Gradient theorem Part of a series of articles about Calculus Fundamental theorem Leibniz integral rule Limits of functionsContinuity Mean value theoremRolle's theorem show Differential show Integral show Series hide Vector GradientDivergenceCurlLaplacianDirectional derivativeIdentities Theorems GradientGreen'sStokes'Divergencegeneralized Stokes show Multivariable show Advanced show Specialized show Miscellaneous vte The gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve. The theorem is a generalization of the second fundamental theorem of calculus to any curve in a plane or space (generally n-dimensional) rather than just the real line.

For φ : U ⊆ Rn → R as a differentiable function and γ as any continuous curve in U which starts at a point p and ends at a point q, then {displaystyle int _{gamma }nabla varphi (mathbf {r} )cdot mathrm {d} mathbf {r} =varphi left(mathbf {q} right)-varphi left(mathbf {p} right)} where ∇φ denotes the gradient vector field of φ.

The gradient theorem implies that line integrals through gradient fields are path-independent. In physics this theorem is one of the ways of defining a conservative force. By placing φ as potential, ∇φ is a conservative field. Work done by conservative forces does not depend on the path followed by the object, but only the end points, as the above equation shows.

The gradient theorem also has an interesting converse: any path-independent vector field can be expressed as the gradient of a scalar field. Just like the gradient theorem itself, this converse has many striking consequences and applications in both pure and applied mathematics.

Contents 1 Proof 2 Examples 2.1 Example 1 2.2 Example 2 2.3 Example 3 3 Converse of the gradient theorem 3.1 Proof of the converse 3.2 Example of the converse principle 4 Generalizations 5 See also 6 References Proof If φ is a differentiable function from some open subset U ⊆ Rn to R and r is a differentiable function from some closed interval [a, b] to U (Note that r is differentiable at the interval endpoints a and b. To do this, r is defined on an interval that is larger than and includes [a, b].), then by the multivariate chain rule, the composite function φ ∘ r is differentiable on [a, b]: {displaystyle {frac {mathrm {d} }{mathrm {d} t}}(varphi circ mathbf {r} )(t)=nabla varphi (mathbf {r} (t))cdot mathbf {r} '(t)} for all t in [a, b]. Here the ⋅ denotes the usual inner product.

Now suppose the domain U of φ contains the differentiable curve γ with endpoints p and q. (This is oriented in the direction from p to q). If r parametrizes γ for t in [a, b] (i.e., r represents γ as a function of t), then {displaystyle {begin{aligned}int _{gamma }nabla varphi (mathbf {r} )cdot mathrm {d} mathbf {r} &=int _{a}^{b}nabla varphi (mathbf {r} (t))cdot mathbf {r} '(t)mathrm {d} t\&=int _{a}^{b}{frac {d}{dt}}varphi (mathbf {r} (t))mathrm {d} t=varphi (mathbf {r} (b))-varphi (mathbf {r} (a))=varphi left(mathbf {q} right)-varphi left(mathbf {p} right),end{aligned}}} where the definition of a line integral is used in the first equality, the above equation is used in the second equality, and the second fundamental theorem of calculus is used in the third equality.[1] Even if the gradient theorem (also called fundamental theorem of calculus for line integrals) has been proved for a differentiable (so looked as smooth) curve so far, the theorem is also proved for a piecewise-smooth curve since this curve is made by joining multiple differentiable curves so the proof for this curve is made by the proof per differentiable curve component.[2] Examples Example 1 Suppose γ ⊂ R2 is the circular arc oriented counterclockwise from (5, 0) to (−4, 3). Using the definition of a line integral, {displaystyle {begin{aligned}int _{gamma }y,mathrm {d} x+x,mathrm {d} y&=int _{0}^{pi -tan ^{-1}!left({frac {3}{4}}right)}((5sin t)(-5sin t)+(5cos t)(5cos t)),mathrm {d} t\&=int _{0}^{pi -tan ^{-1}!left({frac {3}{4}}right)}25left(-sin ^{2}t+cos ^{2}tright)mathrm {d} t\&=int _{0}^{pi -tan ^{-1}!left({frac {3}{4}}right)}25cos(2t)mathrm {d} t = left.{tfrac {25}{2}}sin(2t)right|_{0}^{pi -tan ^{-1}!left({tfrac {3}{4}}right)}\[.5em]&={tfrac {25}{2}}sin left(2pi -2tan ^{-1}!!left({tfrac {3}{4}}right)right)\[.5em]&=-{tfrac {25}{2}}sin left(2tan ^{-1}!!left({tfrac {3}{4}}right)right) = -{frac {25(3/4)}{(3/4)^{2}+1}}=-12.end{aligned}}} This result can be obtained much more simply by noticing that the function {displaystyle f(x,y)=xy} has gradient {displaystyle nabla f(x,y)=(y,x)} , so by the Gradient Theorem: {displaystyle int _{gamma }y,mathrm {d} x+x,mathrm {d} y=int _{gamma }nabla (xy)cdot (mathrm {d} x,mathrm {d} y) = xy,|_{(5,0)}^{(-4,3)}=-4cdot 3-5cdot 0=-12.} Example 2 For a more abstract example, suppose γ ⊂ Rn has endpoints p, q, with orientation from p to q. For u in Rn, let |u| denote the Euclidean norm of u. If α ≥ 1 is a real number, then {displaystyle {begin{aligned}int _{gamma }|mathbf {x} |^{alpha -1}mathbf {x} cdot mathrm {d} mathbf {x} &={frac {1}{alpha +1}}int _{gamma }(alpha +1)|mathbf {x} |^{(alpha +1)-2}mathbf {x} cdot mathrm {d} mathbf {x} \&={frac {1}{alpha +1}}int _{gamma }nabla |mathbf {x} |^{alpha +1}cdot mathrm {d} mathbf {x} ={frac {|mathbf {q} |^{alpha +1}-|mathbf {p} |^{alpha +1}}{alpha +1}}end{aligned}}} Here the final equality follows by the gradient theorem, since the function f(x) = |x|α+1 is differentiable on Rn if α ≥ 1.