# For a curve, find the unit tangent vector and parametric equation of the line tangent to the curve at the given point

For the curve $r(t) = ti+tj+\sqrt{4-t^2}k$ find the unit tangent vector $T ( t )$ and parametric equation of the line tangent to the curve at the point $P(1,1,\sqrt{3}).$

I am not sure what exactly to do here. Any help or suggestion would be greatly appreciated. I think I know how to find the unit tangent vector but I don't know how to find the parametric equation.

• You have to differentiate every component of the curve and then calculate the norm of it. Dividing the derivative vector by its norm will get you the unit tanget vector. The tangent line is exactly $P+r'(t)t$. Recall that to define a line you must have a point and a direction for it. Feb 21, 2017 at 22:05
• @math.h Instead of $P+r'(t)t$, I would say $P+\lambda r'(t)$ for any $\lambda$. Feb 21, 2017 at 22:14
• yeah, better. Using the same variable for different things isn't very clear, right? Feb 21, 2017 at 22:16
• is this correct for just the x, x = 1+sqrt(7/3)t Feb 21, 2017 at 22:58

Assuming you are right about knowing how to find $T(t)$( I believe in you, you can do it!) Recall that a line can be parameterized as $l(t)=v_0+tv$ where $v_0$ is a position point(i.e. the value of your function at the point you want to find the tangent) and $v$ the direction vector(i.e. the "slope" of your line, so the tangent to your curve at the desired point.)

• UHHH I am still confused...so I found T(t) to be <1/sqrt(9-2t^2/4-t^2) , 1/sqrt(9-2t^2/4-t^2), -t/sqrt(9-2t^2)>....now do I plug in point P into this tangent? If I do that, I get (1/sqrt(7/3), 1/sqrt(7/3), -1) Feb 21, 2017 at 22:47
• Would you mind writing the parametric for x or give me some more explanation? Feb 21, 2017 at 22:48
• That is not normal so something is wrong since that vector is not normal, but in principal yes.
– user416426
Feb 21, 2017 at 22:50
• Oh wait you just forgot to decide the last coordinate by $\sqrt{7}$, yeah you got it.
– user416426
Feb 21, 2017 at 22:51
• So that is your " slope" or direction vector.
– user416426
Feb 21, 2017 at 22:51