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Eq. of tangent plane at a point.

  1. Oct 29, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-29_21-37-8.png

    2. Relevant equations


    3. The attempt at a solution
    what is the difference between the form

    z - zo = fx(x-xo) + fy(y-yo)

    and:

    Fx(xo,yo,zo) + Fy(xo,yo,zo) + Fz(xo,yo,zo)=0
    ____________________________________________________________________________


    using the first eq :
    upload_2015-10-29_21-48-20.png

    while the latter gives:
    upload_2015-10-29_21-49-44.png

    Am I doing something wrong? shouldn't they give the same answer?
     
  2. jcsd
  3. Oct 30, 2015 #2

    Student100

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    You're doing something wrong, when you solve for ##z-z_0##, what happens to the term in front of ##(z-z_0)##, and what about your signs?
     
  4. Oct 30, 2015 #3
    are you talking about the first equation z - zo = fx(x-xo) + fy(y-yo)? I thought there aren't suppose to be any terms in front of (z−z0)

    actually, in order to get z - zo = fx(x-xo) + fy(y-yo), didn't we have to divide everything by z so we had something like A/C = Fx and B/C = Fy
     
  5. Oct 30, 2015 #4
    I think I found the problem.

    if I use gradient, the Fz has to equate to -1, right?
     
  6. Oct 30, 2015 #5
    is this correct?
    upload_2015-10-29_22-48-52.png
     

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  7. Oct 30, 2015 #6

    HallsofIvy

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    The difference between "z= f(x, y)" and "F(x, y, z)= 0" is that you are taking F(x, y, z)= f(x, y)- z. Taking the gradient [itex]\nabla F= <f_x, f_y, -1>[[/itex].

    It is correct that [itex]F(1, -1, 4\sqrt{2})= \frac{1}{4}+ \frac{1}{4}+ \frac{32}{64}= 1[/itex] but that does NOT say "z= 1". You are told that at this point [itex]z= 4\sqrt{2}[/itex]. The fact that [itex]F(1, -1, 4\sqrt{2})= 1[/itex] just verifies that the given point is on the given surface.

    The tangent plane is [itex]z- z_0= f_x(x_0, y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0)[/itex] only applies if you are given z= f(x, y) which is NOT the case here. You are given, rather, [itex]F(x, y, z)= x^2/4+ y^2/4+ z^2/64= 1[/itex]. In order to convert that to the "z= f(x,y)" form you need to solve for z: [itex]z^2/64= 1- x^2/4- y^2/4[/itex], [itex]z/8= \sqrt{1- x^2/4- y^2/4}[/itex], [itex]z= 8\sqrt{1- x^2/4- y^2/4}[/itex].

    Much simpler is to use the fact that if F(x, y, z)= 1 (or any constant) then [itex]\nabla F[/itex] is perpendicular to the surface at any point. Since for any point (x, y, z) on the tangent plane to the surface at [itex](x_0, y_0, z_0)[/itex], the vector [itex](x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}[/itex] is in the tangent plane, we must have [itex]\nabla F\cdot [(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}]= 0[/itex] which is the same as [itex]F_x(x_0, y_0, z_0)(x- x_0)+ F_y(x_0, y_0, z_0)(y- y_0)+ F_z(x_0, y_0, z_0)(z- z_0)= 0[/itex] as the equation of the tangent plane.
     
  8. Oct 30, 2015 #7
    upload_2015-10-30_6-11-52.png

    looks correct?
     
  9. Oct 30, 2015 #8

    HallsofIvy

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    What are [itex]x_0[/itex], [itex]y_0[/itex] and [itex]z_0[/itex]?
     
  10. Oct 30, 2015 #9
    I used r⋅n = n ⋅ (1,-1,4sqrt(2))
    to find my previous answer.

    r being (x,y,z)
     
  11. Oct 30, 2015 #10

    HallsofIvy

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    So you do not know what [itex]x_0[/itex], [itex]y_0[/itex], and [itex]z_0[/itex] are? Then you need to talk to your teacher.
     
  12. Oct 30, 2015 #11
    they are the point given, (1,-1,4sqrt(2))
     
  13. Nov 1, 2015 #12

    HallsofIvy

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    Yes. I asked that in post #8 and you did not answer. Now, my question is "why do you not have (x- 1), (y+ 1), and [itex](z- 4\sqrt{2})[/itex] in your answer"?
     
  14. Nov 3, 2015 #13
    upload_2015-10-29_21-49-44-png.91054.png

    that was what I initially had
     
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