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Math Help - can some one please help me finding the tangent and the norm please

  1. #1
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    can some one please help me finding the tangent and the norm please

    i can find the tangent and the norm with some easy when the equation to start is something like
    3x^2+8x+4 when x=1
    but i have a new equation
    3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
    i can differentiate to get
    6x+2x2y+2y^2=6y^2
    but dont get how i use y=mx+c
    if x=2 and y=3 this means
    3=m2+c
    all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you
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  2. #2
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    Quote Originally Posted by alpharebel10 View Post
    i can find the tangent and the norm with some easy when the equation to start is something like
    3x^2+8x+4 when x=1
    but i have a new equation
    3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
    i can differentiate to get
    6x+2x2y+2y^2=6y^2
    but dont get how i use y=mx+c
    if x=2 and y=3 this means
    3=m2+c
    all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you
    You need to be careful when differentiating implicitly. For your function

    3x^2+2xy^2+6=2y^3

    you have 6x + 2y^2 + 4xy y' = 6y^2 y' which solving for y' gives

     <br />
y' =\frac{3x + y^2}{3y^2-2xy}<br />
and at (2,3) this becomes y' = 1, which is the slope of the tangent. Now the equation of the tangent is y - 3 = 1(x-2) or y = x + 1. For the normal, use y - 3 = -1(x-2)
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  3. #3
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    Quote Originally Posted by alpharebel10 View Post
    i can find the tangent and the norm with some easy when the equation to start is something like
    That's "normal", not "norm", which has quite a different meaning in mathmathics. Also it is not clear whether you want the tangent and normal lines or vectors. I will asume you mean lines.

    3x^2+8x+4 when x=1
    but i have a new equation
    3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
    i can differentiate to get
    6x+2x2y+2y^2=6y^2
    No, you don't. Since y is a function of x and you are differentiating with respect to x, you have to use the "chain rule". The derivative of 2xy^3, with respect to x, is, by the product rule 2y^2+ 2x(y^2)'= 2y^2+ 4xyy' and the derivative of 2y^3, with respect to x is 6y y'.
    The derivative of 3x^2+ 2xy^2+ 6= 2y^3 is 6x+ 2y^2+ 4xyy'= 6y^2y'. Solve that for y' (when x= 2 and y= 3) and use that for your m.

    but dont get how i use y=mx+c
    if x=2 and y=3 this means
    3=m2+c
    all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you
    Follow Math Help Forum on Facebook and Google+

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