# Defining tangent line using new parameter to a curve

• Apr 1st 2011, 06:10 AM
olski1
Defining tangent line using new parameter to a curve
Hi,
I have been asked a series of questions on the following curve

r(t) = (1+cost,2sint,0) where t varies from 0 to 2pi

part a) required me to find the orientation of the tangent vector point (pi/3)

I found the tangent vector r'(t)= (-sint,2cost,0) and substituted in (pi/3)

r'(pi/3)= -sqrt(3)/2 i + j

but part b) requires me to use a second parametric variable to define the tangent line to the point (pi/3).

I knwo this isnt to hard , but i dont quite understand what the question is asking. What do they mean define? have i not already defined it at this point? btw the answer given is q(s)=(1.5-sqrt(3)s/2, sqrt(3) + s ,0)

Any help understanding this would be appreciated
• Apr 1st 2011, 07:40 AM
HallsofIvy
You have parametric equations for the given figure (it happens to be an ellipse) but now they want you to write parametric equations for the tangent line to the figure. Yes, the tangent vector to the ellipse at $t= \pi/3$ is $-\sqrt{3}{2}\vec{i}+ \vec{j}$ and one point on the tangent line is, of course, the point on the ellipse, $(3/2, \sqrt{3}, 0)$.

In three dimensions, parametric equations for a line through $(x_0, y_0, z_0)$ with "direction vector" $A\vec{i}+ B\vec{j}+ C\vec{k}$ are $x= As+ x_0$, $y= Bs+ y_0$, $z= Cs+ z_0$ where "s" is the parameter.