# Space curve tangent vector

1. Jun 15, 2010

### roam

1. The problem statement, all variables and given/known data

Here's a worked problem, I can't understand how they have evaluated T at the given point (in part c):

[PLAIN]http://img31.imageshack.us/img31/3725/97856984.gif [Broken]

3. The attempt at a solution

I just substituted $$(0,1, \pi/2)$$ into r'(s) but

$$\frac{1}{\sqrt{2}} cos \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \neq 0$$

$$\frac{1}{\sqrt{2}}. -sin \left(\frac{0}{\sqrt{2}}\right) = 0 \neq \frac{-1}{\sqrt{2}}$$

Why is it that I'm not getting the right answer? Is there something else I need to do here?

Last edited by a moderator: May 4, 2017
2. Jun 15, 2010

### lanedance

to susbtitute into r'(s) you need to find s at that point

3. Jun 17, 2010

### roam

Okay, but still it doesn't work:

Since $$s= \sqrt{2}$$ , so at point 0 for example s=0. Then

$$\frac{1}{\sqrt{2}} . -sin \left( \frac{0}{\sqrt{2}} \right)=0$$

You see, it should equal zero. But how did they get "$$-\frac{1}{\sqrt{2}}$$"??

4. Jun 17, 2010

### HallsofIvy

Staff Emeritus
"at $(0, 1, \pi/2)$" does NOT mean s= 0! It is referring to
$$r(t)= \begin{pmatrix}cos(t)\\ sin(t) \\ t\end{pmatrix}$$
so t= [/itex]\pi/2[/itex].

Last edited by a moderator: May 4, 2017
5. Jun 18, 2010

### roam

How did you get $$t=\frac{\pi}{2}$$ out of that? Because by substituting these values into r(t) I got

$$r(t)= \begin{pmatrix}cos(0)\\ sin(1) \\ \pi/2\end{pmatrix} = \begin{pmatrix}1\\ 0.84 \\ \pi/2\end{pmatrix}$$

And even if I set $$t=\frac{\pi}{2}$$ (therefore $$s= \frac{\pi}{\sqrt{2}}$$) in r'(s), I still don't end up with $$-1/\sqrt{2}$$ in the first row like they have! :(

6. Jun 18, 2010

$$-\sin\left(\frac{\pi/\sqrt{2}}{\sqrt{2}}\right) = -\sin(\pi/2) = -1$$.