Find the directional derivative of the function at \( P \) in the direction of \( \mathbf{v} \). \( \qquad h(x, y, z)=x y z, \quad P(2,6,4), \quad \mathbf{v}=<2,1,2\rangle \)

To find the directional derivative of \( h(x, y, z) = xyz \) at the point \( P(2,6,4) \) in the direction of the vector \( \mathbf{v} = \langle 2, 1, 2 \rangle \), we start by calculating the gradient of \( h \). 1. Calculate the gradient \( \nabla h \): \[ \nabla h = \left( \frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z} \right) \] The partial derivatives are: \[ \frac{\partial h}{\partial x} = yz, \quad \frac{\partial h}{\partial y} = xz, \quad \frac{\partial h}{\partial z} = xy \] Thus, at the point \( P(2, 6, 4) \): \[ \frac{\partial h}{\partial x} \bigg|_P = 6 \cdot 4 = 24, \quad \frac{\partial h}{\partial y} \bigg|_P = 2 \cdot 4 = 8, \quad \frac{\partial h}{\partial z} \bigg|_P = 2 \cdot 6 = 12 \] Therefore, the gradient at point \( P \) is: \[ \nabla h(2, 6, 4) = \langle 24, 8, 12 \rangle \] 2. Normalize the direction vector \( \mathbf{v} \): The magnitude of \( \mathbf{v} = \langle 2, 1, 2 \rangle \) is: \[ \|\mathbf{v}\| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is: \[ \mathbf{u} = \frac{1}{3} \langle 2, 1, 2 \rangle = \langle \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \rangle \] 3. Compute the directional derivative \( D_{\mathbf{u}} h \): The directional derivative is given by the dot product: \[ D_{\mathbf{u}} h = \nabla h \cdot \mathbf{u} \] Plugging in the values: \[ D_{\mathbf{u}} h = \langle 24, 8, 12 \rangle \cdot \langle \frac{2}{3}, \frac{1}{3}, \frac{2}{3} \rangle = 24 \cdot \frac{2}{3} + 8 \cdot \frac{1}{3} + 12 \cdot \frac{2}{3} \] Simplifying this gives: \[ = 16 + \frac{8}{3} + 8 = 16 + \frac{8 + 24}{3} = 16 + \frac{32}{3} = 16 + \frac{32}{3} = \frac{48}{3} + \frac{32}{3} = \frac{80}{3} \] Thus, the directional derivative of the function \( h \) at \( P(2, 6, 4) \) in the direction of \( \mathbf{v} \) is \( \frac{80}{3} \).