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Math Help - Unit Tangent Vector

  1. #1
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    Oct 2009
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    Unit Tangent Vector

    Hello,

    I was just wondering if my solution to the following question looks good, if so, do I set
    r'(t) = 1 and solve for t? Can anyone help???

    Finding the Unit Tangent Vector to the curve give:
    r(t) = (6+10t^7)i+ (6+11t^7)j +(6+2t^7)k


    r ' (t)= 70t^6i +77^6j +14t^6k
    Unit Tagent Vector:

    T(t)= r'(t) / ll r' (t) ll

    = 70t^6i +77^6j +14t^6k / sqrt{ (70t^6i)^2 +(77^6j)^2 + (14t^6k)^2 }

    =70t^6i +77^6j +14t^6k / sqrt { 4900t^12 i + 5929t^12 + 196t^12) }

    * Now I am pretty sure I solve for t if my arithmetic is correct!?!?!
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  2. #2
    Member
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    Dec 2009
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    First, when finding the magnitude of a vector, such as || r'(t) ||, don't square the unit vectors \vec i, \vec j, or \vec k, square only the components.

    Regarding your actual question, you don't need to solve for t. Like the position vector, the unit tangent vector varies from time to time, so naturally it is a function of time. Therefore, your answer of \dfrac{70t^6 \vec i + 77t^6 \vec j + 14t^6 \vec k}{\sqrt{4900t^{12} + 5929t^{12} + 196t^{12}}} is correct (except for the "i" you placed after the 4900t^{12} in the denominator, which I removed).

    Of course, your answer could be simplified. For example, the denominator is equal to 105|t^6|. Assuming t is positive, then T(t) = \dfrac{2}{3} \vec i + \dfrac{11}{15} \vec j + \dfrac{2}{15} \vec k.
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