Find a line tangent to the parametric curve

[flyingpig]

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

 

[LCKurtz]

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

Isn't it in your text?

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

 

[flyingpig]

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]

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].<br /> <br /> 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 <b>1</b> parameter is <b>1</b> dimensional). The parametric equations for the tangent line to <br /> $$\vec{r}(t)= f(t)\vec{i}+ g(t)\vec{j}+ h(t)\vec{k}$$<br /> at $(x_0, y_0, z_0)$ are given by <br /> $$(x_0+ tf'(t_0)\vec{i}+ (y_0+ tg'(t_0))\vec{j}+ (z_0+ th'(t_0))\vec{k}$$<br /> 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.[/itex]

 

[LCKurtz]

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.

 

[flyingpig]

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].[/itex]

[itex]I am not sure what you are getting at here, the first and second looks the same to me<br /> <br /> <blockquote data-attributes="" data-quote="LCKurtz" data-source="post: 3278260" cite="https://www.physicsforums.com/goto/post?id=3278260" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> LCKurtz said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> And to flying pig, just in case you are puzzled by the difference between these two equations of the tangent line:<br /> <br /> $$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)t$$ </div> </div> </blockquote><br /> $$\vec R(t)= \vec R(t_0)+ \vec R'(t_0)(t- t_0)$$<br /> <br /> They both give the same line but with a different parameterization. The first tangent line is at <b>R</b>(t₀) when t = 0 and the second is at <b>R</b>(t₀) when t = t₀.[/QUOTE]<br /> <br /> If t = 0 for the first one, shouldn't we get $$\vec{R(t)} = \vec{R(t_0)} - \vec{R(t_0)}t_0$$?<br /> <br /> Same argument for the other one.<br /> <br /> Anyways taking the form $$r(t) - r(t_0) = r'(t_0)(t - t_0)$$ <br /> <br /> I get q(t) (I will use another letter)<br /> <br /> $$q(t) = <\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3\pi}{4}> + \frac{\pi}{4}<\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}, -3> (t - (- ln\frac{\pi}{4}))$$<br /> <br /> This is the one I thought it was right originally but according to HallsoIvy, it should be<br /> <br /> $$q(t) = <\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3\pi}{4}> + \frac{\pi t}{4}<\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}, -3>$$<br /> <br /> Where I completely removed t₀[/itex]