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Math Help - Calculus help for test -- find the detrivative when there is y-prime invovled

  1. #1
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    Calculus help for test -- find the detrivative when there is y-prime invovled

    Hey guys, its my first post here and I was wondering if you could help.

    I need to show the point on the graph of:

    x^2-3xy+y^2=1

    has a horizontal tangent line.

    I am assume that since I know when its horizontal = 0 for slope and y'=0

    ps (when I am writing y' it = yprime)

    So step one is find the derivative.

    The teacher told us the derivative is 2x-3y-3xy'-2yy'=0

    is this right ? can anyone explain it?
    Last edited by richtree; April 3rd 2009 at 04:48 PM.
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  2. #2
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    Quote Originally Posted by richtree View Post
    Hey guys, its my first post here and I was wondering if you could help.

    I need to show the point on the graph of:

    x^2-3xy+y^2=1

    has a horizontal tangent line.

    I am assume that since I know when its horizontal = 0 for slope and y'=0

    ps (when I am writing y' it = yprime)

    So step one is find the derivative.

    The teacher told us the derivative is 2x-3y-yxy'-2yy'=0

    is this right ? can anyone explain it?
    So 2x- 3y- (yx+ 2y)y'= 0. Solve for y'.
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    hallsof ivy, thanks for responding,

    but my teacher is telling me the derivative is :

    2x-3y-3xy'-2yy'=0


    I don't know how this came from the original equation ?
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  4. #4
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    x^2-3xy+y^2=1
    you are finding an implicit derivative w/r to x ... treat y as a function of x

    \frac{d}{dx} (x^2 - 3xy + y^2 = 1)

    don't forget the product rule for the middle term ...

    2x -3x \frac{dy}{dx} - 3y + 2y\frac{dy}{dx} = 0

    isolate \frac{dy}{dx} ...

    -3x \frac{dy}{dx} + 2y\frac{dy}{dx} = 3y - 2x

    \frac{dy}{dx}(2y - 3x) = 3y - 2x

    \frac{dy}{dx} = \frac{3y-2x}{2y-3x}
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    thanks to both of you, I still just don't understand it, the teacher really confused me by not showing the rational explanation that the book uses.

    If anyone could do a step by step solution with text I would appreciate it.

    I think if I could see the whole problem start to finish maybe I could grasp it better
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    ok, skeeter, I'm with so far far, can you continue?

    thx guys

    -----------------------------------------------------

    ok, so y = 2/3x

    now I take 2/3x and plug it into the original equation.

    So,

    x^2-3xy+y^2=1

    becomes

    x^2-3x(2/3x) + (2/3x)2 = 1

    right??????? I think I am close . help me finish !!!!

    -------------------------------------------------------

    so next is

    x^2 - 2x^2 +4/9x^2 = 1



    then what ???????????

    I'm offically stuck !!!!!
    Last edited by mr fantastic; April 4th 2009 at 02:20 PM.
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  7. #7
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    Quote Originally Posted by richtree View Post
    so next is

    x^2 - 2x^2 +4/9x^2 = 1

    then what ???????????

    I'm offically stuck !!!!!
    x^2 - 3x\left(\frac{2x}{3}\right) + \left(\frac{2x}{3}\right)^2 = 1

    x^2 - 2x^2 + \frac{4x^2}{9} = 1

    combine like terms ...

    -\frac{5x^2}{9} = 1

    x^2 = -\frac{9}{5}

    this equation has no real solution ... so, what does that tell you?
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  8. #8
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    yeah, so x= pos.neg.sqareroot of -9/5

    so there is no point right ?
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  9. #9
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    Quote Originally Posted by richtree View Post
    yeah, so x= pos.neg.sqareroot of -9/5

    so there is no point right ?
    correct ... no horizontal tangent.
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