Gradiant of the function 

2x2−y232x2−y23

 at point 

(1,2)(1,2)

 is 

To find the gradient of the function \( f(x, y) = 2x^2 - \frac{y^2}{3} \) at the point \( (1, 2) \), we need to compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and then evaluate them at the given point.

### Step 1: Compute the partial derivatives

1. **Partial derivative with respect to \( x \):**
   \[
   \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 2x^2 - \frac{y^2}{3} \right)
   \]
   Since \( \frac{y^2}{3} \) is independent of \( x \), it becomes zero when differentiating with respect to \( x \).
   \[
   \frac{\partial f}{\partial x} = 4x
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( 2x^2 - \frac{y^2}{3} \right)
   \]
   Since \( 2x^2 \) is independent of \( y \), it becomes zero when differentiating with respect to \( y \).
   \[
   \frac{\partial f}{\partial y} = -\frac{2y}{3}
   \]

### Step 2: Evaluate the partial derivatives at the point \( (1, 2) \)

1. **Evaluate \( \frac{\partial f}{\partial x} \) at \( (1, 2) \):**
   \[
   \frac{\partial f}{\partial x} \bigg|_{(1, 2)} = 4 \times 1 = 4
   \]

2. **Evaluate \( \frac{\partial f}{\partial y} \) at \( (1, 2) \):**
   \[
   \frac{\partial f}{\partial y} \bigg|_{(1, 2)} = -\frac{2 \times 2}{3} = -\frac{4}{3}
   \]

### Step 3: Write the gradient vector

The gradient of the function at the point \( (1, 2) \) is:

\[
\nabla f(1, 2) = \left( 4, -\frac{4}{3} \right)
\]

So, the gradient of the function \( f(x, y) = 2x^2 - \frac{y^2}{3} \) at the point \( (1, 2) \) is \( \mathbf{4i - \frac{4}{3}j} \).

Do you want more details or have any questions about this?

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**Related Questions:**
1. What does the gradient vector represent in a physical context?
2. How can the gradient be used to find the direction of steepest ascent?
3. What is the significance of the gradient being zero at a point?
4. How do you compute the directional derivative using the gradient?
5. Can the gradient be used to find the level curves of a function?
6. What happens to the gradient as you move away from the critical points?
7. How do you visualize the gradient field for a given function?
8. How is the gradient related to the tangent plane of a surface at a point?

**Tip:** The gradient vector points in the direction of the greatest rate of increase of the function.

Learn how to compute the gradient of the function f(x, y) = 2x^2 - y^2/3 at the point (1, 2). This problem involves multivariable calculus concepts such as partial derivatives and the gradient. Suitable for advanced undergraduate students studying calculus.

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