Question

Find the unit tangent vector of r(t)=

Answer 1

I Need help really bad!

Answer 2

First, let's find the derivative.
\[r(t) = i\ e^t \cos t + j\ e^t \sin t + k\ e^t\\r\prime(t) = -i\ e^t\ \sin t + j\ e^t\ \cos t + k\ e^t\]
Now, the unit tangent vector is defined as the derivative but normalized, so...
\[\hat{r\prime}(t) = \frac{r\prime(t)}{||r\prime(t)||}\\\ \ \ \ \ \ = \frac{ -i\ e^t\ \sin t + j\ e^t\ \cos t + k\ e^t}{\sqrt{e^{2t}\ \sin^2{t} + e^{2t}\ \cos^2{t} + e^{2t}}}\\\ \ \ \ \ \ = \frac{i\sin t + j\cos{t} + k}{\sqrt{2}}\]

Answer 3

how did you find the length of r'(t) that is where i was getting lost.

Answer 4

\[\ \ \ \ ||xi + yj + zk||\\ = \sqrt{x^2 + y^2 + z^2}\]

Answer 5

for the derivative do you have to do the product rule?

Answer 6

@Drums_2014 good point... \[ r'(t) = i\ e^t(\cos{t} - \sin{t}) + j\ e^t(\sin{t} + \cos{t}) + k\ e^t\\\ \ \ \ \ \ = i\ e^t\cos{t} - i\ e^t\sin{t} + j\ e^t\sin{t} + j\ e^t\cos{t} + k\ e^t\\\ \ \ \ \ \ =e^t\]

\[\hat{r'}(t) = \frac{i\ e^t(\cos{t} - \sin{t}) + j\ e^t(\sin{t} + \cos{t}) + k\ e^t}{\sqrt{e^{2t}((\cos{t} - \sin{t})^2 + (\sin{t} + \cos{t})^2 + 1))}}\\ \ \ \ \ \ \ = \frac{i\ e^t(\cos{t} - \sin{t}) + j\ e^t(\sin{t} + \cos{t}) + k\ e^t}{e^t(\cos^2{t} - 2\sin{t}\cos{t} + \sin^2{t} + \sin^2{t} + 2\sin{t}\cos{t} + \cos^2{t} + 1)}\\ \ \ \ \ \ \ = \frac{i(\cos{t} - \sin{t}) + j(\sin{t} + \cos{t}) + k}{2(\cos^2{t} + \sin^2{t}) + 1}\\ \ \ \ \ \ \ = \frac{i(\cos{t} - \sin{t}) + j(\sin{t} + \cos{t}) + k}{3}\]

Answer 7

Ignore those two lines under r'(t)

Answer 8

So then i got that answer and went onto answering the normal vector question and then how do you simplify

Answer 9

\[\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{3}\sqrt{2}}{3} = \frac{\sqrt{6}}{3}\]

Answer 10

can you simplfy this