https://file.aiquickdraw.com/mathbot/file/1729962457319zvpw9ni5.jpg

To solve this problem, we need to follow these steps:

### Step 1: Find the Gradient of \( f(x, y) = x^2 + 3y^2 \)
The gradient vector of \( f(x, y) \), denoted as \( \nabla f \), is given by:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]

1. Compute \( \frac{\partial f}{\partial x} \):
   \[
   \frac{\partial f}{\partial x} = 2x
   \]

2. Compute \( \frac{\partial f}{\partial y} \):
   \[
   \frac{\partial f}{\partial y} = 6y
   \]

Thus, the gradient of \( f \) is:
\[
\nabla f = (2x, 6y)
\]

At the point \( (1, 1) \), we get:
\[
\nabla f(1, 1) = (2 \cdot 1, 6 \cdot 1) = (2, 6)
\]

### Step 2: Find the Gradient of the Constraint \( g(x, y) = x^2 + y^2 - 2 = 0 \)
The problem specifies a direction normal to the circle \( x^2 + y^2 = 2 \), which is represented by \( g(x, y) = x^2 + y^2 - 2 \).

The gradient vector of \( g(x, y) \), denoted as \( \nabla g \), is:
\[
\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)
\]

1. Compute \( \frac{\partial g}{\partial x} \):
   \[
   \frac{\partial g}{\partial x} = 2x
   \]

2. Compute \( \frac{\partial g}{\partial y} \):
   \[
   \frac{\partial g}{\partial y} = 2y
   \]

Thus, the gradient of \( g \) is:
\[
\nabla g = (2x, 2y)
\]

At the point \( (1, 1) \), we get:
\[
\nabla g(1, 1) = (2 \cdot 1, 2 \cdot 1) = (2, 2)
\]

### Step 3: Directional Derivative of \( f \) in the Direction of \( \nabla g \)
The directional derivative of \( f \) in the direction of \( \nabla g \) at \( (1, 1) \) is given by:
\[
D_{\mathbf{u}} f = \frac{\nabla f \cdot \nabla g}{|\nabla g|}
\]
where \( \mathbf{u} = \frac{\nabla g}{|\nabla g|} \) is the unit vector in the direction of \( \nabla g \).

1. Calculate \( \nabla f \cdot \nabla g \):
   \[
   \nabla f \cdot \nabla g = (2, 6) \cdot (2, 2) = 2 \cdot 2 + 6 \cdot 2 = 4 + 12 = 16
   \]

2. Calculate \( |\nabla g| \):
   \[
   |\nabla g| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
   \]

3. Therefore, the magnitude of the directional derivative is:
   \[
   |D_{\mathbf{u}} f| = \frac{|\nabla f \cdot \nabla g|}{|\nabla g|} = \frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}
   \]

### Final Answer:
The magnitude of the directional derivative of \( f(x, y) \) in the direction normal to the circle at \( (1, 1) \) is:
\[
\boxed{4\sqrt{2}}
\]

Would you like more details or have any questions?

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### Related Questions
1. How do we interpret the result of a directional derivative geometrically?
2. How does the gradient of a function relate to level curves in general?
3. What is the significance of taking a derivative in the direction normal to a curve?
4. How can we determine if a directional derivative is maximizing or minimizing the rate of change of the function?
5. How do we find the unit normal vector in cases where the gradient vector is zero?

### Tip
Remember, the gradient vector \( \nabla f \) always points in the direction of the greatest rate of increase of the function.

The magnitude of the directional derivative of the function f(x, y) = x^2 + 3y^2 in a direction normal to the circle x^2 + y^2 = 2, at the point (1, 1), is

Learn how to calculate the magnitude of the directional derivative of the function f(x, y) = x^2 + 3y^2 in a direction normal to the circle x^2 + y^2 = 2, at the point (1, 1). This solution includes steps for finding gradients and applying vector calculus concepts. Suitable for college calculus students.

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