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**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 \( (1, 1) \)?**

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### 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 \( (-1, 1) \).

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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.

Find the direction in which f(x, y) = (x^2)/2 + (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)?

Learn how to find the direction of maximum increase and zero change for the function f(x, y) = (x^2)/2 + (y^2)/2 at the point (1, 1). This problem involves computing the gradient vector and understanding the relationship between the gradient and directional derivatives. Suitable for university-level calculus students.

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