Unit tangent vector

Christian.

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Sep 1, 2018
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I have worked on this problem for quite some time and cannot figure out the solution. Please help me and thanks.
Screen Shot 2018-10-07 at 9.19.58 PM.jpg

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I, also, cannot make out your attempted solution but the problem is to find the tangent vector to the curve given by \(\displaystyle \vec{r}(t)= e^{13t}cos(t)\vec{i}+ e^{13t}sin(t)\vec{j}+ e^{13t}\vec{k}\) at \(\displaystyle t= \frac{\pi}{2}\).

Differentiate "component wise" using the product rule:
\(\displaystyle \vec{r'}(t)= [13e^{13t}cos(t)- e^{13t}sin(t)]\vec{k}+ [13e^{13t}sin(t)+ e^{13t}cos(t)]\vec{j}+ 13e^{13t}\vec{k}\).

At \(\displaystyle x= \frac{\pi}{2}\), cos(x)= 0 and sin(x)= 1 so that is

\(\displaystyle \vec{r'}(\pi/2)= -e^{13\pi/2}\vec{i}+ 13e^{13\pi/2}\vec{j}+ 13e^{13\pi/2}\vec{k}\)

The length of that vector is \(\displaystyle \sqrt{e^{13\pi}+ 169e^{13\pi}+ 169e^{13\pi}}= e^{13\pi/2}\sqrt{339}\)

Divide the tangent vector by its length to get the unit tangent vector.
 
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Found the solution.

I originally took the derivative incorrectly and used the wrong formula. Everything is good now.Screen Shot 2018-10-08 at 1.31.11 PM.jpg
 
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