How to find directional derivative

algebra-precalculusmultivariable-calculus

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

Best Answer

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 $z$, $$ \begin{align} \frac{1}{\sqrt{2}}\cdot f_x(2, 1, 2) + 0 \cdot f_y(2, 1, 2) - \frac{1}{\sqrt{2}}\cdot f_z(2,1,2) &= \\ \frac{1}{\sqrt{2}}\cdot 10 + \frac{0}{\sqrt{2}}\cdot 3 - \frac{1}{\sqrt{2}} \cdot 2 &= \\\frac{10 - 2}{\sqrt{2}} &= \\\frac{8}{\sqrt{2}} \end{align} $$