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 \( (3,1,3) \).

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