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Equation of tangent to curve

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data
    "Find the equation of the plane tangent to the surface [itex] (x^2-y^2)(x^2+y^2)=15 [/itex] at the point [itex] (2,1) [/itex]"

    If only it really were a plane and a surface, I could do that. I have a formula for that. Unfortunately, this is a curve and I'm looking for tangent line.


    2. Relevant equations

    In three dimensions, the formula for the equation of the tangent plane to the surface z=f(x,y) at the point [itex] P(x_0,y_0,z_0) [/itex] is [itex] z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) [/itex]

    where [itex] f_a [/itex] is the partial derivative of f wrt a.


    3. The attempt at a solution

    Well, pretending it's in three variables, I can do
    [tex] f_x=4x^3 [/tex]
    [tex] f_y=-4y^3 [/tex]

    [tex] z-z_0=f_x(2,1)(x-2)+f_y(2,1)(y-1) [/tex]
    [tex] z-z_0=32(x-2)-4(y-1) [/tex]

    So, how do I repair this situation/make the formula work in two dimensions/try something else?
     
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  3. Jan 18, 2012 #2

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    I'm afraid you equation is not of the form z=f(x,y).
    If you want to compare it with a 3D case, it's of the form g(x,y,z)=0.

    So your formula won't work.
    Do you have other methods available?
     
  4. Jan 18, 2012 #3
    I can not thank you enough for all your help! I really need to start paying you ;)

    I can use anything through vector calculus. Something about the gradient? I think I might be supposed to take the gradient and then evaluate at the point. Does that sound reasonable?
     
  5. Jan 18, 2012 #4

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    :)

    Yes, the gradient sounds reasonable.
    But what would you get?
     
  6. Jan 18, 2012 #5
    well, the gradient is [tex] 4x^3i-4y^3j [/tex] isn't it?
     
  7. Jan 18, 2012 #6

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    Yes, it is.
     
  8. Jan 18, 2012 #7
    Well how do I evaluate that at a point? Plug in the x and y values?
    If so, I get 32i-4j
     
  9. Jan 18, 2012 #8

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    Yes.
    This is a vector.
    Can you say anything about its direction and how it relates to the curve?
     
  10. Jan 18, 2012 #9
    Is the gradient normal to the curve perhaps? I can't find it in my calc book.
     
  11. Jan 18, 2012 #10

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    Yes, it is.
     
  12. Jan 18, 2012 #11
    Okay so now I just need a line normal to the normal.

    How do I make a vector that is perpendicular to 32i-4j ?
     
  13. Jan 18, 2012 #12

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    You don't need to.

    Suppose you have a line given by h(x,y)=ax+by-c=0.
    What would its gradient be?


    EDIT: to answer your question, a perpendicular vector would have a dot product that comes out as zero.
     
  14. Jan 18, 2012 #13
    It's gradient would be ai+bj.
    But, if a gradient is normal to the curve, don't I need to find the line perpendicular to the gradient? oh.. wait, for the equation of a line, do I use a point and a normal vector? I think I do... *looking up*

    [EDIT] no no, I see you use a parallel vector. So, what do I do now?

    [EDIT] now I'm reading about normal vectors and equations of lines and checking my notes....
     
  15. Jan 18, 2012 #14

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    A line is also a curve.

    If you have 2 curves that both have a normal vector that point in the same direction, wouldn't they have to be tangential to each other?
     
  16. Jan 18, 2012 #15
    I'm confused. I have my curve and I have my normal. I don't have another curve.
     
  17. Jan 18, 2012 #16

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    The parametric representation of a curve (or a surface), has a (partial) derivative that yields a vector tangential to the curve (or surface).

    The equation of a curve (or surface) has a gradient that is normal to the curve (or surface).
     
  18. Jan 18, 2012 #17

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    The line given by by h(x,y)=ax+by-c=0 is also a curve.
    Its gradient is normal to the line.
     
  19. Jan 18, 2012 #18
    So are you saying I need to parametrize instead and take the partial derivative?
     
  20. Jan 18, 2012 #19
    I'm going to call it quits for tonight, I still have to iron clothes, and rest! I'll pick this back up tomorrow. Thank you so much for your time tonight!
     
  21. Jan 18, 2012 #20

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    Okay. Thanks for telling me.
    See you later.
     
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