
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!?!?!

First, when finding the magnitude of a vector, such as $\displaystyle  r'(t) $, don't square the unit vectors $\displaystyle \vec i$, $\displaystyle \vec j$, or $\displaystyle \vec k$, square only the components.
Regarding your actual question, you don't need to solve for $\displaystyle t$. 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 $\displaystyle \dfrac{70t^6 \vec i + 77t^6 \vec j + 14t^6 \vec k}{\sqrt{4900t^{12} + 5929t^{12} + 196t^{12}}}$ is correct (except for the "i" you placed after the $\displaystyle 4900t^{12}$ in the denominator, which I removed).
Of course, your answer could be simplified. For example, the denominator is equal to $\displaystyle 105t^6$. Assuming $\displaystyle t$ is positive, then $\displaystyle T(t) = \dfrac{2}{3} \vec i + \dfrac{11}{15} \vec j + \dfrac{2}{15} \vec k$.