Question

Let \( (x, y) \in \mathbb{R}^2 \). The rate of change of the real-valued function
\[ V(x, y) = x^2 + x + y^2 + 1 \] at the origin in the direction of the point \( (1, 2) \) is _____________ (round off to the nearest integer). 

Hint:

To find the rate of change of a function \( V(x, y) \) at a point in a specific direction: \[ {Directional derivative } = \nabla V \cdot \hat{u} \] 1. Compute gradient \( \nabla V \) at the point. 2. Normalize the direction vector to get \( \hat{u} \). 3. Take the dot product.

Answer 1

Correct Answer: 0

The directional derivative of a scalar field \( V(x, y) \) at a point \( (x_0, y_0) \) in the direction of a unit vector \( \hat{u} \) is given by: \[ D_{\hat{u}} V = \nabla V \cdot \hat{u} \] First, compute the gradient: \[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) = (2x + 1, 2y) \] At the origin \( (0, 0) \): \[ \nabla V(0, 0) = (1, 0) \] Next, the direction vector from origin to point \( (1, 2) \) is: \[ \vec{v} = (1, 2) \Rightarrow \hat{u} = \frac{1}{\sqrt{1^2 + 2^2}}(1, 2) = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \] Now compute the directional derivative: \[ D_{\hat{u}} V = \nabla V \cdot \hat{u} = (1, 0) \cdot \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) = \frac{1}{\sqrt{5}} \approx 0.447 \] \[ \boxed{{Rounded answer lies between } 0 { and } 1} \]