Evaluate the directional derivative of $f$ for the points and directions specified Evaluate the directional derivative of $f$ for the points and directions specified:
(a) $f(x,y,z)=3x-5y+2z$ at $(2,2,1)$ in the direction of the outward normal to the sphere $x^2+y^2+z^2=9.$
(b) $f(x,y,z)=x^2-y^2$ at a general point of the surface $x^2+y^2+z^2=4$ in the direction of the outward normal at that point.
(c) $f(x,y,z)=x^2+y^2-z^2$ at $(3,4,5)$ along the curve of intersection of the two surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$
The answer for the problems are (a) -2/3 (b) $x^2-y^2$ (c) $0$.
For (a) and (b) I thought the direction of the outward normal was the gradient of the level surface, which would be $(2x,2y,2z)$ in both cases. So to get the directional derivative I would simply have to compute the dot product of grad$f$ and $(2x,2y,2z)$ at the corresponding points, however, the result does not match the answers. 
For (c), I'm stuck on finding the curve of intersection of the two surfaces.
I would greatly appreciate it if anyone could help me out.
 A: For (a) and (b), you are almost right, except that you need to multiply by the unit vector in that direction. So you need to the direction $(2x,2y,2z)$ at the point, and divide by its norm first.
For (c), the intersection can be found by setting the two equations to each other. Since $z^2=x^2+y^2$, plugging this into the first one gives you $z^2=25$. So the intersection happens at the two planes $z=\pm 5$. Now if you plug this into the two equations, you will get $x^2+y^2=25$. So the intersections are the two circles at $z=\pm 5$. Now you need to find the tangent direction by parametrizing it. Notice that $(3,4,5)$ is on the top one. Once you find the tangent vector, the same method as in (a),(b) should give you the answer.
A: (a)
The outward normal of the sphere given is $\left(\frac{x}{3},\frac{y}{3}, \frac{1}{3}\sqrt{9-x^2-y^2}\right)$. The same at the given point is $\overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right).$ The vector of the partial derivatives is $\nabla f =(3,-5,2).$ The directional derivative is then
$$\nabla f \cdot \overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right)\cdot (3,-5,2)=2-\frac{10}{3}+\frac{2}{3}=-\frac{2}{3}.$$
(b)
The outward normal of the sphere given is $\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)$.The vector of the partial derivatives is $\nabla f =(2x,-2y,0)$. So,
$$\nabla f \cdot \overline n=(2x,-2y,0)\cdot\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)=x^2-y^2.$$
(c)
We have two surfaces this time: $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$ These intersect above the circle $x^2+y^2=25$ at $z=\pm 5$. (The intersection line was given by setting $2x^2+2y^2-25=x^2+y^2$. Also, where $x^2+y^2=25$, there $z=\pm 5$.) Now, the vectors pointing to the intersection lines are of the form $(x,\sqrt{25-x^2},\pm 5)$; at $(3,4,5)$ it gives $(3,4,5)$. The vector of the partial derivatives: $(2x,2y,-2z)$, which at the given point is $(6,8,-10)$. So,
$$\nabla f \cdot \overline n= (3,4,5)\cdot(6,8,-10)=0.$$
(c')
I've seen that the other people answering considered the tangent vector to the intersection line. That approach gave the same result at the given point. If "along" means "tangent to" then my answer was only accidentally correct. 
A: For (a) note that the surface is a sphere of radius $3$ and the normal to a sphere at a point is the normalized vector from the origin to the point, in your case $\frac{1}{3}(x,y,z)^T$ so, being $(3,-5,2)^T$ the gradient of the function, the directional derivative is:$(3,-5,2)(\frac{2}{3},\frac{2}{3},\frac{1}{3})^T=-\frac{2}{3}$.
(b) is similar, being the surface a sphere of radius 2. For the gradient you have $(2x,-2y,0)^T$ and for the normal you can use $\frac{1}{2}(x,y,\sqrt{4-x^2-y^2})^T$ and you find the result.
For (c), intersecting the two surface you find $x^2+y^2=z^2=25$ , so the point $(3,4,5)$ stay on a circunference of radius $5$ at $z=5$, and the tangent vector at a point is $(y,-x,0)^T$ ( not normalized), since the gradient of the function is $(2x,2y,-2z)^T$, the directional derivative is $(2x,2y,-2z)(y,-x,0)^T=0$.
