Question 16
Find the directional derivative of the scalar function of in the direction of the outer normal to the surface at the point
.

Ask by Tyler Wells. in India

Jan 24,2025

Question

Question 16 Find the directional derivative of the scalar function of $$ (x, y, z)=x y z $$ in the direction of the outer normal to the surface $$ z=x y $$ at the point $$ (3,1,3) $$.

UpStudy AI · Accepted Answer

To find the **directional derivative** of the scalar function $$ f(x, y, z) = xyz $$ in the direction of the **outer normal** to the surface $$ z = xy $$ at the point $$ (3, 1, 3) $$, follow these steps:

### 1. Compute the Gradient of $$ f $$
The gradient of $$ f $$ is given by:
$$
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
$$
For $$ f(x, y, z) = xyz $$:
$$
\nabla f = \left( yz, xz, xy \right)
$$
At the point $$ (3, 1, 3) $$:
$$
\nabla f(3, 1, 3) = \left(1 \cdot 3, 3 \cdot 3, 3 \cdot 1\right) = (3, 9, 3)
$$

### 2. Determine the Outer Normal to the Surface $$ z = xy $$
First, rewrite the surface equation as $$ F(x, y, z) = z - xy = 0 $$. The normal vector to the surface is given by the gradient of $$ F $$:
$$
\nabla F = \left( -y, -x, 1 \right)
$$
At the point $$ (3, 1, 3) $$:
$$
\mathbf{n} = \nabla F(3, 1, 3) = (-1, -3, 1)
$$
To get the **unit normal vector**, normalize $$ \mathbf{n} $$:
$$
\| \mathbf{n} \| = \sqrt{(-1)^2 + (-3)^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11}
$$
$$
\mathbf{u} = \frac{\mathbf{n}}{\| \mathbf{n} \|} = \left( \frac{-1}{\sqrt{11}}, \frac{-3}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right)
$$

### 3. Compute the Directional Derivative
The directional derivative of $$ f $$ in the direction of $$ \mathbf{u} $$ is:
$$
D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}
$$
Substituting the values:
$$
D_{\mathbf{u}} f = (3, 9, 3) \cdot \left( \frac{-1}{\sqrt{11}}, \frac{-3}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right) 
$$
$$
D_{\mathbf{u}} f = \frac{3 \cdot (-1) + 9 \cdot (-3) + 3 \cdot 1}{\sqrt{11}} = \frac{-3 - 27 + 3}{\sqrt{11}} = \frac{-27}{\sqrt{11}}
$$

### Final Answer
$$
D_{\mathbf{u}} f = \frac{-27}{\sqrt{11}}
$$

**Answer:** Its directional derivative is –27 divided by the square root of 11, that is

D = −27 ⁄ √11
The directional derivative of $$ f(x, y, z) = xyz $$ in the direction of the outer normal to the surface $$ z = xy $$ at the point $$ (3, 1, 3) $$ is $$ -\frac{27}{\sqrt{11}} $$.

Question 16
Find the directional derivative of the scalar function of in the direction of the outer normal to the surface at the point
.

Solución de inteligencia artificial de Upstudy

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Solución

1. Compute the Gradient of

2. Determine the Outer Normal to the Surface

3. Compute the Directional Derivative

Final Answer

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Question 16 Find the directional derivative of the scalar function of in the direction of the outer normal to the surface at the point .