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Math Help - Implicit differentiation

  1. #1
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    Implicit differentiation

    Finding dy/dx for y^2 = 5x^3. Find equation of the tangent line to the graph at the given point. (1, sqrt 5)

    Here's what I did;
    2yy' = 15x^2
    y' = 15x^2 / 2y

    Where did I go wrong? Or, if it's right by some chance, how do I proceed to find the equation?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    Finding dy/dx for y^2 = 5x^3

    Here's what I did;
    2yy' = 15x^2
    y' = 15x^2 / 2y

    Where did I go wrong?
    Do you have to use Implicit Differentiation?

    y^2 = 5x^3

    y = \sqrt{5}x^{\frac{3}{2}}

    \frac{dy}{dx} = \frac{3\sqrt{5}}{2}x^{\frac{1}{2}}

    \frac{dy}{dx} = \frac{3\sqrt{5x}}{2}.


    If you WERE going to use Implicit Differentiation:

    y^2 = 5x^3

    \frac{d}{dx}(y^2) = \frac{d}{dx}(5x^3)

    \frac{d}{dy}(y^2)\,\frac{dy}{dx} = 15x^2

    2y\,\frac{dy}{dx} = 15x^2

    \frac{dy}{dx} = \frac{15x^2}{2y}


    And since y = \sqrt{5x^3}

    \frac{dy}{dx} = \frac{15x^2}{2\sqrt{5x^3}}

     = \frac{15x^2\sqrt{5x^3}}{10x^3}

     = \frac{3\sqrt{5x^3}}{2x}

     = \frac{3x\sqrt{5x}}{2x}

     = \frac{3\sqrt{5x}}{2}

    which is the same as found above...
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    Quote Originally Posted by Prove It View Post
    And since y = \sqrt{5x^3}
    Sorry, I just editted my post because I forgot to include that the problem already had given points.
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    Well if you have (x,y) = (1, \sqrt{5}), then what does

    \frac{dy}{dx} = \frac{15x^2}{2y} equal?
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    Quote Originally Posted by Prove It View Post
    Well if you have (x,y) = (1, \sqrt{5}), then what does

    \frac{dy}{dx} = \frac{15x^2}{2y} equal?
    15/ (2 sqrt5) ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?
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    Quote Originally Posted by Archduke01 View Post
    15/ (2 sqrt5) ... which I'm assuming is the slope of the tangent line. How do I get the rest of the equation?
    Correct, which I would rewrite as \frac{3\sqrt{5}}{2} by rationalising the denominator.


    Now, you have the tangent line y = mx + c.

    You know y = \sqrt{5}, x = 1, m = \frac{3\sqrt{5}}{2}.

    Solve for c.
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    Quote Originally Posted by Prove It View Post
    You know y = 2\sqrt{5}.
    Shouldn't it be y = sqrt {5}?
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    Quote Originally Posted by Archduke01 View Post
    Shouldn't it be y = sqrt {5}?
    Yes it is, typo.

    Will edit now.
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    If I did it correctly, the value of c should be - sqrt 5 / 2...

    So the equation should be y = 3 sqrt 5 /2 x - sqrt 5/2

    Does that seem right? The answer book says 15x - 2 (sqrt5)y - 5 = 0
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    Quote Originally Posted by Archduke01 View Post
    If I did it correctly, the value of c should be - sqrt 5 / 2...

    So the equation should be y = 3 sqrt 5 /2 x - sqrt 5/2

    Does that seem right? The answer book says 15x - 2 (sqrt5)y - 5 = 0
    Yes it is correct. You just need some algebraic manipulation.
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  11. #11
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    Quote Originally Posted by Prove It View Post
    Yes it is correct. You just need some algebraic manipulation.
    The equation is correct, or the value of c?
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  12. #12
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    Both.
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  13. #13
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    Quote Originally Posted by Prove It View Post
    Both.
    Thank you! Sorry for the lengthy back and forth.
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