Jawaharlal Nehru Technological University Hyderabad B.Tech II MATHEMATICS - II Previous Question Paper (July/August - 2021 - R15)

A vector field F = Pi + Qj is conservative if the curl of F is zero, which in two dimensions simplifies to the condition ∂Q/∂x = ∂P/∂y. Given P = x^2 + xy^2 and Q = y^2 + x^2y:

Since ∂P/∂y = ∂Q/∂x, the field is conservative. To find the scalar potential φ such that ∇φ = F, we solve:

Integrating the first equation with respect to x:

Differentiating this with respect to y and equating to Q:

Integrating g'(y) gives g(y) = y^3/3 + C. Thus, the scalar potential is:

Let f = x^2yz. The gradient of f is ∇f = (∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k = (2xyz)i + (x^2z)j + (x^2y)k. At the point (1, 2, -1):

The direction vector is v = i + 2j + 3k. Its magnitude is |v| = sqrt(1^2 + 2^2 + 3^2) = sqrt(14). The unit vector in the direction of v is u = (i + 2j + 3k) / sqrt(14). The directional derivative is the dot product ∇f · u: