Finding dV/dt in V=4*L^3 at t=0.1s

[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

 

[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}

 

[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...

 

[rock.freak667]

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...

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



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

 

[Tymick]

\frac{dV}{dt}=12L^2
You don't need L in terms of t when they tell you that \frac{dL}{dt}=10t

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

 

[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?

 

[Tymick]

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

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

 

[rock.freak667]

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

 

[Doctoress SD]

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

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

 

[Doctoress SD]

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



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

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


I think the chain rule is needed.

 

[rock.freak667]

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


I think the chain rule is needed.

Yes that was a typo on my part.