This is pretty hard to decipher.
Consider the function f (x comma y comma z )equals 1 plus 4 xyz,
[tex]f(x,y,z)=1 +4xyz[/tex]
the point P(negative 1 comma negative 1 comma 1 ),
P(-1, -1, 1)
and the unit vector Bold uequalsleft angle StartFraction 1 Over StartRoot 3 EndRoot EndFraction comma StartFraction 1 Over StartRoot 3 EndRoot EndFraction comma StartFraction 1 Over StartRoot 3 EndRoot EndFraction right angle .
Huh? Oh, I see. We won't stress about the vector notation.
[tex]u =\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}} \right)[/tex]
a. Compute the gradient of f and evaluate it at P.
[tex] \nabla f = \left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right)[/tex]
[tex] \nabla f = (4yz, 4xz, 4xy)[/tex]
P(-1, -1, 1)
[tex]g =(-4, -4, 4)[/tex]
b. Find the unit vector in the direction of maximum increase of f at P.
That's just the normalized version of the above. It has a magnitude
[tex]\sqrt{3(4)^2} = 4\sqrt{3}[/tex]
so we have to divide 4 by that, so 1/√3.
The unit direction vector is
[tex]g =\left( -\dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)[/tex]
Sometimes they seek the teacher friendly
[tex]g =\left( \dfrac{-\sqrt{3}}{3}, -\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3} \right)[/tex]
c. Find the rate of change of the function in the direction of maximum increase at P.
We already did, that's the magnitude of the gradient,
[tex]4\sqrt{3}[/tex]
d. Find the directional derivative at P in the direction of the given vector.
We want the total derivative in the direction of u, so that's the weighted average of the partials. For unit direction u=(a,b,c) we can write it
[tex]D_u f = D_{(a,b,c)} f =\dfrac{\partial f}{\partial z} a + \dfrac{\partial f}{\partial z} b + \dfrac{\partial f}{\partial z} c[/tex]
For us that's
[tex]D_u f = (-4) \dfrac{1}{\sqrt{3}} + (-4) \dfrac{1}{\sqrt{3}}+(4)\dfrac{1}{\sqrt{3}}[/tex]
[tex]D_u f =-\frac 4 3 \sqrt{3}[/tex]
