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mathstina Group Title

Show that every normal line to the sphere x^2+y^2+z^2=r^2,where r is the radius,passes through the centre of the sphere.

  • one year ago
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  1. mathstina Group Title
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    hi

    • one year ago
  2. TuringTest Group Title
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    label a point on the surface of the sphere with the position vector\[\vec P=\langle a,b,c\rangle\]what is the gradient at that point?

    • one year ago
  3. mathstina Group Title
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    ok.

    • one year ago
  4. TuringTest Group Title
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    @mathstina what is the gradient vector for any point on the sphere?

    • one year ago
  5. mathstina Group Title
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    what is point ? is it (0,0,0)?

    • one year ago
  6. mathstina Group Title
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    what do i sub for r value?

    • one year ago
  7. TuringTest Group Title
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    we aren't going top specify the point yet, because we want to prove the statement for the general case the function for the surface of the sphere is\[f(x,y,z)=x^2+y^2+z^2=r^2\]what is\[\nabla f\]?

    • one year ago
  8. mathstina Group Title
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    2x i + 2y j + 2z k = r^2

    • one year ago
  9. TuringTest Group Title
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    r is a constant for a sphere, so this thingy should be =0

    • one year ago
  10. TuringTest Group Title
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    grad f=2x i + 2y j + 2z k = 0 still with me?

    • one year ago
  11. mathstina Group Title
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    yes

    • one year ago
  12. TuringTest Group Title
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    okay, now let the point be some x=a, y=b, and z=c that satisfies the function f and so is on the sphere what is the position vector at that point?

    • one year ago
  13. mathstina Group Title
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    (a,b,c)?

    • one year ago
  14. TuringTest Group Title
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    yes, and what is the gradient at that point ?

    • one year ago
  15. mathstina Group Title
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    is it find the parametric eqn?

    • one year ago
  16. TuringTest Group Title
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    ? no you just need the two vectors to solve this\[P(a,b,c)=\langle a,b,c\rangle\]\[\nabla f(a,b,c)=?\]

    • one year ago
  17. TuringTest Group Title
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    a picture may help us visualize what we're doing|dw:1352129016463:dw|

    • one year ago
  18. mathstina Group Title
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    2a i+2b j+2c k =0

    • one year ago
  19. TuringTest Group Title
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    yes, but be consistent in your notation. Write *all* vectors, including the position vector, the same way now compare the position vector and the gradient vector. What do you see?

    • one year ago
  20. mathstina Group Title
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    multiplied by 2

    • one year ago
  21. TuringTest Group Title
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    yes, so what does that say bout the relationship between the two vectors?

    • one year ago
  22. mathstina Group Title
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    sry, im nt sure

    • one year ago
  23. TuringTest Group Title
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    the scalar multiple of any vector points in the *same direction* as the original the gradient vector is a scalar multiple of the position vector, hence what can we say about their relative orientations?

    • one year ago
  24. mathstina Group Title
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    positive gradient

    • one year ago
  25. mathstina Group Title
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    both pointing in the same dirn; parallel

    • one year ago
  26. TuringTest Group Title
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    yes, now look at what this means visually...

    • one year ago
  27. TuringTest Group Title
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    |dw:1352129640330:dw|

    • one year ago
  28. mathstina Group Title
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    the points are on the level surface

    • one year ago
  29. TuringTest Group Title
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    |dw:1352129723349:dw|since the position vector and grad f are parallel for any point on the sphere, they are collinear. Since gradf is normal to the sphere, so must be the line hence, we can draw a line that contains both. Since the position vector passes through the origin, we know that any such line will also pass through the origin. Hence all normal lines to the surface. Make sense?

    • one year ago
  30. TuringTest Group Title
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    hence all normal lines to the surface pass through the origin*

    • one year ago
  31. mathstina Group Title
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    yes.

    • one year ago
  32. TuringTest Group Title
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    good, then we're done :)

    • one year ago
  33. mathstina Group Title
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    ihw do i answer the qn? mathematically written

    • one year ago
  34. TuringTest Group Title
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    let P be a point on the sphere\[\vec P=\langle a,b,c\rangle\]then\[\nabla f=2\vec P\implies \nabla f\parallel\vec P\]hence they are collinear. since the line containing P passes through the origin, and any line that contains P must also contain grad f, this means that so does any line normal to the sphere passes though the origin. QED I don't think any more math symbols are necessary, sometimes words are important in proofs.

    • one year ago
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