 amps. Is R increasing or decreasing?

I don't think I had any problems with the first 3 parts...

$$a) \frac{dV}{dt}=1 v/s$$
$$b)\frac{dI}{dt}=-\frac{1}{3}amp/sec$$
$$c)\frac{dV}{dt}=(\frac{dI}{dt})(\frac{dR}{dt})$$

I don't quite understand part d. It gives information to be used in the original equation, not the differentiated one. Maybe it is irrelevant and I just need to do this? :

$$\frac{dV}{dt}=(\frac{dI}{dt})(\frac{dR}{dt})$$
$$1v/s=(-\frac{1}{3})(\frac{dR}{dt})$$
$$\frac{dR}{dt}=-3 ohms/s$$

If someone could help me quickly I would appreciate it!

 

[Nylex]

Why don't you need to use the product rule to differentiate the RHS? I would have thought it should be dV/dt = d(IR)/dt = RdI/dt + IdR/dt.

 

[kreil]

You're right, slipped my mind, Thanks.

 

[dextercioby]

Why don't you need to use the product rule to differentiate the RHS? I would have thought it should be dV/dt = d(IR)/dt = RdI/dt + IdR/dt.

Well,Nylex,you're right...Again.He should be differentiating Ohm's law wrt ti time and substitute all known quantities in the new equation and from there to extract dR/dt.

Let's hope he sees his mistake.

Daniel.

 

[kreil]

Yes, I understand my mistake. I had a very similar problem earlier in the homework and did it correctly, the time pressure just made me think a little too fast. When he pointed out I differentiated it incorrectly, I checked my work and realized you actually don't even use product rule, but rather quotient rule since it asks for dR/dt in relation to the others, you need to solve for R then differentiate:

$$R=\frac{V}{I}$$
$$\frac{dR}{dt}=\frac{I{\frac{dV}{dt}}-V{\frac{dI}{dt}}}{I^2}$$

substituting in numbers:

$$\frac{dR}{dt}=\frac{(2)(1)-(12)(-\frac{1}{3})}{4}$$

$$\frac{dR}{dt}=\frac{6}{4}=1.5 ohms/s$$

Thanks!