for a fixed t,the tangent line to a regular curve "alpha" at a point alpha(t) is the staight line u-alpha(t)+alpha'(t),where we delet the point of application of alpha'(t).find the tangent line to the helix alpha(t)=(2cost,2sint,t)at poin 0

Question

for a fixed t,the tangent line to a regular curve "alpha" at a point alpha(t) is the staight line u->alpha(t)+alpha'(t),where we delet the point of application of alpha'(t).find the tangent line to the helix alpha(t)=(2cost,2sint,t)at poin 0

zehanz · Answer

Do you mean at the point with t=0?

anonymous · Answer

yes

zehanz · Answer

So it is the point (2, 0, 0).
alpha'(t)=(-2sint, 2cost, 1). 
What is alpha'(0)?

anonymous · Answer

(
o,2,1) ?

zehanz · Answer

BTW, welcome to OpenStudy!!

If you set t=0 in alpha'(t), you get (0, 2, 1) as tangent vector.

You could use vectors to describe the tangent line in (2, 0, 0):$$\left(\begin{matrix}x \ y \ z\end{matrix}\right)=\left(\begin{matrix}2 \ 0 \0\end{matrix}\right)+\lambda \left(\begin{matrix}0 \ 2\ 1\end{matrix}\right)$$

zehanz · Answer

Or, you could use a system of equations: 
Because $x=2,~ y=2+2\lambda, ~z=\lambda $, you could use this:$$x=2 , ~z=\frac{1}{2}y$$

zehanz · Answer

Sorry, of course, $y=2\lambda$

anonymous · Answer

thankyou zehanz so much
and what is lemda?

anonymous · Answer

can u help me to find a unique curve?

zehanz · Answer

It is lambda, a Greek letter. It is used as a parameter:
For every value of lambda, you get another point of the tangent line.