# Thread: Another Calc III Problem

1. ## Another Calc III Problem

Here's the problem:
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point:

$\displaystyle x = e^-t * cos(t)$
$\displaystyle y = e^-t * sin(t)$
$\displaystyle z = e^-t$
At the point: (1,0,1)

(It's supposed to be e to the negative t, I just can't get the math software to work properly).

I know the answer is supposed to be:
$\displaystyle x = 1-t$
$\displaystyle y = t$
$\displaystyle z = 1-t$

but I have no idea how to get there! Help!

2. Originally Posted by saxyliz
Here's the problem:
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point:

$\displaystyle x = e^-t * cos(t)$
$\displaystyle y = e^-t * sin(t)$
$\displaystyle z = e^-t$
At the point: (1,0,1)

(It's supposed to be e to the negative t, I just can't get the math software to work properly).

I know the answer is supposed to be:
$\displaystyle x = 1-t$
$\displaystyle y = t$
$\displaystyle z = 1-t$

but I have no idea how to get there! Help!
let our curve be given by $\displaystyle r(t) = \left< x(t), y(t), z(t) \right> = \left<e^{-t} \cos t, e^{-t} \sin t , e^{-t} \right>$.

note that the line we want is the line passing through the point (1,0,1) with direction vector given by $\displaystyle r'(t)$.

now we are concerned with when $\displaystyle t = 0$ (since in that case, x = 1, y = 0 and z = 1, hence we are passing through the point (1,0,1))

So, you want the line passing through (1,0,1) that has direction vector $\displaystyle r'(0)$

i trust you can take it from here

3. Thanks! I understand it now!