Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on Differentiation

  1. Jul 2, 2008 #1
    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. jcsd
  3. Jul 2, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    Since V is a function of L.

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

    [tex]\frac{dV}{dt}=\frac{dV}{dL} \times \frac{dL}{dt}[/tex]
     
  4. Jul 2, 2008 #3
    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...
     
  5. Jul 2, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    [tex]\frac{dV}{dt}=12L^2[/tex]



    You don't need L in terms of t when they tell you that [itex]\frac{dL}{dt}=10t[/itex]
     
  6. Jul 2, 2008 #5
    so then this question doesn't have a numerical solution, only one in terms of L?
     
  7. Jul 2, 2008 #6

    rock.freak667

    User Avatar
    Homework Helper

    Is that all the data presented in the question? you don't have the initial length or any length at a particular time?
     
  8. Jul 2, 2008 #7
    that's about it, my first post states out the entire question, no initial values, at all...and thanks by the way.
     
  9. Jul 2, 2008 #8

    rock.freak667

    User Avatar
    Homework Helper

    Well without a length at any particular time, you can't find a numerical solution.
     
  10. Jul 15, 2008 #9
    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
  11. Jul 15, 2008 #10
    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.
     
  12. Jul 16, 2008 #11

    rock.freak667

    User Avatar
    Homework Helper

    Yes that was a typo on my part.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question on Differentiation
Loading...