Question

A direction in which the function \( f(x) = 2x + y \) remains constant at the point (1,1) is .........

• A. \( \frac{\hat{i}}{\sqrt{5}} - \frac{2\hat{j}}{\sqrt{5}} \)
• B. \( \frac{-2\hat{i}}{\sqrt{5}} + \frac{\hat{j}}{\sqrt{5}} \)
• C. \( \frac{-\hat{i}}{\sqrt{5}} - \frac{\hat{j}}{\sqrt{5}} \)
• D. \( \frac{\hat{i}}{\sqrt{5}} + \frac{2\hat{j}}{\sqrt{5}} \)

Hint:

To find the direction in which a function remains constant, compute the gradient of the function and then find a vector perpendicular to the gradient. Normalize this vector to get the unit vector in the required direction.

Answer 1

The Correct Option is A

The function \( f(x) = 2x + y \) is a linear function of two variables, \( x \) and \( y \). To find the direction in which the function remains constant, we need to find the gradient of \( f(x) \) at the point \( (1, 1) \), because the gradient points in the direction of maximum increase of the function. The gradient of \( f(x) \) is given by: \[ \nabla f(x) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] For \( f(x) = 2x + y \), the partial derivatives are: \[ \frac{\partial f}{\partial x} = 2, \frac{\partial f}{\partial y} = 1 \] So, the gradient vector is: \[ \nabla f(x) = (2, 1) \] The direction in which the function remains constant is perpendicular to the gradient. Therefore, we need a vector perpendicular to \( (2, 1) \). A vector perpendicular to \( (2, 1) \) can be obtained by swapping the components and changing the sign of one of them: \[ {Perpendicular vector} = (-1, 2) \] Now, we normalize this vector to make it a unit vector: \[ {Magnitude} = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \] Thus, the unit vector in the direction perpendicular to \( (2, 1) \) is: \[ \frac{-\hat{i}}{\sqrt{5}} + \frac{2\hat{j}}{\sqrt{5}} \] So, the correct answer is option (1) \( \frac{\hat{i}}{\sqrt{5}} - \frac{2\hat{j}}{\sqrt{5}} \).