Question

Find the greatest value of the directional derivative of the function \( f = x^2 y z^3 \) at (2,1,-1)

• A. \( 5\sqrt{11} \)
• B. \( 4\sqrt{11} \)
• C. \( 3\sqrt{11} \)
• D. \( 2\sqrt{11} \)

Hint:

Directional Derivative. Max value occurs in the direction of the gradient \(\nabla f\). Max value = \(|\nabla f|\). Calculate \(\nabla f = (\partial f/\partial x, \partial f/\partial y, \partial f/\partial z)\), evaluate at the point, then find the magnitude.

Answer 1

The Correct Option is B

The directional derivative of a function \(f\) in the direction of a unit vector \(\hat{u}\) is given by \( D_{\hat{u}}f = \nabla f \cdot \hat{u} \).
The directional derivative has its greatest value in the direction of the gradient vector \(\nabla f\), and the magnitude of this greatest value is equal to the magnitude of the gradient vector, \( |\nabla f| \).
First, find the gradient of \( f = x^2 y z^3 \): $$ \nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} $$ $$ \frac{\partial f}{\partial x} = 2x y z^3 $$ $$ \frac{\partial f}{\partial y} = x^2 z^3 $$ $$ \frac{\partial f}{\partial z} = x^2 y (3z^2) = 3x^2 y z^2 $$ $$ \nabla f = (2xyz^3)\hat{i} + (x^2z^3)\hat{j} + (3x^2yz^2)\hat{k} $$ Next, evaluate the gradient at the point (2, 1, -1): $$ \frac{\partial f}{\partial x} = 2(2)(1)(-1)^3 = 4(1)(-1) = -4 $$ $$ \frac{\partial f}{\partial y} = (2)^2(-1)^3 = 4(-1) = -4 $$ $$ \frac{\partial f}{\partial z} = 3(2)^2(1)(-1)^2 = 3(4)(1)(1) = 12 $$ So, at (2, 1, -1), \(\nabla f = -4\hat{i} - 4\hat{j} + 12\hat{k}\).
The greatest value of the directional derivative is the magnitude of this gradient vector: $$ |\nabla f| = \sqrt{(-4)^2 + (-4)^2 + (12)^2} $$ $$ |\nabla f| = \sqrt{16 + 16 + 144} = \sqrt{176} $$ Simplify the square root: \(176 = 16 \times 11\).
$$ |\nabla f| = \sqrt{16 \times 11} = \sqrt{16} \sqrt{11} = 4\sqrt{11} $$ The greatest value is \( 4\sqrt{11} \).