I am given the following function $$F(x,y,z)=x^2y+y^2-z^4$$ and point $(2,1,1)$ and I am asked to find the max directional derivative at the point
Here is my attempt to solve it:
$$F_x=2xy, F_y =x^2+2y, F_z=-4z^3$$
Therefore
$$\nabla f(x,y,z)\mid_{2,1,1}= 4i+6j-4k \text{ and the unit vector } a = 1/2 i+4/3j-1/2k$$
Hence the maximum value of $\nabla_v f$ is given by
$$4/2+18/4-4/2= 9/2$$
Can anyone confirm that all my steps are correct and my solution is correct to if not please show me the right solution
Best Answer
The directional derivative in the direction of the unit vector $\hat{\mathbf{d}}$ is $$\nabla{F}\cdot\hat{\mathbf{d}}= |\nabla{F}||\hat{\mathbf{d}}|\cos(\theta)$$
Since $|\hat{\mathbf{d}}|=1$ and the maximum value of $\cos(\theta)=1$ (achieved when $\theta =0$ and $\hat{\mathbf{d}}$ points in the same direction as the gradient) the maximum directional derivative is in the direction of $\nabla{F}$ and has value $|\nabla{F}|$. Only the gradient of $F$ is required, $\hat{\mathbf{a}}$ does not come in to it.