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?

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.