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.