# Thread: unit tangent vector

1. ## unit tangent vector

im trying to find the unit tangent vector of an elipse of the form
$\displaystyle x = acos(t) y= asin(t)$ at $\displaystyle t = \pi$

Do i firstly need to get the coordinates in non-parametric form i.e.
use the equation of an elipse?

2. For ellipse x = a*cosφ, and y = a*sinφ is not possible. I think it should be
y = b*sinφ.
To find the tangent vector, write the equation of the ellipse in the standard form and proceed.

3. Well, if you consider a circle to be a special case of an ellipse?

If you meant $\displaystyle x= a cos(t)$, $\displaystyle y= b sin(t)$ then you can do the following: [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}= \frac{b cos(t)}{-a sin(t)}[tex]. At $\displaystyle t= \pi$, [tex]$\displaystyle \frac{dy}{dt}= cos(t)= cos(\pi)= -1$ and $\displaystyle \frac{dx}{dt}= -a sin(t)= -a sin(\pi)= 0$ so the slope does not exist. The tangent line at $\displaystyle t= \pi$ is the vertical line x= -a. The unit tangent vector is $\displaystyle -\vec{j}$.