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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
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