EDUCA ION b) Find the directional derivative \( D_{u} f(x, y, z) \) for \( f(x, y, z)=\sin (y z)+\ln x^{2} \) in the direction of the vector \( v=\langle 1,1,-1\rangle \). Hence find \( D_{u} f(1,1, \pi) \). [12 marks]

To find the directional derivative \( D_{u} f(x, y, z) \) of the function \( f(x, y, z) = \sin(yz) + \ln(x^2) \) in the direction of the vector \( v = \langle 1, 1, -1 \rangle \), we will follow these steps: 1. **Calculate the gradient \( \nabla f \)**. 2. **Normalize the direction vector \( v \)** to obtain the unit vector \( u \). 3. **Compute the directional derivative using the formula** \( D_{u} f = \nabla f \cdot u \). ### Step 1: Calculate the Gradient \( \nabla f \) The gradient \( \nabla f \) is given by the vector of partial derivatives: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] **Partial Derivative with respect to \( x \)**: \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \sin(yz) + \ln(x^2) \right) = 0 + \frac{2}{x} = \frac{2}{x} \] **Partial Derivative with respect to \( y \)**: \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( \sin(yz) + \ln(x^2) \right) = z \cos(yz) + 0 = z \cos(yz) \] **Partial Derivative with respect to \( z \)**: \[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left( \sin(yz) + \ln(x^2) \right) = y \cos(yz) + 0 = y \cos(yz) \] Thus, the gradient is: \[ \nabla f = \left( \frac{2}{x}, z \cos(yz), y \cos(yz) \right) \] ### Step 2: Normalize the Direction Vector \( v \) The direction vector \( v = \langle 1, 1, -1 \rangle \) needs to be normalized to find the unit vector \( u \): \[ \|v\| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \] The unit vector \( u \) is: \[ u = \frac{1}{\sqrt{3}} \langle 1, 1, -1 \rangle = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right\rangle \] ### Step 3: Compute the Directional Derivative Now we need to evaluate \( \nabla f \) at the point \( (1, 1, \pi) \): 1. **Calculate \( \nabla f(1, 1, \pi) \)**: \[ \nabla f(1, 1, \pi) = \left( \frac{2}{1}, \pi \cos(1 \cdot \pi), 1 \cdot \cos(1 \cdot \pi) \right) = \left( 2, \pi \cos(\pi), \cos(\pi) \right) \] Since \( \cos(\pi) = -1 \): \[ \nabla f(1, 1, \pi) = \left( 2, \pi(-1), -1 \right) = \left( 2, -\pi, -1 \right) \] 2. **Compute the directional derivative**: \[ D_{u} f(1, 1, \pi) = \nabla f(1, 1, \pi) \cdot u \] Calculating the dot product: \[ D_{u} f(1, 1, \pi) = \left( 2, -\pi, -1 \right) \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right\rangle \] \[ = 2 \cdot \frac{1}{\sqrt{3}} + (-\pi) \cdot \frac{1}{\sqrt{3}} + (-1) \cdot \left(-\frac{1}{\sqrt{3}}\right) \] \[ = \frac{2}{\sqrt{3}} - \frac{\pi}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2 - \pi + 1}{\sqrt{3}} = \frac{3 - \pi}{\sqrt{3}} \] ### Final Result Thus, the directional derivative \( D_{u} f(1, 1, \pi) \) is: \[ \boxed{\frac{3 - \pi}{\sqrt{3}}} \]