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Math Help - calculus help please

  1. #1
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    Exclamation calculus help please

    hi there, a friend told me this site helped her to understand well so i figured i'd give it a try! i have a few things i'm confused about on my calculus II homework. any tips or pointers would be super

    f(x)= (x)/(x-3). find the derivative of the inverse of f(x) at x=2
    a) -3
    b) -1/3
    c) 1/3
    d) 1
    e) 3

    i found the inverse to be x= y/y-3 but i'm stuck..

    additionally,

    An ellipse with equation x^2 + 9y^2= 9 is the base of a solid in which each cross section perpendicular to the x axis is a square. find the volume of the solid.
    a) 4
    b) 8
    c) 16
    d) 32
    e) 64

    i drew the graph of the ellipse and now i'm stuck

    anyone have any ideas for either of these problems? thank you
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    Quote Originally Posted by mathrox View Post
    f(x)= (x)/(x-3). find the derivative of the inverse of f(x) at x=2
    a) -3
    b) -1/3
    c) 1/3
    d) 1
    e) 3

    i found the inverse to be x= y/y-3 but i'm stuck..
    x=\frac y{y-3}

    \Rightarrow x(y-3)=y\Rightarrow (x-1)y=3x

    \Rightarrow y=\frac{3x}{x-1}

    So f^{-1}(x)=\frac{3x}{x-1}.

    additionally,

    An ellipse with equation x^2 + 9y^2= 9 is the base of a solid in which each cross section perpendicular to the x axis is a square. find the volume of the solid.
    a) 4
    b) 8
    c) 16
    d) 32
    e) 64

    i drew the graph of the ellipse and now i'm stuck
    The volume of a solid with cross sections, perpendicular to the x-axis, of known area A(x) on [a,b] is

    V=\int_a^bA(x)\,dx.

    If we solve for y in your equation, we get y=\pm\sqrt{1-\frac{x^2}9}. So, the area of the cross section at x is

    A(x)=\left[2\sqrt{1-\frac{x^2}9}\right]^2 (squaring because the cross section is a square).

    The total volume is then

    V=4\int_{-3}^3\left(1-\frac{x^2}9\right)\,dx.

    I use -3 and 3 for the limits because the major axis of the ellipse is of length 6 and it is centered at the origin.
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  3. #3
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    thank you!! i can do it from there now
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