# Find a line tangent to the parametric curve

## Homework Statement

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

## Homework Equations

Okay, in just normal Cartesian Coord, we have $$y - y_0 = f'(x)(x - x_0)$$

So I suspect in parametric form we have something like $$r - r_0 = r'(t)(t - t_0)$$

I am a bit unsure about the t - t0 part.

## The Attempt at a Solution

$$r'(t) = <-e^{-t}\sin(e^{-t}), -e^{-t}\cos(e^{-t}), -3e^{-t}>$$

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

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

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

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

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

## Answers and Replies

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LCKurtz
Homework Helper
Gold Member
So I suspect in parametric form we have something like $$r - r_0 = r'(t)(t - t_0)$$
Isn't it in your text?

$$\vec R(t) = \vec R(t_0) + t\vec R'(t_0)$$

Isn't it in your text?

$$\vec R(t) = \vec R(t_0) + t\vec R'(t_0)$$
No, mine had a minus sign

HallsofIvy
Homework Helper
If you are taking multiple variable calculus, you should be able to recognize that
$$r- r_0= r'(t_0)(t- t_0)$$
(NOT "r'(t)") is the same as
$$r= r_0+ r'(t_0)(t- t_0)$$
though NOT necessarily the same as
$$r= r_0+ r'(t_0)t$$
unless $t_0= 0[/tex]. You should have learned by now that a curve or line (one-dimensional figure) cannot be written in a single equation. A single equation reduces from 3 dimensions to 3- 1 dimensions- a surface. To write a curve you have to have either two equations, reducing to 3- 2= 1 dimension, or three parametric equations (so you have 4 variables with 3 equations: 4- 3= 1- or just think that depending on 1 parameter is 1 dimensional). The parametric equations for the tangent line to $$\vec{r}(t)= f(t)\vec{i}+ g(t)\vec{j}+ h(t)\vec{k}$$ at [itex](x_0, y_0, z_0)$ are given by
$$(x_0+ tf'(t_0)\vec{i}+ (y_0+ tg'(t_0))\vec{j}+ (z_0+ th'(t_0))\vec{k}$$
where $t_0$ is, of course, the value of the parameter t that gives $(x_0, y_0, z_0)$ in the original vector function.

Last edited by a moderator:
LCKurtz
Homework Helper
Gold Member
And to flying pig, just in case you are puzzled by the difference between these two equations of the tangent line:

$$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)t$$

$$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)(t- t_0)$$

They both give the same line but with a different parameterization. The first tangent line is at R(t0) when t = 0 and the second is at R(t0) when t = t0.

Let me correct myself. I think the tangent line should have the form of

$$r(t) - r(t_0) = r'(t_0)(t - t_0)$$

If you are taking multiple variable calculus, you should be able to recognize that
$$r- r_0= r'(t_0)(t- t_0)$$
(NOT "r'(t)") is the same as
$$r= r_0+ r'(t_0)(t- t_0)$$
though NOT necessarily the same as
$$r= r_0+ r'(t_0)t$$
unless [itex]t_0= 0[/tex].

I am not sure what you are getting at here, the first and second looks the same to me

And to flying pig, just in case you are puzzled by the difference between these two equations of the tangent line:

$$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)t$$
$$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)(t- t_0)$$

They both give the same line but with a different parameterization. The first tangent line is at R(t0) when t = 0 and the second is at R(t0) when t = t0.[/QUOTE]

If t = 0 for the first one, shouldn't we get $$\vec{R(t)} = \vec{R(t_0)} - \vec{R(t_0)}t_0$$?

Same argument for the other one.

Anyways taking the form $$r(t) - r(t_0) = r'(t_0)(t - t_0)$$

I get q(t) (I will use another letter)

$$q(t) = <\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3\pi}{4}> + \frac{\pi}{4}<\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}, -3> (t - (- ln\frac{\pi}{4}))$$

This is the one I thought it was right originally but according to HallsoIvy, it should be

$$q(t) = <\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3\pi}{4}> + \frac{\pi t}{4}<\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}, -3>$$

Where I completely removed t0