Question

Consider the function \( f(x, y, z) = x^4 + 2y^3 + z^2 \). The directional derivative of the function at the point \( P(-1, -1, -1) \) along \( (\mathbf{i} + \mathbf{j}) \), where \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the \( x \)- and \( y \)-directions, respectively, rounded off to 2 decimal places, is:

Hint:

To compute directional derivatives, always normalize the direction vector and calculate the gradient at the given point.

Answer 1

Given: \[ f(x, y, z) = x^4 + 2y^3 + z^2, \quad \vec{a} = \hat{i} + \hat{j}. \] Gradient of the function \( f \): \[ \nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}. \] \[ \nabla f = 4x^3 \hat{i} + 6y^2 \hat{j} + 2z \hat{k}. \] At point \( (-1, 1, -1) \): \[ \nabla f|_{(-1,1,-1)} = -4\hat{i} + 6\hat{j} - 2\hat{k}. \] Unit normal vector: \[ \hat{a} = \frac{\vec{a}}{|\vec{a}|}. \] Given \( \vec{a} = \hat{i} + \hat{j} \): \[ |\vec{a}| = \sqrt{1^2 + 1^2} = \sqrt{2}. \] \[ \hat{a} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}. \] Directional derivative of \( f \) along \( \vec{a} \): \[ DD = \nabla f \cdot \hat{a}. \] \[ DD = (-4\hat{i} + 6\hat{j} - 2\hat{k}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right). \] Simplify: \[ DD = \frac{-4}{\sqrt{2}} + \frac{6}{\sqrt{2}}. \] \[ DD = \frac{-4 + 6}{\sqrt{2}}. \] \[ DD = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414. \] Conclusion: The directional derivative of the function at the point \( P(-1, 1, -1) \) along \( \hat{i} + \hat{j} \) is \( 1.41 \).