When Can You Use the Fundamental Theorem of Line Integrals

Evaluates a line integral through a gradient field using the original scalar field

The gradient theorem, besides known as the central 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 key theorem of calculus to any bend in a plane or space (generally n-dimensional) rather than but the real line.

Let φ : U ⊆ R ^{due north} → R be a continuously differentiable function and γ any curve in U which starts at p and ends at 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 i of the ways of defining a conservative force. By placing φ as potential, ∇φ is a bourgeois field. Work done past conservative forces does not depend on the path followed by the object, simply simply the terminate points, every bit the higher up equation shows.

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

Proof [edit]

If φ is a differentiable office from some open up subset U (of R ^{due north} ) to R , and if r is a differentiable function from some closed interval [a, b] to U, then by the multivariate chain rule, the blended function φ ∘ r is differentiable on (a, b) and

$${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}(\varphi \circ \mathbf {r} )(t)=\nabla \varphi (\mathbf {r} (t))\cdot \mathbf {r} '(t)}$$

for all t in (a, b). Hither the ⋅ denotes the usual inner product.

Now suppose the domain U of φ contains the differentiable curve γ with endpoints a and b , (oriented in the management from a to b ). If r parametrizes γ for t in [a, b], then the above shows that ^[1]

$${\displaystyle {\begin{aligned}\int _{\gamma }\nabla \varphi (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} &=\int _{a}^{b}\nabla \varphi (\mathbf {r} (t))\cdot \mathbf {r} '(t)\mathrm {d} t\\&=\int _{a}^{b}{\frac {d}{dt}}\varphi (\mathbf {r} (t))\mathrm {d} t=\varphi (\mathbf {r} (b))-\varphi (\mathbf {r} (a))=\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right),\end{aligned}}}$$

where the definition of the line integral is used in the showtime equality, the to a higher place equation is used in the second equality, and the 2nd fundamental theorem of calculus is used in the 3rd equality.

Examples [edit]

Example ane [edit]

Suppose γ ⊂ R ^two is the circular arc oriented counterclockwise from (5, 0) to (−4, 3). Using the definition of a line integral,

$${\displaystyle {\begin{aligned}\int _{\gamma }y\,\mathrm {d} ten+10\,\mathrm {d} y&=\int _{0}^{\pi -\tan ^{-ane}\!\left({\frac {3}{4}}\right)}((5\sin t)(-5\sin t)+(5\cos t)(v\cos t))\,\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {three}{4}}\right)}25\left(-\sin ^{two}t+\cos ^{2}t\right)\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-i}\!\left({\frac {3}{4}}\right)}25\cos(2t)\mathrm {d} t\ =\ \left.{\tfrac {25}{ii}}\sin(2t)\correct|_{0}^{\pi -\tan ^{-1}\!\left({\tfrac {3}{4}}\right)}\\[.5em]&={\tfrac {25}{2}}\sin \left(2\pi -2\tan ^{-i}\!\!\left({\tfrac {3}{4}}\right)\right)\\[.5em]&=-{\tfrac {25}{2}}\sin \left(2\tan ^{-1}\!\!\left({\tfrac {three}{iv}}\correct)\right)\ =\ -{\frac {25(three/4)}{(3/4)^{two}+1}}=-12.\end{aligned}}}$$

This event can be obtained much more but by noticing that the function ${\displaystyle f(10,y)=xy}$ has slope ${\displaystyle \nabla f(10,y)=(y,x)}$ , so by the Gradient Theorem:

$${\displaystyle \int _{\gamma }y\,\mathrm {d} ten+x\,\mathrm {d} y=\int _{\gamma }\nabla (xy)\cdot (\mathrm {d} x,\mathrm {d} y)\ =\ xy\,|_{(5,0)}^{(-4,iii)}=-four\cdot iii-5\cdot 0=-12.}$$

Case 2 [edit]

For a more abstract example, suppose γ ⊂ R ⁿ has endpoints p , q , with orientation from p to q . For u in R ^northward , let | u | denote the Euclidean norm of u . If α ≥ 1 is a existent number, then

$${\displaystyle {\begin{aligned}\int _{\gamma }|\mathbf {ten} |^{\alpha -ane}\mathbf {x} \cdot \mathrm {d} \mathbf {x} &={\frac {1}{\alpha +1}}\int _{\gamma }(\alpha +one)|\mathbf {x} |^{(\alpha +1)-two}\mathbf {x} \cdot \mathrm {d} \mathbf {x} \\&={\frac {1}{\blastoff +1}}\int _{\gamma }\nabla |\mathbf {10} |^{\blastoff +one}\cdot \mathrm {d} \mathbf {x} ={\frac {|\mathbf {q} |^{\alpha +one}-|\mathbf {p} |^{\alpha +ane}}{\blastoff +1}}\terminate{aligned}}}$$

Hither the final equality follows past the slope theorem, since the part f(x) = | x | ^α+one is differentiable on R ⁿ if α ≥ 1.

If α < i then this equality will withal hold in most cases, but circumspection must be taken if γ passes through or encloses the origin, because the integrand vector field | 10 | ^{α − 1} 10 will fail to be defined there. However, the case α = −1 is somewhat dissimilar; in this example, the integrand becomes | 10 |⁻² x = ∇(log | x |), then that the final equality becomes log | q | − log | p |.

Note that if n = i, and then this example is just a slight variant of the familiar power rule from single-variable calculus.

Example 3 [edit]

Suppose there are n bespeak charges bundled in three-dimensional space, and the i-thursday point accuse has charge Q _i and is located at position p _i in R ^three . We would similar to calculate the work done on a particle of charge q equally it travels from a point a to a indicate b in R ^three . Using Coulomb's law, we tin hands make up one's mind that the force on the particle at position r will be

$${\displaystyle \mathbf {F} (\mathbf {r} )=kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{iii}}}}$$

Here | u | denotes the Euclidean norm of the vector u in R ³ , and k = 1/(4πε ₀), where ε ₀ is the vacuum permittivity.

Allow γ ⊂ R ³ − {p ₁, ..., p _n } exist an arbitrary differentiable curve from a to b . Then the piece of work washed on the particle is

$${\displaystyle W=\int _{\gamma }\mathbf {F} (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\int _{\gamma }\left(kq\sum _{i=1}^{n}{\frac {Q_{i}(\mathbf {r} -\mathbf {p} _{i})}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{3}}}\right)\cdot \mathrm {d} \mathbf {r} =kq\sum _{i=1}^{north}\left(Q_{i}\int _{\gamma }{\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\correct|^{3}}}\cdot \mathrm {d} \mathbf {r} \right)}$$

Now for each i, directly ciphering shows that

$${\displaystyle {\frac {\mathbf {r} -\mathbf {p} _{i}}{\left|\mathbf {r} -\mathbf {p} _{i}\right|^{iii}}}=-\nabla {\frac {1}{\left|\mathbf {r} -\mathbf {p} _{i}\right|}}.}$$

Thus, continuing from to a higher place and using the gradient theorem,

$${\displaystyle Westward=-kq\sum _{i=ane}^{north}\left(Q_{i}\int _{\gamma }\nabla {\frac {i}{\left|\mathbf {r} -\mathbf {p} _{i}\correct|}}\cdot \mathrm {d} \mathbf {r} \right)=kq\sum _{i=1}^{n}Q_{i}\left({\frac {ane}{\left|\mathbf {a} -\mathbf {p} _{i}\right|}}-{\frac {one}{\left|\mathbf {b} -\mathbf {p} _{i}\correct|}}\right)}$$

We are finished. Of course, we could have easily completed this calculation using the powerful language of electrostatic potential or electrostatic potential free energy (with the familiar formulas W = −ΔU = −qΔ5 ). However, we take not yet defined potential or potential energy, because the converse of the gradient theorem is required to show that these are well-defined, differentiable functions and that these formulas hold (see below). Thus, nosotros have solved this problem using merely Coulomb's Law, the definition of piece of work, and the gradient theorem.

Converse of the gradient theorem [edit]

The gradient theorem states that if the vector field F is the gradient of some scalar-valued office (i.due east., if F is conservative), and so F is a path-independent vector field (i.e., the integral of F over some piecewise-differentiable bend is dependent only on end points). This theorem has a powerful antipodal:

Theorem  — If F is a path-independent vector field, and then F is the gradient of some scalar-valued function.^[2]

It is straightforward to show that a vector field is path-independent if and only if the integral of the vector field over every closed loop in its domain is zero. Thus the converse tin can alternatively be stated as follows: If the integral of F over every closed loop in the domain of F is zero, and then F is the gradient of some scalar-valued function.

Proof of the converse

Suppose U is an open, path-connected subset of R ^north , and F : U → R ⁿ is a continuous and path-contained vector field. Fix some element a of U, and define f : U → R by

$${\displaystyle f(\mathbf {x} ):=\int _{\gamma [\mathbf {a} ,\mathbf {x} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }$$

Hither γ[a, x] is whatever (differentiable) curve in U originating at a and terminating at 10 . We know that f is well-defined because F is path-contained.

Permit v be any nonzero vector in R ⁿ . By the definition of the directional derivative,

$${\displaystyle {\begin{aligned}{\frac {\partial f}{\partial \mathbf {5} }}(\mathbf {ten} )&=\lim _{t\to 0}{\frac {f(\mathbf {ten} +t\mathbf {v} )-f(\mathbf {x} )}{t}}\\&=\lim _{t\to 0}{\frac {\int _{\gamma [\mathbf {a} ,\mathbf {ten} +t\mathbf {v} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} -\int _{\gamma [\mathbf {a} ,\mathbf {ten} ]}\mathbf {F} (\mathbf {u} )\cdot d\mathbf {u} }{t}}\\&=\lim _{t\to 0}{\frac {1}{t}}\int _{\gamma [\mathbf {10} ,\mathbf {x} +t\mathbf {five} ]}\mathbf {F} (\mathbf {u} )\cdot \mathrm {d} \mathbf {u} \end{aligned}}}$$

To calculate the integral within the concluding limit, we must parametrize γ[x, ten + t five]. Since F is path-independent, U is open, and t is approaching cipher, we may presume that this path is a direct line, and parametrize it as u(s) = 10 + southward five for 0 < due south < t . Now, since u'(southward) = v , the limit becomes

$${\displaystyle \lim _{t\to 0}{\frac {one}{t}}\int _{0}^{t}\mathbf {F} (\mathbf {u} (s))\cdot \mathbf {u} '(south)\,\mathrm {d} s={\frac {\mathrm {d} }{\mathrm {d} t}}\int _{0}^{t}\mathbf {F} (\mathbf {ten} +due south\mathbf {v} )\cdot \mathbf {v} \,\mathrm {d} s{\bigg |}_{t=0}=\mathbf {F} (\mathbf {10} )\cdot \mathbf {5} }$$

Thus we accept a formula for ∂ _v f , where five is arbitrary. Let 10 = (x ₁, x _two, ..., x_{due north} ) and let e _i announce the i-thursday standard basis vector, so that

$${\displaystyle \nabla f(\mathbf {x} )={\bigg (}{\frac {\fractional f(\mathbf {x} )}{\partial x_{i}}},{\frac {\fractional f(\mathbf {10} )}{\partial x_{2}}},\dots ,{\frac {\partial f(\mathbf {ten} )}{\fractional x_{n}}}{\bigg )}=(\mathbf {F} (\mathbf {x} )\cdot \mathbf {eastward} _{1},\mathbf {F} (\mathbf {x} )\cdot \mathbf {east} _{ii},\dots ,\mathbf {F} (\mathbf {x} )\cdot \mathbf {e} _{n})=\mathbf {F} (\mathbf {x} )}$$

Thus we have constitute a scalar-valued function f whose slope is the path-independent vector field F , as desired.^[2]

Instance of the converse principle [edit]

To illustrate the power of this converse principle, we cite an case that has pregnant physical consequences. In classical electromagnetism, the electric force is a path-independent force; i.e. the piece of work done on a particle that has returned to its original position within an electric field is aught (assuming that no changing magnetic fields are nowadays).

Therefore, the above theorem implies that the electric force field F _east  : Due south → R ³ is conservative (hither Due south is some open up, path-connected subset of R ⁱⁱⁱ that contains a charge distribution). Following the ideas of the above proof, we can gear up some reference point a in S, and ascertain a office U_e : S → R by

$${\displaystyle U_{e}(\mathbf {r} ):=-\int _{\gamma [\mathbf {a} ,\mathbf {r} ]}\mathbf {F} _{e}(\mathbf {u} )\cdot \mathrm {d} \mathbf {u} }$$

Using the above proof, we know U _e is well-defined and differentiable, and F _e = −∇U_e (from this formula we can utilise the slope theorem to easily derive the well-known formula for calculating work done by conservative forces: Westward = −ΔU ). This function U _eastward is oft referred to as the electrostatic potential energy of the system of charges in Due south (with reference to the zero-of-potential a ). In many cases, the domain South is assumed to exist unbounded and the reference point a is taken to be "infinity", which tin can be made rigorous using limiting techniques. This function U _e is an indispensable tool used in the analysis of many physical systems.

Generalizations [edit]

Many of the critical theorems of vector calculus generalize elegantly to statements nearly the integration of differential forms on manifolds. In the language of differential forms and outside derivatives, the gradient theorem states that

$${\displaystyle \int _{\partial \gamma }\phi =\int _{\gamma }\mathrm {d} \phi }$$

for any 0-form, ϕ, divers on some differentiable curve γ ⊂ R ⁿ (here the integral of ϕ over the purlieus of the γ is understood to be the evaluation of ϕ at the endpoints of γ).

Notice the striking similarity between this argument and the generalized version of Stokes' theorem, which says that the integral of whatever compactly supported differential form ω over the boundary of some orientable manifold Ω is equal to the integral of its exterior derivative dω over the whole of Ω, i.e.,

$${\displaystyle \int _{\partial \Omega }\omega =\int _{\Omega }\mathrm {d} \omega }$$

This powerful statement is a generalization of the gradient theorem from ane-forms divers on one-dimensional manifolds to differential forms defined on manifolds of arbitrary dimension.

The converse statement of the slope theorem also has a powerful generalization in terms of differential forms on manifolds. In particular, suppose ω is a course defined on a contractible domain, and the integral of ω over any closed manifold is zero. Then there exists a course ψ such that ω = dψ . Thus, on a contractible domain, every closed form is verbal. This result is summarized by the Poincaré lemma.

See besides [edit]

• State part
• Scalar potential
• Jordan curve theorem
• Differential of a function
• Classical mechanics
• Line integral § Path independence
• Conservative vector field § Path independence

References [edit]

1. ^ Williamson, Richard and Trotter, Hale. (2004). Multivariable Mathematics, Quaternary Edition, p. 374. Pearson Instruction, Inc.
2. ^ ^{a b} "Williamson, Richard and Trotter, Unhurt. (2004). Multivariable Mathematics, Fourth Edition, p. 410. Pearson Education, Inc."

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Source: https://en.wikipedia.org/wiki/Gradient_theorem

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