# Question on Differentiation

1. Jul 2, 2008

### Tymick

I'm having some trouble with the following question, it was on a test previously and I haven't been able to figure it out :/

Let V=4*L^3 cm^3, where dl/dt=10*t cm/s. Find dV/dt at t=0.1 second

2. Jul 2, 2008

### rock.freak667

Since V is a function of L.

What you can do is use the chain rule and say that

$$\frac{dV}{dt}=\frac{dV}{dL} \times \frac{dL}{dt}$$

3. Jul 2, 2008

### Tymick

Right, following the chain rule I do get dv/dt=120*L^2, I tried before to get L in terms of t, via integration but of course that gives 5t^2+C whereas C would pose as a problem since I've no way of obtaining it, so I can't get a numerical solution I just get a solution in terms of L...

4. Jul 2, 2008

### rock.freak667

$$\frac{dV}{dt}=12L^2$$

You don't need L in terms of t when they tell you that $\frac{dL}{dt}=10t$

5. Jul 2, 2008

### Tymick

so then this question doesn't have a numerical solution, only one in terms of L?

6. Jul 2, 2008

### rock.freak667

Is that all the data presented in the question? you don't have the initial length or any length at a particular time?

7. Jul 2, 2008

### Tymick

that's about it, my first post states out the entire question, no initial values, at all...and thanks by the way.

8. Jul 2, 2008

### rock.freak667

Well without a length at any particular time, you can't find a numerical solution.

9. Jul 15, 2008

### Doctoress SD

I would think that as you are given dl/dt and you can work out dv/dl and you are looking for dv/dt

dv/dl [12L^3] dL = 12L^2 dl/dt = 10t

dv/dl x dl/dt... the dl's cancel to leave dv/dt

Wouldn't that be the chain rule?

Perhaps u'v + v'u
or in this case v't + t'v

NO WAIT!!!!! So much simpler.

Look at it as...

the change in v over the change in time. The change in V is simply 12L^3 , so dv/dt = 12L^3 / dt ( it would seem something is missing as one is not sure the intial time, I would have to assume time began at zero and so dt would be 0.1 s,
thus dv/dt = 12L^3/0.1 = 120L^3 (I think).

eta: oh I missed he t=0.1 s

Last edited: Jul 16, 2008
10. Jul 15, 2008

### Doctoress SD

Isn't dv/dt the differential of v with respect to t? Then surely the answer shouldnt be given in 'L' Also you have worked out dv/dl not dv/dt

I think the chain rule is needed.

11. Jul 16, 2008

### rock.freak667

Yes that was a typo on my part.