Question

Consider a vector \( \Bar{u} = 2\hat{x} + \hat{y} + 2\hat{z} \), where \( \hat{x}, \hat{y}, \hat{z} \) represent unit vectors along the coordinate axes \( x, y, z \) respectively. The directional derivative of the function \( f(x, y, z) = 2 \ln(xy) + \ln(yz) + 3 \ln(xz) \) at the point \( (x, y, z) = (1, 1, 1) \) in the direction of \( \mathbf{u} \) is:

• A. \( 0 \)
• B. \( \frac{7}{5\sqrt{2}} \)
• C. \( 7 \)
• D. \( 21 \)

Hint:

The directional derivative gives the rate of change of a scalar field in the direction of a given vector. To compute it, always normalize the direction vector to ensure the magnitude is 1.

Answer 1

The Correct Option is C

Step 1: Compute the gradient of \( f(x, y, z) \). The gradient of a scalar function \( f(x, y, z) \) is given by: \[ \nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}. \] Given: \[ f(x, y, z) = 2 \ln(xy) + \ln(yz) + 3 \ln(xz), \] we compute the partial derivatives of \( f \) with respect to \( x \), \( y \), and \( z \): \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \big( 2 \ln(xy) + \ln(yz) + 3 \ln(xz) \big) = \frac{2}{x} + \frac{3}{x}, \] \[ \frac{\partial f}{\partial x} = \frac{5}{x}. \] Similarly: \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \big( 2 \ln(xy) + \ln(yz) + 3 \ln(xz) \big) = \frac{2}{y} + \frac{1}{y}, \] \[ \frac{\partial f}{\partial y} = \frac{3}{y}. \] And: \[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \big( 2 \ln(xy) + \ln(yz) + 3 \ln(xz) \big) = \frac{1}{z} + \frac{3}{z}, \] \[ \frac{\partial f}{\partial z} = \frac{4}{z}. \] Thus, the gradient is: \[ \nabla f = \frac{5}{x} \hat{i} + \frac{3}{y} \hat{j} + \frac{4}{z} \hat{k}. \] Step 2: Evaluate \( \nabla f \) at \( (x, y, z) = (1, 1, 1) \). Substituting \( x = 1, y = 1, z = 1 \): \[ \nabla f = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}. \] Step 3: Compute the unit vector in the direction of \( \mathbf{u} \). The given vector \( \mathbf{u} = 2\hat{i} + \hat{j} + 2\hat{k} \). The magnitude of \( \mathbf{u} \) is: \[ |\mathbf{u}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \] The unit vector in the direction of \( \mathbf{u} \) is: \[ \hat{\mathbf{u}} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{1}{3} (2\hat{i} + \hat{j} + 2\hat{k}). \] Step 4: Compute the directional derivative. The directional derivative is given by: \[ \text{Directional Derivative} = \nabla f \cdot \hat{\mathbf{u}}. \] Substitute \( \nabla f = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} \) and \( \hat{\mathbf{u}} = \frac{1}{3} (2\hat{i} + \hat{j} + 2\hat{k}) \): \[ \nabla f \cdot \hat{\mathbf{u}} = \frac{1}{3} \big( (5)(2) + (3)(1) + (4)(2) \big). \] Simplify: \[ \nabla f \cdot \hat{\mathbf{u}} = \frac{1}{3} \big( 10 + 3 + 8 \big) = \frac{1}{3} (21) = 7. \] Thus, the directional derivative is: \[ \text{Directional Derivative} = 7. \]