Question

If $f(x, y)$ is a real-valued function of two variables, then $\vec{\nabla} f(x, y)$ is:

• A. $\frac{\partial f}{\partial y} \hat{i} - \frac{\partial f}{\partial x} \hat{j}$
• B. $\frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j}$
• C. $\frac{\partial f}{\partial x} \hat{i} - \frac{\partial f}{\partial y} \hat{j}$
• D. $\frac{\partial f}{\partial y} \hat{i} + \frac{\partial f}{\partial x} \hat{j}$

Hint:

Geometric Property: The gradient vector $\vec{\nabla} f(x, y)$ is always perpendicular (orthogonal) to the level curves $f(x, y) = c$ of the function!

Answer 1

The Correct Option is B

Step 1: Concept
By definition, the gradient of a scalar field $f(x, y)$ in two dimensions is a vector field pointing in the direction of the maximum rate of increase of $f$, whose components are the first-order partial derivatives of $f$.

Step 2: Key Formulas and Approach
The Del operator in 2D Euclidean space is defined as: \[ \vec{\nabla} = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} \]

Step 3: Step-by-step Explanation

• Apply the Del operator directly to the scalar function $f(x, y)$: \[ \vec{\nabla} f(x, y) = \left( \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} \right) f(x, y) \] \[ = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} \]
• Note that the $x$-partial derivative is paired with $\hat{i}$ and the $y$-partial derivative is paired with $\hat{j}$, both added together.

Step 4: Final Answer
The 2D gradient is $\frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j}$. Thus, Option (B) is correct.