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Math Help - Differentiation Problem

  1. #1
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    Differentiation Problem

    Hello, I ran across this problem and want to make sure my answers are right.

    Consider the curve defined by
    x^2 + xy + y^2 = 27.

    a. Write an expression for the slope of the curve at any point (x,y).
    b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leadds to your conclusion.
    c. Find the points on the curve where the lines tangent to the curve are vertical.


    Thanks for your help!
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  2. #2
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    Quote Originally Posted by kittykat616 View Post
    Hello, I ran across this problem and want to make sure my answers are right.

    Consider the curve defined by
    x^2 + xy + y^2 = 27.

    a. Write an expression for the slope of the curve at any point (x,y).
    b. Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leadds to your conclusion.
    c. Find the points on the curve where the lines tangent to the curve are vertical.


    Thanks for your help!
    so ... what are your answers?
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  3. #3
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    I've currently working on B but for A i got 1/xy.
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  4. #4
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    Quote Originally Posted by kittykat616 View Post
    I've currently working on B but for A i got 1/xy.
    that is incorrect ... what do you know about finding an implicit derivative?
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  5. #5
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    I just redid it and got -3x - y / 2y.

    my work:
    2x + [(x)(1)+(1)(y)] + 2y (dy/dx) = 0

    then solved for dy/dx
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  6. #6
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    Quote Originally Posted by kittykat616 View Post
    I just redid it and got -3x - y / 2y.

    my work:
    2x + [(x)(1)+(1)(y)] + 2y (dy/dx) = 0
    ..........no
    then solved for dy/dx
    \frac{d}{dx}[x^2 + xy + y^2 = 27]

    2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0<br />

    now solve for \frac{dy}{dx}
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  7. #7
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    Quote Originally Posted by skeeter View Post
    \frac{d}{dx}[x^2 + xy + y^2 = 27]

    2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0<br />

    now solve for \frac{dy}{dx}
    thank you for catching my mistake. for the final answer i got -4x-y/2y+x
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  8. #8
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    Quote Originally Posted by kittykat616 View Post
    thank you for catching my mistake. for the final answer i got -4x-y/2y+x
    ...............no again
    ...
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  9. #9
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    Quote Originally Posted by skeeter View Post
    ...
    x(dy/dx) + 2y(dy/dx) = -2x - y
    divide both sides by x
    (dy/dx) + 2y(dy/dx) = -2x - y / x
    divide both sides by 2y
    1(dy/dx) + 1(dy/dx) = -2x -y/ 2xy
    2(dy/dx) = -2xy -y/2xy
    divide by 2 or times by 2/1.

    = -4x -2y/2xy.
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  10. #10
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    2x + x \cdot \frac{dy}{dx} + y \cdot 1 + 2y \cdot \frac{dy}{dx} = 0<br />

    x \cdot \frac{dy}{dx} + 2y \cdot \frac{dy}{dx} = -2x - y

    \frac{dy}{dx}(x + 2y) = -2x - y

    \frac{dy}{dx} = \frac{-2x-y}{x+2y}

    you need some more practice with the basics
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  11. #11
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    it always makes sense when i see it done correctly.. you're right i do need more practice. Thank you for your help. How would i go about part b?
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  12. #12
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    Quote Originally Posted by kittykat616 View Post
    it always makes sense when i see it done correctly.. you're right i do need more practice. Thank you for your help. How would i go about part b?
    find the x-intercepts of the original curve ... parallel tangent lines should tell you that the respective slopes have to be equal.
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  13. #13
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    So i would set -2x - y = 0, right?
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  14. #14
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    Quote Originally Posted by kittykat616 View Post
    So i would set -2x - y = 0, right?
    no ... x-intercepts are the points on the curve x^2 + xy + y^2 = 27 where y = 0
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  15. #15
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    Gosh thanks for all your help.

    So it would be x^2= 27, or x = 5.19 and -5.19
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