- #1

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## Homework Statement

Let [tex]r(t) = <\cos(e^{-t}),\sin(e^{-t}),3e^{-t}>[/tex], find the equation of the line tangent to r(t) at the point [tex]\left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{3\pi}{4} \right)[/tex]

## Homework Equations

Okay, in just normal Cartesian Coord, we have [tex]y - y_0 = f'(x)(x - x_0)[/tex]

So I suspect in parametric form we have something like [tex] r - r_0 = r'(t)(t - t_0)[/tex]

I am a bit unsure about the t - t

_{0}part.

## The Attempt at a Solution

[tex]r'(t) = <-e^{-t}\sin(e^{-t}), -e^{-t}\cos(e^{-t}), -3e^{-t}>[/tex]

So I must find a t corresponding to the point [tex]\left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{3\pi}{4} \right)[/tex].

I set [tex]r(t) = \left ( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{3\pi}{4} \right)[/tex]

Solving (well I only need to solve for one of them), I get t = [tex]-ln{\frac{\pi}{4}}[/tex]

Then I evaluated [tex]r'(-ln{\frac{\pi}{4}}) = \frac{-\pi}{4}<\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 3>[/tex]

Now the problem is, I don't know how to set up my line. I know I am in Calculus right now and I should know it by now, but I don't so that's why I am here.

Thank you