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anonymous

  • 5 years ago

An equation of the tangent plane to the parametrized surface Φ(u,v) = (3u, −3u2 + 2v, −5v2) at the point (6, −8, −20) is (in the variables x, y, z). To normalize the answer, make sure your coefficient of x is 240. = 0.

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  1. amistre64
    • 5 years ago
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    oh im just simply salivating over this one lol

  2. anonymous
    • 5 years ago
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    please take your time, thanks

  3. amistre64
    • 5 years ago
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    we take the point; and apply the gradient to it right?

  4. amistre64
    • 5 years ago
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    but the initial form seems to be in a vector position equation...

  5. anonymous
    • 5 years ago
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    hmm i am trying to find something in my book

  6. amistre64
    • 5 years ago
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    (3u) i + (3u^2 +2v) j + (-5v^2) k

  7. amistre64
    • 5 years ago
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    the derivative of a vector position is the tangent line .... if i recall correctly

  8. amistre64
    • 5 years ago
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    tangent vector that is...

  9. anonymous
    • 5 years ago
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    i believe so

  10. amistre64
    • 5 years ago
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    3u = 6 3u^2 +2v = -8 -5v^2 = -20 should give us values for u and v maybe?

  11. anonymous
    • 5 years ago
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    it's worth a try

  12. amistre64
    • 5 years ago
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    u = 2 and v = 2, -2 3(2^2) + 2v = -8 12 + 2v = -8 2v = -20 v = -10 doesnt seem to fit the point in question..

  13. amistre64
    • 5 years ago
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    so the point is on the tangent plane itself, and not the original equation

  14. anonymous
    • 5 years ago
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    yes

  15. amistre64
    • 5 years ago
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    then lets find a tangent vector that matches this contraption lol

  16. amistre64
    • 5 years ago
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    Φ(u,v) = (3u, −3u^2 + 2v, −5v^2) Φ'(u,v) = (3, m', −10v) m = −3u^2 + 2v m' = -6u u' + 2 v' if i see it right

  17. amistre64
    • 5 years ago
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    can we assume u' and v' = 1?

  18. anonymous
    • 5 years ago
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    i don't know

  19. amistre64
    • 5 years ago
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    im thinking yes ..... and im usually right about this unless im wrong lol

  20. amistre64
    • 5 years ago
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    du/du = 1 and dv/dv = 1 is what im thinking...

  21. amistre64
    • 5 years ago
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    Φ'(u,v) = (3, (-6u +2), −10v) would be our tangent vector; now lets point that to our point and see the vector we get....

  22. amistre64
    • 5 years ago
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    <3,-6u+2,-10v> <a, b , c > ----------------- <6, -8, -20> right?

  23. anonymous
    • 5 years ago
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    you mean add them?

  24. amistre64
    • 5 years ago
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    thats my thought; we have a vector going from whereever the point on the curve and to point it to the point outside of it we have to add something to it to get to there..

  25. amistre64
    • 5 years ago
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    its in the same plane; just have to account for rotation...

  26. amistre64
    • 5 years ago
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  27. amistre64
    • 5 years ago
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    -10 means -20; i got no short term memory apparently lol

  28. anonymous
    • 5 years ago
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    the approach looks resonable

  29. amistre64
    • 5 years ago
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    with 2 vectors on the plane we can cross product them to get the nomal right?

  30. amistre64
    • 5 years ago
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    <3,-6u+2,-10v> -<6, -8, -20> ----------------- < a, b, c> is an equivalent statement right?

  31. anonymous
    • 5 years ago
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    the cross product will give the normal i think

  32. amistre64
    • 5 years ago
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    it will; but first we have to determine what the vectors should be :)

  33. amistre64
    • 5 years ago
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    perhaps we can dot product this to an unknown vector to get an orthogonal in the plane to establish a good plave equation?

  34. amistre64
    • 5 years ago
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    <3,-6u+2,-10v> < r , s, t> --------------- = 0

  35. amistre64
    • 5 years ago
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    watchmath will surely let me know when my stupidity shows tho right ? :)

  36. amistre64
    • 5 years ago
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    < 3, -6u+2, -10v > <-6, 8, 20 > --------------------- <-3, -6u+10, -10v+20> if my original thought is correct

  37. amistre64
    • 5 years ago
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    lets cross these and see what we get... ;)

  38. amistre64
    • 5 years ago
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    (-6u+2)(-10v+20) - (8)(-10v) -[(3.20) - (-6.-10v)] (3.8) - (-6(-6u+2))

  39. amistre64
    • 5 years ago
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    <(60uv -20v -120u +40) , (-60 - 60v) ,(-36u +36)> maybe ....

  40. amistre64
    • 5 years ago
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    if we had a surface instead of a vector function; that would be easier i think lol ... I gotta wonder if I interpreted that first part correctly

  41. amistre64
    • 5 years ago
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    <(60uv -20v -120u +40+80v) , (-60 - 60v) ,(-36u +36)> <(60uv -60v -120u +40) , (-60 - 60v) ,(-36u +36)>

  42. amistre64
    • 5 years ago
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    (60uv -60v -120u +40)(x-6) +(-60 - 60v)(y+8) +(-36u +36)(z+20) = 0 is what I come up with.....

  43. amistre64
    • 5 years ago
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    can we come up with a surface equation from the parametrics used? im sure we can....

  44. amistre64
    • 5 years ago
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    x = 3u y = −3u^2 + 2v z = −5v^2

  45. amistre64
    • 5 years ago
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    u = 3/x y = -27/x^2 + 2v y + 27/x^2 = 2v v = (y + 27)/ 2x^2 gonna have to research this one some more maybe lol

  46. amistre64
    • 5 years ago
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    iv often wondered if there can be a tangent plane to a curve..... it seems like trying;

  47. amistre64
    • 5 years ago
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    i spose if it can be done for a tangent line to a point; its possible for a plane on a curve :)

  48. amistre64
    • 5 years ago
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    ok..... the derivatives for df/du and df/dv give us the vectors for our plane at the point (x,y,z) f(u,v)=<3u, −3u^2 + 2v, −5v^2> df/du = <3, −6u, 0> and df/dv=<0, 2, −10v> ; we cross these to get a normal to the plane at f(u,v) then

  49. amistre64
    • 5 years ago
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    3 -6u 0 0 2 -10v (60uv - 2)i + (30v)j + (6)k <(60uv-2),30v,6> is our normal to the plane whenever we determiine a u and v to input

  50. amistre64
    • 5 years ago
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    to include the point (6,-8,-20) lets put together the plane equation (60uv-2)(x-6) + (30v)(y+8) + (6)(z+20) = 0

  51. amistre64
    • 5 years ago
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    in order for the coeff of x = 240, then 60uv -2 = 240 60uv = 242 uv = 242/60 if i did it right

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