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anonymous

  • 5 years ago

Find the parametric equations for the normal line to the surface z=e^(2x^2+8y^2) at the point (0,0,1)?

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  1. amistre64
    • 5 years ago
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    normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?

  2. anonymous
    • 5 years ago
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    I think its the unit normal vector

  3. anonymous
    • 5 years ago
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    You first need to find the equation for the tangent plane at that point. Then you can use the equation of the plane to find the normal to the plane.

  4. anonymous
    • 5 years ago
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    if it's parametric equations, then it's probably the line, not the vector, i.e., the line inthe direction of the vector

  5. amistre64
    • 5 years ago
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    ...do we find the normal plane by partial derivatives?

  6. anonymous
    • 5 years ago
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    Yes.

  7. amistre64
    • 5 years ago
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    i was reading up on some of the vector stuff last night as my eyes were glossing over :)

  8. amistre64
    • 5 years ago
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    I would soo love to be able to answer this question as easily and stupididly as I do the algebra stuff lol

  9. anonymous
    • 5 years ago
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    if i remember correctly, you don't need to actually find the plane, just the gradient

  10. anonymous
    • 5 years ago
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    I think its the unit normal vector

  11. anonymous
    • 5 years ago
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    That's true. \(<\frac{\delta z}{\delta x}|_{x=x_0,y=y_0}, \frac{\delta z}{\delta y}|_{x=x_0,y=y_0}, -1> \) is the normal vector.

  12. anonymous
    • 5 years ago
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    how would I find the equation of the tangent plane at that point?

  13. amistre64
    • 5 years ago
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    you a good wizard or a bad wizard? I can never get inline equations....

  14. anonymous
    • 5 years ago
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    it's all points perpendicular to that vector

  15. anonymous
    • 5 years ago
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    dz/dx(x-x0) + dz/dy(y-y0) + -1(z-z0) = 0, where (x0,y0,z0) is your initial point

  16. anonymous
    • 5 years ago
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    ok, so the unit tangent vector is the derivative over the magnitude of the derivative?

  17. anonymous
    • 5 years ago
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    The tangent vector is the gradient. The unit tangent is just the gradient divided by its magnitude. Just like all unit vectors are the vector divided by its magnitude.

  18. anonymous
    • 5 years ago
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    so the normal vector is just the gradient of the tangent vector?

  19. anonymous
    • 5 years ago
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    No. The normal is the gradient of the surface.

  20. anonymous
    • 5 years ago
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    ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

  21. anonymous
    • 5 years ago
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    so using the normal vector, how do i convert it to the parametric equation? sorry i just have trouble grasping this stuff

  22. anonymous
    • 5 years ago
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    use slope intercept form, y = mx + b, or x = mt + b

  23. anonymous
    • 5 years ago
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    where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint

  24. anonymous
    • 5 years ago
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    and do the same with y and b

  25. anonymous
    • 5 years ago
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    so if the gradient of x is \[4xe ^{{2x^2}+{8y^2}}\] i just plug in the point (0,0,1) and solve it for x0?

  26. anonymous
    • 5 years ago
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    correct

  27. anonymous
    • 5 years ago
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    that will give you the slope of the x coordinate for the parametric equation

  28. anonymous
    • 5 years ago
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    that will give me x0 or will it give me a?

  29. anonymous
    • 5 years ago
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    what is a?

  30. anonymous
    • 5 years ago
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    x=x0 +at

  31. anonymous
    • 5 years ago
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    the parametric equations will take the form of x = (dz/dx@x0)*t + x0

  32. anonymous
    • 5 years ago
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    yes, that will give you a

  33. anonymous
    • 5 years ago
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    ok thanks, and x0 is just equal to 0 correct?

  34. anonymous
    • 5 years ago
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    since the point is (0,0,1)

  35. anonymous
    • 5 years ago
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    correct

  36. anonymous
    • 5 years ago
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    thank you very much

  37. anonymous
    • 5 years ago
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    no problem

  38. anonymous
    • 5 years ago
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    ok, so when i find the gradient where do i plug the coordinates of the point given into the parametric equation?

  39. anonymous
    • 5 years ago
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    x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate

  40. anonymous
    • 5 years ago
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    so be would be 0, for (0,0,1)

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