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I am trying to find the directional derivative of the following problem

$F(x,y,z) = 4x^2+ 3y−3xz+ 2z^2$

at the point $(2,1,2)$ in the direction $i−k$;

I worked out the derivatives of $F(x,y,z)$ as

$f_x = 8x -3z$

$f_y = 3$

$f_z = -3x+4z$

But I don't know to do next; can someone explain how to find the directional derivative here please?

Thank you

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  • $\begingroup$ evaluate partial derivatives at the given point, and dot product with direction vector $\endgroup$ Apr 28 '20 at 16:00
  • $\begingroup$ @J.W.Tanner The inner product of $\nabla F$ should be taken with the UNIT direction vector. $\endgroup$
    – Mark Viola
    Apr 28 '20 at 16:03
  • $\begingroup$ @MarkViola: I hedged; the Wikipedia page suggests that some authors don't require a UNIT direction vector $\endgroup$ Apr 28 '20 at 16:11
  • $\begingroup$ @J.W.Tanner It's more likely that the OP is using the definition in which the directional vector has unit magnitude. This vector space is, after all, Euclidean. $\endgroup$
    – Mark Viola
    Apr 28 '20 at 16:38
  • $\begingroup$ Perhaps you might review the course material that preceded this exercise. I’m pretty sure that you’ll find the next steps in there somewhere. $\endgroup$
    – amd
    Apr 28 '20 at 17:37
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Let the gradient $\nabla F$ of $F$ be $$\nabla F = (f_x, f_y, f_z).$$ Then the gradient evaluated at $(2,1,2)$, $$\nabla F(2, 1, 2) = (8 \cdot 2 - 3 \cdot 2, 3, -3 \cdot 2 + 4 \cdot 2) = (10, 3, 2)$$ can be dotted with the normalized direction $\frac{1}{\sqrt{2}}(1, 0, -1) = \frac{1}{\sqrt{2}}(i - k)$ to arrive at the directional derivative you are searching for

$$ \nabla F(2, 1, 2) \bullet \frac{1}{\sqrt{2}}(1, 0,-1) = (10, 3, 2)\bullet \frac{1}{\sqrt{2}} (1, 0, -1) = \frac{1}{\sqrt{2}}(10 - 2) = \frac{8}{\sqrt{2}}. $$

Using the method described by J.W. Tanner, we notice that the same conclusion is found by computing it simply as a weighting of the directional derivatives in directions $x, y$ and $z$, $$ \begin{align} \frac{1}{\sqrt{2}}\cdot f_x(2, 1, 2) + 0 \cdot f_y(2, 1, 2) - \frac{1}{\sqrt{2}}\cdot f_z(2,1,2) &= \\ \frac{1}{\sqrt{2}}\cdot 10 + \frac{0}{\sqrt{2}}\cdot 3 - \frac{1}{\sqrt{2}} \cdot 2 &= \\\frac{10 - 2}{\sqrt{2}} &= \\\frac{8}{\sqrt{2}} \end{align} $$

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  • $\begingroup$ I suppose that the directional vector one take the dot product against should be a unit vector? $\Vert (1,0,-1) \Vert = \sqrt 2$ $\endgroup$ Apr 28 '20 at 16:16
  • $\begingroup$ Sure, I altered my answer to reflect this. $\endgroup$
    – Mikal
    Apr 28 '20 at 16:32
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Firstly, you need to figure out the gradient of $F$ at the point, $$ \nabla F = ( \frac {\partial F} {\partial x}, \frac {\partial F} {\partial y}, \frac {\partial F} {\partial z})$$.

Say that the direction vector of interest is $\vec l$. Take the dot product between $\nabla F$ and the unit vector in the direction of $\vec l$, which should give $$\frac {\partial F} {\partial \vec l} = \nabla F \cdot \frac {\vec l} {\Vert \vec l \Vert}$$

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