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Math Help - rate of change

  1. #1
    Faz
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    rate of change

    (a) At what rate is the radius of a spherical balloon changing at the time that the radius is 10
    cm, if air is being pumped in at a rate of 40 cm3s1?

    (b) Find an expression for
    dy/dx

    given that 3
    xy + ey = 3.

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  2. #2
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    Quote Originally Posted by Faz View Post
    (a) At what rate is the radius of a spherical balloon changing at the time that the radius is 10
    cm, if air is being pumped in at a rate of 40 cm3s1?

    (b) Find an expression for
    dy/dx

    given that 3
    xy + ey = 3.

    1) The volume of a sphere is  V = \frac{4}{3} \pi r^3 . We know that the volume and radius are functions of time here, hence, using implicit differentiation:

     \frac{dV}{dt} = \frac{4}{3} \pi \times \frac{d}{dt}(r^3)

     \frac{dV}{dt} = \frac{4}{3} \pi \times 3r^3 \times \frac{dr}{dt}

    Hence:

     \frac{dr}{dt} = \frac{dV}{dt} \times \frac{1}{4\pi r^2}

    You know what  \frac{dV}{dt} is, and you know what the radius is at our point of interest. So plug them in!

    2)  xy+ey = 3

    You could take out a factor of y on the LHS, and then move all other terms to the RHS, giving you a nice explicit function to differentiate. Or you could go about it using implicit differentiation!

     y(x+e) = 3

     y = \frac{3}{x+e}

    Then use the quotient rule!

    Or

     \frac{d}{dx}(yx) + \frac{d}{dx}(ye) = \frac{d}{dx}(3)

     x\frac{dy}{dx} + y\frac{dx}{dx} + e\frac{d}{dx}(y) = 0

     x\frac{dy}{dx} + y + e\frac{dy}{dx} = 0

     \frac{dy}{dx}(x+e) + y = 0

    Rearrange and remember to substitute in y again from the fact that  y = \frac{3}{x+e} .
    Last edited by Mush; January 15th 2009 at 05:27 PM.
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  3. #3
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    Thanks & noted

    Thank you
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