Line Integrals: Curve Parameterization, Geometry & Scalar Evaluation

To compute a line integral with respect to arc length, we first need to parameterize the curve C. The line from (0,0) to (1,1) can be parameterized as r(t) = (t, t) where t goes from 0 to 1. The derivative of this parameterization is r'(t) = (1, 1). The differential arc length ds is equal to the magnitude of r'(t) times dt, which is ds = √(1² + 1²) dt = √2 dt. The integrand x + y becomes t + t = 2t when we substitute our parameterization. Therefore, the integral becomes ∫₀¹ 2t √2 dt. We can factor out the √2 constant to get √2 ∫₀¹ 2t dt. The integral of 2t is t². Evaluating this from 0 to 1 gives us (1² - 0²) = 1. Multiplying by the √2 factor gives us √2 · 1 = √2. So the value of the line integral is √2.

We start by parameterizing the quarter circle. Since x² + y² = 1, we can use trigonometric functions: r(t) = (cos t, sin t) where t ranges from 0 to π/2 for the first quadrant. The derivative is r'(t) = (-sin t, cos t). To find ds, we calculate the magnitude of r'(t): √[(-sin t)² + (cos t)²] = √(sin² t + cos² t) = √1 = 1, so ds = dt. The integrand x becomes cos t in our parameterization. The integral is therefore ∫₀^{π/2} cos t dt. The antiderivative of cos t is sin t. Evaluating this from 0 to π/2 gives sin(π/2) - sin(0) = 1 - 0 = 1. The value of the line integral is 1.

The line integral of a vector field F along a curve C is the integral of the dot product F·T ds, where T is the unit tangent vector to the curve. In this scenario, the vector field F points in the positive x-direction, while the motion is in the positive y-direction. Since the force vector and the direction of motion are perpendicular (orthogonal) to each other at every point, their dot product is zero. The integral of zero over any interval is zero.

We parameterize the unit circle as r(t) = (cos t, sin t) for 0 ≤ t ≤ 2π. The derivative is r'(t) = (-sin t, cos t), so dr = (-sin t, cos t) dt. The vector field evaluated on the curve is F(r(t)) = (cos t, sin t). The dot product F·dr is (cos t)(-sin t) + (sin t)(cos t) = -cos t sin t + sin t cos t = 0. Since the dot product simplifies to zero, the integrand is identically zero. Therefore, the line integral ∫₀^{2π} 0 dt = 0.

This problem uses the Fundamental Theorem of Line Integrals. When a vector field is the gradient of a potential function f, the line integral ∫C ∇f·dr depends only on the endpoints, not the path. First, we compute ∇f: ∂f/∂x = y and ∂f/∂y = x, so ∇f = (y, x). According to the Fundamental Theorem, ∫C ∇f·dr = f(end point) − f(start point). The start point is (0,0) and the end point is (1,1). Evaluating: f(1,1) = 1·1 = 1 and f(0,0) = 0·0 = 0. Subtracting gives 1 − 0 = 1. The integral equals 1 regardless of which path we take from (0,0) to (1,1).

To find a potential function φ such that ∇φ = F, we integrate the components. From φ_x = 2x, we integrate with respect to x to get φ = x² + g(y), where g(y) is an arbitrary function of y. Taking the partial derivative with respect to y gives φ_y = g'(y). But we know φ_y must equal the second component of F, which is 2y. So g'(y) = 2y, which integrates to g(y) = y² + C. We can take C = 0, so φ(x,y) = x² + y². Now using the Fundamental Theorem of Line Integrals, ∫C F·dr = φ(1,2) − φ(0,0) = (1² + 2²) − (0² + 0²) = (1 + 4) − 0 = 5.

Arc length is a special case of a scalar line integral where we integrate 1 ds. We can parameterize the line as r(t) = (t, t) for 0 ≤ t ≤ 2. The derivative is r'(t) = (1, 1). The magnitude is ||r'(t)|| = √(1² + 1²) = √2. The arc length formula is L = ∫C ds = ∫₀² ||r'(t)|| dt. Substituting gives L = ∫₀² √2 dt = √2 ∫₀² dt = √2 [t]₀² = √2 (2 − 0) = 2√2.

First, we parameterize the curve. The simplest parameterization is to let x = t, which means y = t². So r(t) = (t, t²) for 0 ≤ t ≤ 1. From this, we get dx = dt and dy = 2t dt. The integral ∫C x dx + y dy becomes ∫₀¹ [t · dt + t² · (2t dt)]. Simplifying inside the integral gives ∫₀¹ (t + 2t³) dt. We integrate term by term: the integral of t is t²/2 and the integral of 2t³ is t⁴/2. Evaluating from 0 to 1 gives (1²/2 + 1⁴/2) − (0 + 0) = (1/2 + 1/2) = 1.

A conservative vector field has the property that its line integral around any closed path is always zero. This follows from the Fundamental Theorem of Line Integrals. For a conservative field, there exists a potential function φ such that F = ∇φ. For a closed curve C, the start point A and end point B are the same point. Therefore, ∮C F·dr = φ(B) − φ(A) = φ(A) − φ(A) = 0.

A vector field F = (P, Q) is conservative if and only if it satisfies the condition ∂P/∂y = ∂Q/∂x throughout a simply connected region. This condition ensures that the field has zero curl in two dimensions. If F = ∇φ, then P = φ_x and Q = φ_y. Differentiating gives ∂P/∂y = φ_xy and ∂Q/∂x = φ_yx. Since mixed partial derivatives are equal for smooth functions, we get ∂P/∂y = ∂Q/∂x. This condition, along with simple connectivity (no holes in the domain), guarantees that the field is conservative.

Path independence is the defining property of conservative vector fields. When a line integral between two points gives the same result regardless of which path you take, it means the work done by the field depends only on the endpoints, not the route taken. This can only happen if F is the gradient of some scalar potential function φ (F = ∇φ). By the Fundamental Theorem of Line Integrals, ∫C ∇φ·dr = φ(B) − φ(A), which clearly depends only on the endpoints A and B. Therefore, if all paths from A to B give the same integral value, F must be conservative on the region containing these paths.

In calculus, the terms 'conservative vector field' and 'gradient vector field' are equivalent. A vector field F is called conservative precisely when there exists a scalar potential function φ such that F = ∇φ. This means every conservative field is a gradient field, and every gradient field is conservative. The potential function φ represents the potential energy associated with the field. The equivalence holds in any dimension.

We parameterize the line segment from (0,0) to (1,0). Since y is always 0 and x goes from 0 to 1, we can use r(t) = (t, 0) for 0 ≤ t ≤ 1. The derivative is r'(t) = (1, 0) with magnitude ||r'(t)|| = √(1² + 0²) = 1. Therefore, ds = dt. On the curve, x = t, so e^x becomes e^t. The line integral becomes ∫₀¹ e^t dt. The antiderivative of e^t is e^t. Evaluating from 0 to 1 gives e¹ − e⁰ = e − 1.

We evaluate the line integral using the parameterization. First, we find dr by differentiating r(t): r'(t) = (2t, 3t²), so dr = (2t, 3t²) dt. Next, we evaluate the vector field F on the curve: F(r(t)) = (t³, t²). Now we compute the dot product F·dr = (t³)(2t) + (t²)(3t²) = 2t⁴ + 3t⁴ = 5t⁴. The line integral becomes ∫₀¹ 5t⁴ dt. Integrating gives 5 · (t⁵/5) = t⁵ evaluated from 0 to 1, which yields 1 − 0 = 1.

Direct parameterization of this curve would be difficult due to the sine term, but we can check if F is conservative. Compute the 2D curl: ∂(2y)/∂x − ∂(2x)/∂y = 0 − 0 = 0. Since the domain is simply connected, F is conservative. We find a potential function f such that ∇f = (2x, 2y). Integrating 2x with respect to x gives x², and integrating 2y with respect to y gives y², so a potential function is f(x,y) = x² + y² (up to an additive constant). By the Fundamental Theorem of Line Integrals, the work depends only on the endpoints. The start point is (0,0) and the end point is (1,0). The integral is f(1,0) − f(0,0) = (1² + 0²) − (0² + 0²) = 1 − 0 = 1.