Questions: Compute the directional derivative of the following function at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. f(x, y) = ln(2+x^2+2y^2); P(3,-1); <3,1> Find the gradient of f(x, y) = ln(2+x^2+2y^2). ∇f(x, y) = square, square The directional derivative is square (Type an exact answer, using radicals as needed.)

Solution Steps

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

1. Find the gradient of the function \( f(x, y) \): The gradient is a vector of partial derivatives with respect to \( x \) and \( y \).
2. Normalize the direction vector: Convert the given direction vector to a unit vector.
3. Compute the directional derivative: Take the dot product of the gradient at the given point \( P \) and the unit direction vector.

1. Compute the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \).
2. Evaluate the gradient at the point \( P(3, -1) \).
3. Normalize the direction vector \( \langle 3, 1 \rangle \).
4. Compute the dot product of the gradient at \( P \) and the unit direction vector.

To find the gradient of the function \( f(x, y) = \ln(2 + x^2 + 2y^2) \), we calculate the partial derivatives:

\[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]

Calculating the partial derivatives at the point \( P(3, -1) \):

\[ \frac{\partial f}{\partial x} = \frac{2x}{2 + x^2 + 2y^2} \quad \text{and} \quad \frac{\partial f}{\partial y} = \frac{4y}{2 + x^2 + 2y^2} \]

Evaluating these at \( P(3, -1) \):

\[ \nabla f(3, -1) = \left( 0.4615, -0.3077 \right) \]

The given direction vector is \( \langle 3, 1 \rangle \). We normalize it to obtain the unit vector:

\[ \text{Magnitude} = \sqrt{3^2 + 1^2} = \sqrt{10} \]

Thus, the unit vector \( \mathbf{u} \) is:

\[ \mathbf{u} = \left( \frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} \right) \approx \left( 0.9487, 0.3162 \right) \]

The directional derivative \( D_{\mathbf{u}} f \) at point \( P \) in the direction of the unit vector \( \mathbf{u} \) is given by the dot product:

\[ D_{\mathbf{u}} f = \nabla f(3, -1) \cdot \mathbf{u} \]

Calculating this gives:

\[ D_{\mathbf{u}} f \approx 0.3406 \]

Final Answer