Finding the Tangent Vector of a Space Curve at a Given Point

[roam]

Homework Statement

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

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?

 

[lanedance]

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

 

[roam]

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

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}}"??

 

[HallsofIvy]

Homework Statement

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

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?

"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].

 

[roam]

"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].

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! :(

 

[Deadleg]

You should get that,
-\sin\left(\frac{\pi/\sqrt{2}}{\sqrt{2}}\right) = -\sin(\pi/2) = -1.

t = pi/2 comes from r(t) = (0, 1, pi/2) = (cos t, sin t, t) and looking at the last entry.