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Parametric Equations of Tangent Line

  1. Mar 7, 2016 #1

    njo

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    1. The problem statement, all variables and given/known data
    z = 2x^2 + 5y^2 +2
    C is cut by the plane x = 2
    Find parametric eqns of the line tangent to C @ P(2, 1, 15)

    2. Relevant equations
    z = 5y^2 + 10
    dz/dx = 10y
    dz/dx (1) = 10

    3. The attempt at a solution
    z = 10y + 15
    y = t + 1

    if the slope is 10/1 then delta z = 10 and delta y = 1. Therefore those values are multiples in front of t for the z and y parametric eqns respectively. Is this correct?

    Thanks in advance.
     
    Last edited: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2

    Mark44

    Staff: Mentor

    Why does z appear on both sides? Is the equation actually w = 2x^2 + 5y^2 + z? Or is the last z a typo?
     
  4. Mar 7, 2016 #3

    njo

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    z = 2x^2 + 5y^2 +2

    That's the equation. Typo.
     
  5. Mar 7, 2016 #4

    Mark44

    Staff: Mentor

    That would be dz/dy, not dz/dx.

    The tangent line is a line in ##\mathbb{R}^3##. Parametric equations would be of the form x = f(t), y = g(t), z = h(t). Since the tangent line is in the plane z = 2, x will be just 2.
     
  6. Mar 7, 2016 #5

    njo

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    dz/dy = 10y

    z = 10y + 15
    y = t + 1
    x = 2

    Is this correct?
     
  7. Mar 7, 2016 #6

    Mark44

    Staff: Mentor

    The equations for y and z should be in terms of a parameter. Typically, t is used.
     
  8. Mar 7, 2016 #7

    njo

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    Oops, I know better than that.

    z = 10t + 15
    y = t + 1
    x = 2
     
  9. Mar 7, 2016 #8

    Mark44

    Staff: Mentor

    That's not what I get. The trace of the surface ##z = 2x^2 + 5y^2 + 2## in the plane x = 2 is ##z = 5y^2 + 10##, a parabola. The slope of the tangent line to this parabola at the point (2, 1, 15) is 10, which you have, but I get a different equation for the tangent line. Use the point-slope form of the equation of the line, with m = 10, and the point (1, 15) -- (y, z) coordinates.
     
  10. Mar 8, 2016 #9
    You want to find the tangent line as the intersection of plane ## P_1 |\ x = 2 ## and plane ## P_2 |\ a x + by + cz + d = 0 ##,
    where ## (a,b,c) = \text{grad }F(2,1,15) ##, ##F(x,y,z) = 2x^2 + 5y^2 -z + 2 ##, and ##d## determined by the fact that ##(2,1,15)## belongs to ##P_2##. In cartesian coordinates, I get that the tangent line is ## 10y - z + 5 = 0 ##.
     
  11. Mar 8, 2016 #10

    njo

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    So I find the gradient of w = 2x^2 + 5y^2 - z + 2 which would be <4x, 10y, -1>. I substitute P into the gradient? and that's how I get the coefficient in front of the parameter?
     
  12. Mar 8, 2016 #11

    Mark44

    Staff: Mentor

    That's what I get as well.
     
  13. Mar 9, 2016 #12

    njo

    User Avatar

    The tangent line is z = 10y+5 therefore the parametric equations of this line are:
    y = t +1
    z = 10t + 15
     
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