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Math Help - Application of derivative

  1. #1
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    Application of derivative

    Hi guys, i have got a few question that needs help before my exams next week

    The surface area of a cube with side x cm, is increasing at a rate of 16cm2/s

    Whnt the total surface area of the cube reaches 216cm2
    a)show that the volume of the cube is 216cm3
    b) find the rate which its volume is increasing

    I completed part a) but for part b, they wants dv/dt so i figure out since da/dt is given i need to find dv/da. But I am stuck ! anyone could explain to me ? THANKS IN ADVANCE !
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Application of derivative

    Let A be the surface area and s be the side of the cube.

    You have \frac{dA}{dt}=16.

    So A=16t+C.

    Taking C=0, A=16t.

    \begin{align*}  A &=16t \\ \implies 6s^2 &=16t \\ \implies s^2 &=\frac{8}{3}t \\ \implies s^3 &=\left( \frac{8}{3}t \right)^{\frac{3}{2}} \\ \implies V &=\left( \frac{8}{3}t \right)^{\frac{3}{2}} \\ \implies \frac{dV}{dt} &=\frac{d\left( \frac{8}{3}t \right)^{\frac{3}{2}}}{dt} \\ \implies \frac{dV}{dt}&= \left(\frac{8\sqrt{2}}{\sqrt{3}} \right)\sqrt{t} \end{align*}
    Last edited by sbhatnagar; December 3rd 2011 at 01:49 AM.
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  3. #3
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    Re: Application of derivative

    A = 6x^2

    \frac{dA}{dt} = 12x \cdot \frac{dx}{dt}

    when A = 216 , x = 6

    16 = 12(6) \cdot \frac{dx}{dt}

    \frac{dx}{dt} = \frac{2}{9}

    V = x^3

    \frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}

    finish it
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