Unit Tangent Vector

**sderosa518** · March 17th 2011, 04:30 PM

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!?!?!

**NOX Andrew** · March 17th 2011, 07:40 PM

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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