Line Integrals: Applications in Physics, Work, Energy & Geometry

The vector field F(x, y) = <2x, 2y> is the gradient of the potential function f(x, y) = x² + y². The path of motion is the curve x² + y² = 25, which is a level curve (contour line) of the potential function f. Since the gradient vector is always perpendicular to the level curves of its potential function, the force F is perpendicular to the path of motion at every point. Therefore, the work done (the integral of the tangential component) is zero. Alternatively, using the Fundamental Theorem of Line Integrals, the work is f(0,5) - f(5,0) = 25 - 25 = 0.

The integral of a constant k ds is k times the length of the path. The length of the line segment from (0,0) to (2,3) is √(2² + 3²) = √(4 + 9) = √13. Therefore the integral is 5 times √13.

The property of additivity for line integrals states that ∫C = ∫C1 + ∫C2. We are given ∫C = 4 and ∫C1 = 12. Solving for ∫C, we get 4 - 12 = -8.

The work done by a constant force is the dot product of the force vector and the displacement vector. The displacement is <5, 12>. Therefore the work is 3(5) + 4(12) = 15 + 48 = 63 Joules. The path being a straight line is necessary for constant force, but since it's straight, the line integral reduces to the dot product.

Parameterize x = 2t, y = 2t, t from 0 to 1. Then dx = 2 dt, dy = 2 dt. Substitute: (2t + 2t) *2 dt + (2t - 2t)*2 dt = (4t)*2 dt + 0 = 8t dt. The integral is ∫01 8t dt = [4 t²]01 = 4.

The fundamental theorem states that the line integral of a gradient field is the difference in the potential function at the endpoints. Compute f(2,1) = 2*1 + 8/3 = 2 + 8/3 = 14/3. f(0,0) = 0. Therefore the integral is 14/3 - 0 = 14/3.

If the line integral of a vector field between two points is the same for every path connecting those points, the field is path independent. Path independence is equivalent to the field being conservative, meaning F = ∇f for some potential function f.

If a vector field is conservative, it is the gradient of a potential function f. The line integral over a closed curve is f at starting point minus f at starting point, which is 0. This is a direct consequence of the fundamental theorem for line integrals applied to closed paths.

Compute ∂P/∂y = -e^x sin y, ∂Q/∂x = -e^x sin y. They are equal, so conservative. Integrate P = e^x cos y with respect to x: e^x cos y + g(y). Differentiate with respect to y: e^x (-sin y) + g'(y) = -e^x sin y. So g'(y) = 0, g(y) = constant. The potential is e^x cos y.

F = <2x, 2y> is the gradient of x² + y², so it is conservative. In 2D, a field is conservative if and only if it is irrotational, meaning ∂P/∂y - ∂Q/∂x = 0, which holds here (0 - 0 = 0). Therefore the line integral is path independent.

The line integral of a gradient field is the potential at the end point minus the potential at the starting point. The potential function is f(x,y) = x e^y. At the end point (2,3), f(2,3) = 2 e³. At the starting point (1,1), f(1,1) = 1 * e^1 = e. Therefore the integral is 2 e³ - e.

Parameterize the path as x = cos t, y = sin t, t from 0 to π/2. Then dx = -sin t dt, dy = cos t dt. F = < -sin t, cos t>. F · dr = (-sin t)(-sin t dt) + cos t (cos t dt) = sin² t dt + cos² t dt = (sin² t + cos² t) dt = dt. The integral is ∫ from 0 to π/2 dt = π/2.

A vector field is conservative if it is the gradient of a scalar function. This property implies that the line integral between two points is the same for all paths connecting those points (path independence) and that the line integral over any closed path is zero. The two conditions are equivalent in simply connected domains.