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Thread: Calculus questions

  1. #1
    Member kezman's Avatar
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    Calculus questions

    1 I need to know the convergence of this series

     \sum_{k=1}^{\infty} \frac{2k+5}{\sqrt{3{k^3} - 1/2}}

    2 I have to obtain the expression of dy/dx for the following function

    3xy - 2x^2y + y^3 =0

    3 5x^2 + 3y^2 = 16
    The x radius is \sqrt{5} ?
    Thanks
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  2. #2
    Super Member 11rdc11's Avatar
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    Use the integral test for the 1st problem to see if it converges
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  3. #3
    Moo
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    Quote Originally Posted by 11rdc11 View Post
    Use the integral test for the 1st problem to see if it converges
    It's better to compare :

    As n approaches infinity, 2k+5 ~ k and sqrt(3k-1/2) ~ k^(3/2)

    Therefore, (2k+5)/sqrt(3k-1/2) ~ k/k^(3/2)=1/k^(1/2)

    By comparison with the Riemann series, since 1/2<1, the series is not convergent.

    2 I have to obtain the expression of dy/dx for the following function

    3xy - 2x^2y + y^3 =0
    Implicit differentiation, do you know it ?

    3 5x^2 + 3y^2 = 16
    The x radius is \sqrt{5} ?
    Thanks
    This is the equation of an ellipse, which is x/a + y/b = 1

    So transform it like it should be.
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  4. #4
    Member kezman's Avatar
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    on 2 ive got this

    dy/dx = (2x^2 - 3x)/2y
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  5. #5
    Moo
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    Quote Originally Posted by kezman View Post
    on 2 ive got this

    dy/dx = (2x^2 - 3x)/2y
    Hmmm I don't think you applied the chain rule properly ! o.O

    Would you mind showing your working ?

    I get something like dy/dx=(4xy-3y)/(3x-2x+3y) (I did it quickly)
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  6. #6
    Member kezman's Avatar
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    y^3 = y(2x^2 - 3x)
    y^2 = x(2x -3) (is this legal?)
    and then I derive the sqrt
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  7. #7
    Moo
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    Quote Originally Posted by kezman View Post
    y^3 = y(2x^2 - 3x)
    y^2 = x(2x -3) (is this legal?)
    and then I derive the sqrt
    This is the problem !

    The derivative of y^3 wrt x is 3y'y, by the chain rule. Remember, y is a function of x ! The RHS is weird too.
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