The image contains a mathematical problem, and here is what it says:

Find the direction in which $f(x, y) = \frac{x^2}{2} + \frac{y^2}{2}$:

(i) Increases most rapidly at the point $(1, 1)$.

(ii) What are the directions of zero change in $f$ at the point $(1, 1)$?

Solution:

(i) Direction of Maximum Increase:

The function given is $f(x, y) = \frac{x^2}{2} + \frac{y^2}{2}$. The direction of maximum increase is given by the gradient vector $\nabla f(x, y)$.

1. Compute the gradient $\nabla f(x, y)$:

   $\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

   First, we calculate the partial derivatives:

   • $\frac{\partial f}{\partial x} = x$
   • $\frac{\partial f}{\partial y} = y$

   Therefore, the gradient is: $\nabla f(x, y) = (x, y)$

2. Evaluate at the point $(1, 1)$:

   $\nabla f(1, 1) = (1, 1)$

   So, the direction of maximum increase at $(1, 1)$ is along the vector $(1, 1)$.

(ii) Directions of Zero Change:

The directions of zero change are orthogonal to the gradient vector. To find the directions where $f(x, y)$ does not change, we find the vector that is perpendicular to $\nabla f(1, 1) = (1, 1)$.

Any vector $\mathbf{v} = (a, b)$ that satisfies the condition $\nabla f(1, 1) \cdot \mathbf{v} = 0$ will be a direction of zero change. The dot product is: $(1, 1) \cdot (a, b) = a + b$ For zero change, we need: $a + b = 0$ Thus, the directions of zero change are along the vector $(-1, 1)$ (or any scalar multiple of it).

Summary:

• Direction of maximum increase: Along $(1, 1)$.
• Directions of zero change: Along $(-1, 1)$.

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. How would the gradient change if the function had a $\frac{z^2}{2}$ term added?
2. What is the magnitude of the gradient vector at any general point $(x, y)$?
3. How do you compute the direction of steepest descent for the function?
4. What is the rate of change of the function in the direction $(1, -1)$?
5. How does the level curve of the function look near the point $(1, 1)$?

Tip: The gradient always points in the direction of the steepest ascent, while the direction of zero change is perpendicular to the gradient vector.