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

The rate of change of a function in the direction of a vector is given by the directional derivative. The formula for the directional derivative of \( f(x, y, z) \) at a point \( (x_0, y_0, z_0) \) in the direction of a unit vector \( \mathbf{u} \) is: \[ D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}, \] where \( \nabla f \) is the gradient of \( f \), and \( \mathbf{u} \) is the unit vector in the direction of interest. To compute this, we first need to find the gradient of the function \( f(x, y, z) = x^2 + 2y^2 + z \). The gradient is the vector of partial derivatives with respect to each variable: \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2x, 4y, 1). \] At the point \( (1, 1, 1) \), the gradient becomes: \[ \nabla f(1, 1, 1) = (2 \times 1, 4 \times 1, 1) = (2, 4, 1). \] Now, we need the direction vector \( 3i + 4k \), which corresponds to the vector \( (3, 0, 4) \). To compute the directional derivative, we normalize this vector to obtain the unit vector: \[ |\mathbf{u}| = \sqrt{3^2 + 0^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5, \] so the unit vector is: \[ \mathbf{u} = \left( \frac{3}{5}, 0, \frac{4}{5} \right). \] Finally, the directional derivative is: \[ D_{\mathbf{u}} f = \nabla f(1, 1, 1) \cdot \mathbf{u} = (2, 4, 1) \cdot \left( \frac{3}{5}, 0, \frac{4}{5} \right) = 2 \times \frac{3}{5} + 4 \times 0 + 1 \times \frac{4}{5} = \frac{6}{5} + \frac{4}{5} = 2. \] Thus, the rate of change of the function in the direction of \( 3i + 4k \) is \( 2 \).