Gradient vector of parametric curve I have ellipse $$(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$$
Gradient is $$(\frac{2x}{a^2}, \frac{2y}{b^2})$$
How I can obtain this vector from parametrization of my curve? Let I know only
$$(x, y) = (a \cdot cos \phi, b \cdot sin \phi)$$
I want have vector-function by φ, and if I choosed value, for example, φ=0, and set in vector, I will have same result if I set φ in $$(x, y) = (a \cdot cos \phi, b \cdot sin \phi)$$ and resulting (x, y) in $$(\frac{2x}{a^2}, \frac{2y}{b^2})$$
I can compute function $$\frac{dy}{dx}=\frac{\frac{dy}{d\phi}}{\frac{dx}{d\phi}}$$ But its give only angle between tangent and abscissa, I dont't find information for norm of gradient and can't just rotate tangent on π/2 Thanks for help!
 A: $\newcommand{\Reals}{\mathbf{R}}$There are three "pieces of data" in your question:


*

*A plane curve $C$:
$$
C = \left\{(x, y) \in \Reals^{2} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\right\}.
$$

*A real-valued function $f$ on the plane, one of whose levels is $C$:
$$
f(x, y) = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}};\qquad
C = f^{-1}(\{1\}).
$$

*A parametrization of $C$:
$$
\gamma(\phi) = (a\cos\phi, b\sin\phi);\qquad
\gamma\bigl([0, 2\pi]\bigr) = C.
$$
You're asking how to retrieve the gradient $\nabla f$ from $\gamma$. This is not possible in general.
To see why, note that if $g:\Reals \to \Reals$ is increasing and differentiable, then the function $F = g \circ f$ has the same level sets as $f$, but not the same gradient:
$$
C = F^{-1}\bigl(\{g(1)\}\bigr);\qquad
\nabla F = g'(f) \cdot \nabla f.
$$
The direction of the gradient of $f$ can be recovered from the parametrization just as you say, but unless you have additional information, the norm cannot be. The best you can do is conclude that the gradient of $f$ is proportional to the velocity $\gamma'(\phi)$ rotated one-quarter turn.
A: Using the parametric equations $(x,y) = (a\cos \phi, b\sin \phi)$, the tangent vector at parameter value $\phi$ is $(-a \sin \phi, b \cos \phi)$. Rotating this by 90 degrees, we get the normal vector $(b \cos \phi, a \sin \phi)$. But this normal vector is just $(\tfrac{b}{a}x, \tfrac{a}{b}y)$.
