Find a unit vector that is parallel to $\nabla f(\cos\theta,\sin\theta)$ Suppose $f(x,y)$ is differentiable for all $(x,y),f(x,y)=17$ on the unit circle $x^2+y^2=1$, and $\nabla f$ is never zero on the unit circle. For any real number $\theta$, I have to find a unit vector that is parallel to $\nabla f(\cos\theta,\sin\theta)$.

My thoughts: When I say this question I feel that it relates to level curves in some way. However not sure how it helps in solving this problem.
If $f$ is a given function of $\theta$ only, then we have the following:

$\nabla f(\cos \theta,\sin \theta)=\frac{df}{d\theta}\nabla \theta=\frac{df}{d\theta} \hat \theta$

Where $\hat \theta$ is a unit vector that points in the direction of increasing $\theta$.  Thus a unit vector parrallel to $\nabla f$ is indeed $\hat \theta$.  In Cartesian coordinates this unit vector is given by $(-\hat y x+\hat y x)/(x^2+y^2)=-\hat x y+\hat y x$ since we are on the unit circle.



However, it seems that $f$ is actually a general function of both $r$ and $\theta$, with $f(1,\theta)=17$.  Then, we note that the directional derivative $\frac{df}{ds}$, where $s$ is the arc length parameter along the unit circle, is zero since $f=17$ is constant on the unit circle.  We can write

$$\frac{df}{ds}=\nabla f\cdot \frac{d\vec r}{ds}=0.$$

But, $\frac{d\vec r}{ds}=\hat \theta$.  Thus, $\nabla f$ is perpendicular to $\hat \theta$ and parallel to $\hat r=\hat x \cos \theta+\hat y \sin \theta$.
i do not think there is sufficient info to find gradient of f for f(x,y) where x and y are on unit circle.

we only know f(x,y) = 17 on unit circle, but there is no info for f(x,y) infinitely close to the unit circle.