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For the function $f(x,y,z) = x^2+y^2+z^2$ I have find the directional derivative in the point $(3,4,5)$ in the direction of the intersection of 2 surfaces. The 2 surfaces are:

$$2x^2+2y^2-z^2 = 25$$

and

$$x^2+y^2=z^2$$

First I got the Partial derivatives of the function $f$ and inserted the values $(3,4,5)$ for $x$,$y$ and $z$. This gave me $(6,8,10)$. Then I tried calculating the intersection line of the 2 surfaces and I got this:

$$x^2 + y^2 - 25 = 0$$

Then I calculated the partial derivatives of that intersection line, inserted the values $3$,$4$,$5$ and normalized the result which gave me $(6/10, 8/10,0)$.

And then I multiplied $(6,8,10)$ with $(6/10, 8/10,0)$ But this didn't give me the correct result, what am I doing wrong here?

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    The two paraboloid surfaces intersect in two circles with z-coordinates $\pm5$ and $x,y$ coordinates satisfying $x^2+y^2=25$. – 2017-01-14
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    Since the point $(3,4,5)$ lies in the intersection of the two surfaces it is not clear what is meant by "the directional derivative in the point $(3,4,5)$ in the direction of the intersection of the 2 surfaces." Could you clarify the question? – 2017-01-14
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    It looks like you computed a normal vector to (the projection of) the intersection instead of a tangent to it. By the way, the equation $x^2+y^2-25=0$ is the equation of a *cylinder*, not that of a curve in $\mathbb R^3$, but in this case you get the same tangents and normals. Also, you’d usually need to specify an orientation in order to compute directional derivatives, but for this specific problem orientation isn’t going to make any difference. – 2017-01-15

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