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Math Help - Differential calculus help/check....

  1. #1
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    Differential calculus help/check....

    Hi I have some questions, I'm not sure if I did them correctly..

    1.
    a.) The Cissoid of Diocles is a classical curve given by the equation
    y^2 = x^3/2-x
    Find the slope of the tangent line at the point (1,1)

    b.) Consider the point (3,-4) on the circle centred at the origin. Find the equation of the tangent line at this point

    Okay this is what I did
    y^2 = x^3/2-x
    = 2y y' = x^3/2-x
    = y' = (1/2y)(x^3)/(2-x)
    Sub the point (1,1) into the equation I get
    m = 1/2
    Then use the point slope equation I get y = 1/2x +1/2???

    b.) x^2 + y^2 = 25
    2x + 2y y' = 0
    y' = -2x-2y
    sub the points (3,-4) in I get m = 2
    then use point slope equation I get y = 2x-10....

    Someone taught me how to do the implicit differentiation for y', thats what I did above not sure If thats how you do it though...........help would be appreciated it

    Also one more question

    2. Find the constants c and K such that the functions f(x) = e^cx + k satisfies the differential equation
    3f''(x) + 13f'(x) = 10f(x)

    What I did was...
    f(x) = e^cx + k
    f'(x) = ce^cx
    f''(x) = c^2e^cx
    Sub these into the equation.
    3(c^2e^cx) + 13(ce^cx) = 10(e^cx + k)
    3c^2e^cx + 13ce^cx - 10e^cx = 10k
    e^cx[3c^2 + 13c - 10] = 10k/e^cx
    ....then im not sure what to do anymore....
    any help would be appreciated , thanks guys
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  2. #2
    Senior Member apcalculus's Avatar
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    your implicit differentiation (first step) is incorrect as you haven't differentiated the right hand side.

    Use the quotient rule for the right hand side.

    The result should be:

    y'=\frac{3 x^2 - x^3}{(-2 + x)^2 y}

    If you show your work here I'll be happy to check.

    Good luck!
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  3. #3
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    Hi, thanks for replying okay

    I'm guessing you are replying to Part a of the first question

    because I'm not sure if i did the implicit differentiation correctly, and based on your result...

    y' = 3x^2 - x^3/(-2 + x)^2y)

    I did
    y' = (1/2y)(x^3/2-x)
    = f'g-g'f/g2
    = (3x^2)(2-x) - (-1)(x^3)/(2-x)^2
    = 6x^2 - 2x^3/(2-x)^2(2y)
    Then simplifly 2
    = 3x^2 - x^3/(2-x)^2y

    my answer is slightly different from yours though...in the denominator it is 2-x instead of -2+x.
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  4. #4
    Senior Member apcalculus's Avatar
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    You're right... it's the same answer, if you pull out a negative sign, then square it from the bottom, you can see that the results are the same.

    y^2 = \frac{x^3}{2-x}

    2yy' = \frac{(2-x) 3x^2 - (-) x^3}{(2-x)^2}

    2yy' =\frac{6x^2-2x^3}{(2-x)^2}

    y' =\frac{3x^2-x^3}{y(2-x)^2}
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  5. #5
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    Okay, thanks, so in this case I can plug in the points and find the equation? what about part b?

    is there anything I need to change, anything I did wrong?
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