# Find the unit tangent, the principal normal and the curvature of $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $(\frac{-3}{\sqrt{2}},\frac{2}{\sqrt{2}})\in C$

I need help with this problem:

Sketch the semi ellipse $$C$$ defined by $$\frac{x^2}{9}+\frac{y^2}{4^2}=1 \quad\quad\quad\quad x\leq0$$Assuming a "clockwise" orientation of $$C$$ (from $$(0,-2)$$ to $$(0,2)$$), find the unit tangent, the principal normal and the curvature to C at $$\left(\frac{-3}{\sqrt{2}},\frac{2}{\sqrt{2}}\right)\in C$$.

I sketched the semi ellipse and I ended up with this:

Then I parametrized the ellipse like this: $$x=3\cos t$$ and $$y=2\sin t$$, but I don't know if it is clockwise. I sketched the paramtrization on desmos and it seems that it isn't clockwise. If I write $$x=3\sin t$$ and $$y=2\cos t$$ it seems to work, it also works with $$x=3\cos t$$ and $$y=-2\sin t$$, which one should I use?

If I pick the first one, the I must find the norm of the derivative, $$f'(t)=(3\cos t, -2\sin t)$$, then $$\Vert f'(t)\Vert=\sqrt{(3\cos t)^2+(-2\sin t)^2}=\sqrt{9\cos^2 t+ 4\sin^2 t}$$. Then I need tried to calculate the path-length parametrization of $$f$$, $$\lambda(t)=\int_{0}^t \Vert f'(t)\Vert dt= \int_{0}^t (\sqrt{9\cos^2 t+ 4\sin^2 t}) \ dt$$, but I'm struggling with integrating that. How can I integrate that and, which of the two parametrizations is correct? Please help me.