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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Hey guys can you help me out with this:
    Given the ellipse 4x^2 + 9y^2 = 36,
    a) Find \frac{dy}{dx}.
    is this correct:
    \frac{d(4x^2)}{dy}+\frac{d(9y^2)}{dx}*\frac{dy}{dx  } = \frac{d(36)}{dx}
    8x+18y*\frac{dy}{dx}=0
    \frac{dy}{dx}=-\frac{4x}{9y}


    b) Give the equations of all tangent lines to the graph at x = 1.


    at this point I plug in 1 in the original function to get y right?


    I got this weird number for y when I did that: y=\pm \frac{4\sqrt{2}}{3}


    are my calculations correct?
    Last edited by RezMan; May 10th 2011 at 10:24 PM.
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by RezMan View Post
    Hey guys can you help me out with this:
    Given the ellipse 4x^2 + 9y^2 = 36,
    a) Find \frac{dy}{dx}.
    is this correct:
    \frac{d(4x^2}{dy}+\frac{9y^2}{dx}*\frac{dy}{dx} = \frac{d(36)}{dx}
    8x+18y*\frac{dy}{dx}=0
    \frac{dy}{dx}=-\frac{4x}{9y}


    b) Give the equations of all tangent lines to the graph at x = 1.


    at this point I plug in x in the original function to get y right?


    I got this weird number for y when I did that: y=\pm \frac{4\sqrt{2}}{3}


    are my calculations correct?
    a) yes
    b) yes, but you're not done. Now you have points and can find slopes (dy/dx), so it's time to actually write down the equations!
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  3. #3
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    I agree with your answer for dy/dx, but I don't understand your middle step...

    I agree with your y value. Now you need to find the gradient of the tangent using your (x,y) coordinate.
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  4. #4
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    so the point is actually (1, \pm \frac{4\sqrt{2}}{3})?

    so I'm correct in pluggin in 1 in the original function?

    so
    -\frac{4(1)}{9(\frac{4\sqrt{2}}{3})}

    slope= \pm \frac{1}{3\sqrt{2}}

    ?
    Last edited by RezMan; May 10th 2011 at 10:23 PM.
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  5. #5
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    First of all, you need to work out each point separately...

    No, don't plug each point into the original function. Plug each point into the derivative function to find the gradient of each tangent line.

    Once you have the gradient, then you have x, y, m for the y = mx + c equation, solve for c. Then you have the equation of the tangent line.
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  6. #6
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    right but to get y don't I have to plug 1 into the original function. Where then I end up with that wierd number. And then I plug 1 and that number into the dy/dx that I got. And then that other weird number is my slope.
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  7. #7
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    thanks so much guys! I'm not sure if I'm on the right track so I emailed my teacher. When she gets back to me I'll get back to this.
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  8. #8
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    hey guys I was able to solve most of it except I'm stuck again. It asks what is the tangent line to the graph at y=0. Give its equation.

    tangent line at y=0 is
    -4x/9(0)= undefined, right?
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