# how to find directional derivative

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

• evaluate partial derivatives at the given point, and dot product with direction vector Apr 28 '20 at 16:00
• @J.W.Tanner The inner product of $\nabla F$ should be taken with the UNIT direction vector. Apr 28 '20 at 16:03
• @MarkViola: I hedged; the Wikipedia page suggests that some authors don't require a UNIT direction vector Apr 28 '20 at 16:11
• @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. Apr 28 '20 at 16:38
• 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.
– amd
Apr 28 '20 at 17:37

## 2 Answers

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}

• I suppose that the directional vector one take the dot product against should be a unit vector? $\Vert (1,0,-1) \Vert = \sqrt 2$ Apr 28 '20 at 16:16
• Sure, I altered my answer to reflect this. Apr 28 '20 at 16:32

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 $$\frac {\partial F} {\partial \vec l} = \nabla F \cdot \frac {\vec l} {\Vert \vec l \Vert}$$