Find the directional derivative of f(x, y, z) = xy2z3 at p(4, 1, 1) in the direction of q(0, −7, 9).

Answer:

Step-by-step explanation:

Since we have been given two points, P(4,1,1) and Q(0,-7,9), we can take the vector between the two toget the directional vector.

So, vector PQ = v = <-4, -8, 8>. Now, we find its UNIT VECTOR.

Since the UNIT VECTOR is just the vector divided by its magnitude, we get that the unit vector u = v / |v|

|v| = [tex]\sqrt{(-4)^2+(-8)^2+(8^2)}[/tex]  = [tex]\sqrt{144}[/tex] = 12

Dividing each part of the original vector v by 12 gives us the unit vector.

Now, we take the partial derivatives, Fx, Fy, and Fz.

Fx = [tex]y^2z^3[/tex]  

Fy = [tex]x(2y)z^3[/tex]

Fz = [tex]xy^2(3z^2)[/tex]

plugging in the original point P(4,1,1) gives us the following values for the gradient, which is just the vector of the partial derivatives.

ΔF = <1,8,12>

Now we just use the equation Duf = ΔF * u, where we take the DOT PRODUCT of the gradient at the given point with the direction unit vector.

This gives us <1,8,12> * <-1/3, -2/3, 2/3> = 7/3

The directional derivative of [tex]\( f \)[/tex] at point [tex]\( P(4, 1, 1) \)[/tex] in the direction of [tex]\( Q(0, -7, 9) \)[/tex] is [tex]\( \frac{7}{\sqrt{6}} \).[/tex]

To find the directional derivative of the function [tex]\(f(x, y, z) = xy^2z^3\)[/tex] at the point [tex]\(P(4, 1, 1)\)[/tex] in the direction of [tex]\(Q(0, -7, 9)\)[/tex], we'll use the formula for directional derivative:

[tex]\[ D_{\mathbf{u}} f(x, y, z) = \nabla f \cdot \mathbf{u} \][/tex]

where [tex]\( \nabla f \)[/tex] is the gradient vector of [tex]\( f \)[/tex] and [tex]\( \mathbf{u} \)[/tex] is the unit vector in the direction of [tex]\( Q \).[/tex]

First, let's find the gradient vector [tex]\( \nabla f \):[/tex]

[tex]\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \][/tex]

[tex]\[ = \left( y^2z^3, 2xyz^3, 3xy^2z^2 \right) \][/tex]

Now, evaluate [tex]\( \nabla f \)[/tex] at the point [tex]\( P(4, 1, 1) \):[/tex]

[tex]\[ \nabla f(P) = \left( (1)^2(1)^3, 2(4)(1)(1)^3, 3(4)(1)^2(1)^2 \right) \][/tex]

[tex]\[ = (1, 8, 12) \][/tex]

Next, find the unit vector in the direction of [tex]\( Q \):[/tex]

[tex]\[ \mathbf{u} = \frac{\mathbf{Q} - \mathbf{P}}{\|\mathbf{Q} - \mathbf{P}\|} \][/tex]

[tex]\[ = \frac{(0 - 4, -7 - 1, 9 - 1)}{\sqrt{(0 - 4)^2 + (-7 - 1)^2 + (9 - 1)^2}} \][/tex]

[tex]\[ = \frac{(-4, -8, 8)}{\sqrt{16 + 64 + 64}} \][/tex]

[tex]\[ = \frac{(-4, -8, 8)}{4\sqrt{6}} \][/tex]

[tex]\[ = \left( -\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \][/tex]

Now, calculate the dot product [tex]\( \nabla f \cdot \mathbf{u} \):[/tex]

[tex]\[ D_{\mathbf{u}} f(P) = \nabla f(P) \cdot \mathbf{u} \][/tex]

[tex]\[ = (1, 8, 12) \cdot \left( -\frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \][/tex]

[tex]\[ = -\frac{1}{\sqrt{6}} + \left(-\frac{16}{\sqrt{6}}\right) + \frac{24}{\sqrt{6}} \][/tex]

[tex]\[ = \frac{7}{\sqrt{6}} \][/tex]

Therefore, the directional derivative of [tex]\( f \)[/tex] at point [tex]\( P(4, 1, 1) \)[/tex] in the direction of [tex]\( Q(0, -7, 9) \)[/tex] is [tex]\( \frac{7}{\sqrt{6}} \).[/tex]

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