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Math Help - [SOLVED] How to solve this question?

  1. #1
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    [SOLVED] How to solve this question?

    Find equation of normal at point (3,1) on curve 2x^2 - xy + 3y^2 = 18. The problem is with differentiating this 2x^2 - xy + 3y^2 = 18 part. How? Thanks
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  2. #2
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    Quote Originally Posted by mark1950 View Post
    Find equation of normal at point (3,1) on curve 2x^2 - xy + 3y^2 = 18. The problem is with differentiating this 2x^2 - xy + 3y^2 = 18 part. How? Thanks
    Read this: http://www.csu.edu.au/division/studs...rentiation.pdf
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    Thanks, Mr. Fantastic but I have a problem. Here is my take on the question :

    I differentiate the whole thing to get,

     4x - (x\frac{dy}{dx} + y) + 6y = 0
     \frac{4x + 5y}{x} = \frac{dy}{dx}

    Then, I solved it to find the gradient :

     \frac{4(3) + 5(1)}{3} = \frac{dy}{dx}

    to get 17/3. Since they want me to find the normal, I solved the gradient to -3/17.

    But when I try to find the equation, I got

    17y = -3x + 26

    which is way different then the answer, which is, 11y = 3x + 2. Am I wrong or is it the answer that's wrong? Thanks.
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  4. #4
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    Quote Originally Posted by mark1950 View Post
    Thanks, Mr. Fantastic but I have a problem. Here is my take on the question :

    I differentiate the whole thing to get,

     4x - (x\frac{dy}{dx} + y) + 6y {\color{red} \frac{dy}{dx}} = 0 Mr F says: Note the correction in red (so all that follows is wrong). This follows from the chain rule.

     \frac{4x + 5y}{x} = \frac{dy}{dx}

    Then, I solved it to find the gradient :

     \frac{4(3) + 5(1)}{3} = \frac{dy}{dx}

    to get 17/3. Since they want me to find the normal, I solved the gradient to -3/17.

    But when I try to find the equation, I got

    17y = -3x + 26

    which is way different then the answer, which is, 11y = 3x + 2. Am I wrong or is it the answer that's wrong? Thanks.
    Furthermore, since you only need the value of dy/dx and not the actual rule for dy/dx, my advice is to substitute (3, 1) into the line I corrected and then solve for dy/dx to get the required value.
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  5. #5
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    Oh! How careless...lol. Thanks, Mr. F!
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