Directional derivative along the intersection of two surfaces How can i find the intersection curve between these two surfaces
$$
\left\{
\begin{array}{cc}
 2x^2 + 2y^2 − z^2 &= 50\\
 x^2 + y^2 -z^2 &= 0
\end{array}
\right.
$$
I need it to find the directional derivative of $f(x, y, z) = x^2 + y^2 − z^2$, with this point $(3, 4, 5)$ and along the intersection curve mentioned above. I know how to get the gradient, I just don't know how to aproach finding that intersection.
 A: You can do this without finding an intersection. Just find normal vectors to both surfaces (i.e. compute gradients) and then compute theirs cross product to obtain curve tangent vector.
For the first surface the gradient is
$$
(4x, 4y, -2z).
$$
The second one gives us
$$
(2x, 2x, -2z).
$$
So, we can take $(2x, 2y, -z)$ and $(x, y, -z)$ as normal vectors and compute their cross product to obtain tangent vector
$$
\boldsymbol{v} = \begin{vmatrix}
\boldsymbol{i} & \boldsymbol{j} &\boldsymbol{k} \\
2x & 2y & -z \\
x & y & -z
\end{vmatrix} = -yz \boldsymbol{i} +xz\boldsymbol{j}.
$$
Let's substitute $(3,4,5)$
$$
\boldsymbol{v} = -4\cdot 5\boldsymbol{i} + 3\cdot{5}\boldsymbol{j}.
$$
It's easy to see that normalizing this vector we get
$$
\boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}.
$$
Now you need to find the directional derivative along $\boldsymbol{v}_1$ or $-\boldsymbol{v}_1$ (both will be tangent vectors indicating two different directions on the curve).
The other way for solving this problem (as suggested in the comments) was to eliminate $z$ from the equations and get the curve (WLOG we take positive $z$ considering the case $z = -5\sqrt{2}$ by analogy)
$$
x^2 + y^2 = 50, \; z = 5\sqrt{2}.
$$
So, we see that the curve is a circle $x^2 + y^2 = 50$ located at the "height" $z=5\sqrt{2}$ (this actually explains why $k$ coordinate is equal to zero in tangent vector). The tangent vector to the circle
$$
x^2 + y^2 = 50
$$
is equal to $(-y, x)$ and after substitution we get again
$$
\boldsymbol{v} = (-4, 5) \implies \boldsymbol{v}_1 = \frac{\boldsymbol{v}}{|\boldsymbol{v}|} = -\frac{4}{5} \boldsymbol{i} + \frac{3}{5}\boldsymbol{j}.
$$
All you need now is to compute the directional derivative (i.e. to find dot product of gradient of the function and obtained unit vector).
