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Math Help - find an equation to r(t)

  1. #1
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    Question find an equation to r(t)

    Given : r(t) = t^2 i + t j +3t^3 k

    Find an equation for the tangent line to r(t) at (1,1,3).

    Thank you very much.
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  2. #2
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    Hello, Jenny20!

    Given: .r(t) .= .ti + tj +3tk

    Find an equation for the tangent line to r(t) at (1,1,3).

    The tangent vector is the derivative of r(t):
    . . r'(t) .= .2ti + j + 9tk

    At t = 1: .r'(1) .= .2i + j + 9k .= .< 2,1,9 >

    The parametric equations of the tangent line are:
    . . x .= .1 + 2p
    . . y .= .1 + p . . for some parameter p
    . . z .= .3 + 9p


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  3. #3
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    Hi Soroban,

    Thank you very much.

    What about this part? How should I solve it?

    Given : rsub1(t) =t^2i + tj + 3t^3k
    and rsub2(t) = (t-1)i + (1/4)t^2j + (5-t)k.

    Find cos theta , where theta is an acute angle between the tangent lines to rsub1 and rsub2 at (1,1,3).
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  4. #4
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    Hello again, Jenny!



    Given: .r_1(t) .= .ti + tj + 3tk
    . .and: .r_2(u) .= .(u-1)i + uj + (5 - u)k

    Find cos θ, where θ is an acute angle between the tangent lines to r_1 and r_2 at (1,1,3).

    I changed the parameter in r_2 to u to avoid confusion.

    When t = 1: .r_1(1) = (1,1,3)

    When u = 2: .r_2(2) = (1,1,3)


    Tangent to r_1: .r'_1(t) .= .2ti + j + 9tk
    . . When t = 1: .r'_1(1) .= .2i + j + 9k

    Tangent to r_2: .r'_2(t) .= .i + uj - k
    . . When u = 2: .r'_2(2) .= .i + j - k


    The angle θ between two vectors u and v is given by:

    . . . . . . . . . . | uv |
    . . cos θ . = . ---------
    . . . . . . . . . . |u| |v|


    The numerator is: .|uv| .= .|<2,1,9>.<1,1,-1>| .= .|2 + 1 -9| .= .6

    The denominator is: .√(2 + 1 + 9) .= .√86
    . . and: .√[1 + 1 + (-1)] .= .√3

    . . . . . . . . . . . . . . . . . . . . 6
    And we have: . cos θ . = . ------- . = . 0.373543684
    . . . . . . . . . . . . . . . . . . .√258

    Therefore: .θ . .68

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