Question

If \( i \), \( j \), and \( k \) are the orthogonal unit vectors in Cartesian x-y-z coordinate system, the rate of change of the function \( f(x, y, z) = x^2 + 2y^2 + z \) at point \( (1, 1, 1) \) in the direction of \( 3i + 4k \) is .............

Hint:

To compute the rate of change of a function in a given direction, always compute the gradient of the function and take the dot product with the unit vector in the desired direction.

Answer 1

Correct Answer: 2

To find the rate of change of the function \( f(x, y, z) = x^2 + 2y^2 + z \) at the point \( (1, 1, 1) \) in the direction of the vector \( 3i + 4k \), we must compute the directional derivative. First, calculate the gradient \(\nabla f\) of the function: \[\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (2x, 4y, 1)\]

Substitute the point \( (1, 1, 1) \) into \(\nabla f\): \[\nabla f(1, 1, 1) = (2 \cdot 1, 4 \cdot 1, 1) = (2, 4, 1)\]

The direction vector \( \mathbf{v} = 3i + 4k \) is \((3, 0, 4)\). Normalize it to get the unit direction vector \(\mathbf{u}\):

\[\|\mathbf{v}\| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 16} = 5\]

\(\mathbf{u} = \left(\frac{3}{5}, 0, \frac{4}{5}\right)\)

The directional derivative is the dot product \(\nabla f \cdot \mathbf{u}\):

\[\nabla f \cdot \mathbf{u} = (2, 4, 1) \cdot \left(\frac{3}{5}, 0, \frac{4}{5}\right) = 2 \cdot \frac{3}{5} + 4 \cdot 0 + 1 \cdot \frac{4}{5}\]

\[= \frac{6}{5} + 0 + \frac{4}{5} = \frac{10}{5} = 2\]

The rate of change of the function in the specified direction is 2. Since this value is within the given range [2, 2], it confirms the correctness of the solution.