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Math Help - equation of the tangent line to the curve

  1. #1
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    equation of the tangent line to the curve

    x^3+3xy+y^3=1 at the point (1,0)

    i solved to get y =sqrt(x^3+3x-1)

    then i think you take the derivative of that and set it equal to 0

    im not sure how i procede
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    Quote Originally Posted by dat1611 View Post
    x^3+3xy+y^3=1 at the point (1,0)

    i solved to get y =sqrt(x^3+3x-1)

    this would be the case if the function was x^3+3x-y^2=1

    then i think you take the derivative of that and set it equal to 0

    you would do that to find a local maximum or minimum.

    im not sure how i procede
    You need "implicit" differentiation.

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

    Differentiate both sides, realising that the derivative of a constant is zero.

    3x^2+3\frac{d}{dx}(xy)+\frac{d}{dx}y^3=0

    xy is a product and you need the chain rule for the "y-cubed" term.

    Then, when you've got that, use x=1 and y=0.
    The derivative at that point is the slope of the tangent.

    Then the equation of the tangent is

    y-0=m(x-1)

    y=\frac{dy}{dx}(x-1)
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    i dont understand how to do the derivative
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  4. #4
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    Quote Originally Posted by dat1611 View Post
    i dont understand how to do the derivative
    If you do not understand the derivative, then why are you trying to do a problem that involves derivatives?
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    Quote Originally Posted by dat1611 View Post
    x^3+3xy+y^3=1 at the point (1,0)

    i solved to get y =sqrt(x^3+3x-1)

    then i think you take the derivative of that and set it equal to 0

    im not sure how i procede
    You should reivew the chain rule to get \frac{dy}{dx} here.
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    how do you take d/dx of y^3
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    Quote Originally Posted by dat1611 View Post
    how do you take d/dx of y^3
    is 3y^2\frac{dy}{dx}

    Follow this example, you will find it very helpful.

    Implicit Differentiation
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    i got -x^2/y^2 but if i plug in x and y it undefined
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    Quote Originally Posted by dat1611 View Post
    i got -x^2/y^2 but if i plug in x and y it undefined
    your derivative is incorrect. it should be ...

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


    you need to review the method for taking implicit derivatives ...

    Implicit Differentiation
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  10. #10
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    Quote Originally Posted by dat1611 View Post
    i got -x^2/y^2 but if i plug in x and y it undefined
    You forgot to differentiate the "product term" 3xy using the product rule of differentiation.
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  11. #11
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    so i plug in the values for x any y and got slope being negative 1 so is the answer just y-0=-1(x-1)
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  12. #12
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    Quote Originally Posted by dat1611 View Post
    so i plug in the values for x any y and got slope being negative 1 so is the answer just y-0=-1(x-1)
    Yep,

    here's a way to "think of" the chain rule.

    \displaystyle\huge\frac{d}{dy}y^3=3y^2

    \displaystyle\huge\frac{d}{dx}y^3=\frac{dy}{dy}\fr  ac{d}{dx}y^3=\frac{dy}{dx}\frac{d}{dy}y^3
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