find vectors T, N, and B at the given point

**khuezy** · November 5th 2008, 03:28 PM

Hi,
find vectors T, N, and B at the given point
r(t)= <cost, sint, lncost>, (1,0,0)

I found T(t) = <-sint/tant, cost,tant, -1>
Is this correct?
Finding N would be messy 'quotient rule' or could I multiply T(t) by tant to remove it from the denominator?
I don't know what to do with the point (1,0,0)

Thanks.

**khuezy** · November 5th 2008, 08:00 PM

please

**Chris L T521** · November 5th 2008, 09:07 PM

Originally Posted by khuezy

Hi,
find vectors T, N, and B at the given point
r(t)= <cost, sint, lncost>, (1,0,0)

I found T(t) = <-sint/tant, cost,tant, -1>
Is this correct?
Finding N would be messy 'quotient rule' or could I multiply T(t) by tant to remove it from the denominator?
I don't know what to do with the point (1,0,0)

Thanks.

Recall the definitions of $\bold T(t),~\bold N(t),\textit{ and }\bold B(t)$ :

$\bold T(t)=\frac{\bold r'(t)}{||\bold r'(t)||}$

We need to find $\bold r'(t)\textit{ and }||\bold r'(t)||$

$\bold r'(t)=\left<-\sin t,~\cos t,~-\tan t\right>$

$||\bold r'(t)||=\sqrt{\sin^2t+\cos^2t+\tan^2t}=\sqrt{1+\ta n^2t}=\sqrt{\sec^2t}=\sec t=\frac{1}{\cos t}$

Therefore, $\bold T(t)=\left<-\tfrac{1}{2}\sin(2t),~\cos^2t,~-\sin t\right>$

What do we do with the point?

After a little examination, this point corresponds to the value of $t=0$ [To see why this is, compare the x,y,z coordinates to the corresponding counterpart defined in the vector $\bold r(t)$ ; i.e. when is $\cos t=1$ , $\sin t=0$ and $\ln (\cos t)=0$ ]

Thus, $\bold T(0)=\left<0,~1,~0\right>$

-------------------------------------------------------------------------

$\bold N(t)=\frac{\bold T'(t)}{||\bold T'(t)||}$

We need to find $\bold T'(t)\textit{ and }||\bold T'(t)||$

$\bold T'(t)=\left<-\cos (2t),~-\sin(2t),~-\cos t\right>$

$||\bold T'(t)||=\sqrt{\sin^2(2t)+\cos^2(2t)+\cos^2t}=\sqrt {1+\cos^2t}$

Therefore, $\bold N(t)=\left<-\frac{\cos(2t)}{\sqrt{1+\cos^2t}},~-\frac{\sin(2t)}{\sqrt{1+\cos^2t}},~-\frac{\cos t}{\sqrt{1+\cos^2t}}\right>$

Thus, $\bold N(0)=\left<-\frac{1}{\sqrt{2}},~0,~-\frac{1}{\sqrt{2}}\right>$

-------------------------------------------------------------------------

Since I found $\bold T(t)$ and $\bold N(t)$ for you, I would like you to find $\bold B(t)$ , given that

$\bold B(t)=\bold T(t)\times \bold N(t)$

-------------------------------------------------------------------------

Does this make sense?

--Chris

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