1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Connected rates of change question

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    At time t seconds the surface area of a cube is A cm² and the volume is V cm³. The surface area of the cube is expanding at a constant rate 2 cm²/s
    Show that
    [tex]\frac{dV}{dt} = \frac{1}{2}V^{\frac{1}{3}}[/tex]

    2. Relevant equations
    This is a rates of change question and therefore I will need to dervie differential equations from the information presented.
    I can say V = x³
    A = 6x²
    Where x is the length of a side

    3. The attempt at a solution

    [tex]\frac{dA}{dT} = A + 2[/tex]
    [tex]\frac{dA}{dT} = x^{2} + 2[/tex]

    Am I going in the right direction?

    Thanks :)
  2. jcsd
  3. Nov 23, 2008 #2

    The rate of change of A being constantly two means
    You wrote down a formula for V(x) and for A(x). Combine these two to get a formular V(A) for the volume in terms of the area. Then ther ate of change for the volume would be
    Then in this expression, express the area A again in terms of the volume V, and you will be done.
  4. Nov 23, 2008 #3
    ahh cool. how do i got about combining these formulae? Should I use x as the parameter (like in parametrics)

    Thanks :)
  5. Nov 23, 2008 #4
    Yes you know A(x). Make x the subject to get x(A). Plug it into V(x) and viola you have V(A).
  6. Nov 23, 2008 #5
    x = sqrt(a)
    v = a^{3/2}

    however when i diff that [tex]\frac{dV}{dA} = \frac{3}{2} V^{1}{2}[/tex]

    that is not the correct answer

    Where am I messing up?

    Thanks :)
  7. Nov 23, 2008 #6


    Staff: Mentor

    You are asked to show that [tex]\frac{dV}{dt} = \frac{1}{2} V^\frac{1}{3}[/tex]
    You stopped with [tex]\frac{dV}{dA}[/tex]

    How can you get [tex]\frac{dV}{dt}[/tex] from [tex]\frac{dV}{dA}[/tex]?
  8. Nov 23, 2008 #7
    multiply by dA/dt which is 2?
    which gives me 3V ^ {3/2}

    that isn't the right answer though?

    Thanks :)
  9. Nov 23, 2008 #8
    A=6x2, you missed that 6, didn't you?:smile:

    Moreover, your final exponent of V is wrong. You have [itex]V\sim A^{3/2}[/itex], so
    [tex]\frac{dV}{dA}\sim A^{1/2}\sim V^{1/3}[/tex]

    You should be able to get the factos right.
  10. Nov 24, 2008 #9
    right im getting this:
    yeh i missed a = 6x²

    dv/da = 3/2 * 6 & A ^(1/2)
    = 9A^(1/2)

    now multiply this by da/dt giving us 18A^(1/2)

    now remeber a = 6 x 3root(V) ^ 2
    = 6V^{2/3}
    and a is to the power of a half


    6^(1/2) x 6V^(1/3)
    now im almost there but I cant get the 1/2 at the front of V

    interestingly 6^2 = 36 and 18/36 is 1/2 so im guessing it's somthing round there?

    Thanks :)
  11. Nov 24, 2008 #10
    No, what are you doing there. A=6x2, so
    Plug this into V=x3, and you get
    Differentiating gives
    Together with dA/dt=2, this means
    which is the correct result
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook