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anonymous

  • 5 years ago

A curve given parametrically by (x,y,z)=(−1+3t, 2−2t^2, −t+2t^3). There is a unique point P on the curve with the property that the tangent line at P passes through the point (8,−8,11). The point P has coordinates... ? Hmm not quite sure where to start.

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  1. anonymous
    • 5 years ago
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    Something not right, write word for word, verbatim the problem.

  2. amistre64
    • 5 years ago
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    a surface would see m more reasonable than a curve; but maybe i havent drawn the curve out anough to see it

  3. amistre64
    • 5 years ago
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    the tangent to the postion vector is just the derivatives of the components..

  4. amistre64
    • 5 years ago
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    <3,-4t,6t> is te tangent vector at any given point

  5. anonymous
    • 5 years ago
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    That is the correct wording of the problem. I don't think it would matter if its a curve or a surface, assuming it does actually intersect...? Hmm does that makes sense 3 is never going to pass through 8, am i right?

  6. amistre64
    • 5 years ago
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    the line equation for that vector is: x = Px +3 y = Py -4t z = Pz +6t right?

  7. anonymous
    • 5 years ago
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    OK, just clumsy wording, it is saying that at some point the tangent line passes through line (8, -8, 11) and that is not the coordinates of point P, I think.

  8. amistre64
    • 5 years ago
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    the line equation for the tangent vector i meant to say

  9. anonymous
    • 5 years ago
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    @ changuanas I understood it as passing through the point (8,-8,11) ? The tangent that is.

  10. amistre64
    • 5 years ago
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    when the vector from P(x,y,z) matches the same vector from the given point; they are in line right?

  11. anonymous
    • 5 years ago
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    I believe so.

  12. amistre64
    • 5 years ago
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    or... simply equate the vector between points maybe

  13. amistre64
    • 5 years ago
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    x = Px +8 y = Py -8 z = Pz +11 = the vector in line with P(x,y,z) and (8,-8,11) right?

  14. amistre64
    • 5 years ago
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    my tangent vector is wrong, i didnt see the -t in the z spot

  15. anonymous
    • 5 years ago
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    lol Hold on I'm solving something, think I got it!

  16. amistre64
    • 5 years ago
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    the vector(x,y,z) from P(-1+3t, 2-2t, -2+2t^3) to (8,-8,11) is equal to; -1+3t +Vx = 8 2-2t + Vy = -8 -2+2t^3 + Vz = 11 right?

  17. amistre64
    • 5 years ago
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    V =: x = 7 -3t y = -10 +2t z = 11 +t -2t^3 right?

  18. anonymous
    • 5 years ago
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    Umm yes, granted I'm not sure how that's helping.. ?

  19. amistre64
    • 5 years ago
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    im thinking on it lol

  20. amistre64
    • 5 years ago
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    this gives us the vector from P to the given(g) when this vector matches the tangent vector, we might bea ble to relate them for an answer

  21. anonymous
    • 5 years ago
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    They are describing the point P (x,y,z) where the tangent vector r (evaluated at that point), and the given curve r' is the tangent line evaluated at point P.

  22. amistre64
    • 5 years ago
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    the tangent vector(t) = <3,-4t, 6t-1> right?

  23. anonymous
    • 5 years ago
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    Nope. :), well yes but that isnt the answer yet. :P

  24. amistre64
    • 5 years ago
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    so lets see if we can equate these vectors and get a result :) it may or maynot work; but im game lol

  25. amistre64
    • 5 years ago
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    7-3t = 3 -3t = -4 t = 3/4 maybe? ..................

  26. amistre64
    • 5 years ago
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    well, i see a possibility that <t> has a scalar; so let apply that as well; say 's'

  27. amistre64
    • 5 years ago
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    7 -3t = 3s -10+2t = -4st 11+t - 2t^3 = 6st - s got no idea where im going yet :)

  28. anonymous
    • 5 years ago
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    :P Well I'm not really sure what I did... but I got the answer which was subbing in t as 1. Forgot what work I did to show that.

  29. amistre64
    • 5 years ago
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    s = (7-3t)/3 -4(7-3/t)(t) = 10 +2t -28 +12 = 10 +2t -28 +12 -10 = 2t -26 = 2t t = -13; so..... s = 7-3(-13)/3 = 7+39/3 = 46/3 = 15' 1/3 gong for broke here....

  30. anonymous
    • 5 years ago
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    err I g2g to class but I'll explain what I did when I get back... in 3 hours ):

  31. anonymous
    • 5 years ago
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    Good luck, finals week. Good work, Amistre, i'm a bit lost, figure it out later.

  32. amistre64
    • 5 years ago
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    11+(-13) -2(-13)^3 ?=? 6(46/3)(-13) - 46/3 4392 ?=? 92 - 46/3 ..... taht dint work for me :)

  33. amistre64
    • 5 years ago
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    maybe determine the tangent plane the the given is on? :)

  34. anonymous
    • 5 years ago
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    There are several relations: first the curve is given in parametric equations. That is usually formed from a point and a vector. So we can work backwards.

  35. amistre64
    • 5 years ago
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    right, the position vector is determined by the vector function; and the tangent is then given by the derivative of the components right?

  36. amistre64
    • 5 years ago
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    if it were the F(x,y,z) = 0 then the derivatives would give us a gradient/normal to the plane...

  37. amistre64
    • 5 years ago
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    the line from a one point to another in R^3 is the vector from one point to the other right?

  38. anonymous
    • 5 years ago
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    Another relation is the tangent line containing (8, -8, 11) is a vector let's call it r' This vector is the tangent to the unique point P. Then there is the tangent vector which is the integral of that tangent line. That is the tangent vector which shoots from the origin to point P.

  39. amistre64
    • 5 years ago
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    right; the backwards of what i stated is what that looks like ... if i read it right

  40. amistre64
    • 5 years ago
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    or the backwards of what im thinking :)

  41. anonymous
    • 5 years ago
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    A place to start i guess is build a vector from point (8, -8, 11) and P (xyz) That represents the tangent line.

  42. amistre64
    • 5 years ago
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    that vector should be: <8+3, -8-4t, 11 + 6t-1> right?

  43. anonymous
    • 5 years ago
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    Yeah, I'm still thinking how to proceed.

  44. amistre64
    • 5 years ago
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    i also did that to the point itself; so those 2 lines should at least be parallel... is my thought

  45. amistre64
    • 5 years ago
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    that should also equal the vector from teh given to the point right? or at least some scalar of it?

  46. anonymous
    • 5 years ago
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    It is just matter of putting this seeming complicated problem into simple Cal I: find the point of tangency.

  47. anonymous
    • 5 years ago
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    froggy click this link http://www.twiddla.com/537628

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