# 6.5 Vector fields, differential forms, and line integrals

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We motivate our third definition of an integral over a curve by returning to physics.This definition is very much a real variable one, so that we think of the plane as ${R}^{2}$ instead of $C.$ A connection between this real variable definition and the complex variable definition of a contour integral will emerge later.

We motivate our third definition of an integral over a curve by returning to physics.This definition is very much a real variable one, so that we think of the plane as ${R}^{2}$ instead of $C.$ A connection between this real variable definition and the complex variable definition of a contour integral will emerge later.

By a vector field on an open subset $U$ of ${R}^{2},$ we mean nothing more than a continuous function $\stackrel{\to }{V}\left(x,y\right)\equiv \left(P\left(x,y\right),Q\left(x,y\right)\right)$ from $U$ into ${R}^{2}.$ The functions $P$ and $Q$ are called the components of the vector field $\stackrel{\to }{V}.$

We will also speak of smooth vector fields, by which we will mean vector fields $\stackrel{\to }{V}$ both of whose component functions $P$ and $Q$ have continuous partial derivatives

$\frac{tialP}{tialx},\frac{tialP}{tialy},\frac{tialQ}{tialx}and\frac{tialQ}{tialy}$

on $U.$

The idea from physics is to think of a vector field as a force field, i.e., something thatexerts a force at the point $\left(x,y\right)$ with magnitude $|\stackrel{\to }{V}\left(x,y\right)|$ and acting in the direction of the vector $\stackrel{\to }{V}\left(x,y\right).$ For a particle to move within a force field, “work” must be done, that is energy must be provided to move the particle against the force,or energy is given to the particle as it moves under the influence of the force field. In either case, the basicdefinition of work is the product of force and distance traveled. More precisely, if a particle is moving in a direction $\stackrel{\to }{u}$ within a force field, then the work done on the particle is the product of the component of the force field in the direction of $\stackrel{\to }{u}$ and the distance traveled by the particle in that direction. That is, we must compute dot products of the vectors $\stackrel{\to }{V}\left(x,y\right)$ and $\stackrel{\to }{u}\left(x,y\right).$ Therefore, if a particle is moving along a curve $C,$ parameterized with respect to arc length by $\gamma :\left[0,L\right]\to C,$ and we write $\gamma \left(t\right)=\left(x\left(t\right),y\left(t\right)\right),$ then the work $W\left({z}_{1},{z}_{2}\right)$ done on the particle as it moves from ${z}_{1}=\gamma \left(0\right)$ to ${z}_{2}=\gamma \left(L\right)$ within the force field $\stackrel{\to }{V},$ should intuitively be given by the formula

$\begin{array}{ccc}\hfill W\left({z}_{1},{z}_{2}\right)& =& {\int }_{0}^{L}⟨\stackrel{\to }{V}\left(\gamma \left(t\right)\right)\mid {\gamma }^{\text{'}}\left(t\right)⟩\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & =& {\int }_{0}^{L}P\left(x\left(t\right),y\left(t\right)\right){x}^{\text{'}}\left(t\right)+Q\left(x\left(t\right),y\left(t\right)\right){y}^{\text{'}}\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt\hfill \\ & \equiv & {\int }_{C}P\phantom{\rule{0.166667em}{0ex}}dx+Q\phantom{\rule{0.166667em}{0ex}}dy,\hfill \end{array}$

where the last expression is explicitly defining the shorthand notation we will be using.

The preceding discussion leads us to a new notion of what kind of object should be “integrated” over a curve.

A differential form on a subset $U$ of ${R}^{2}$ is denoted by $\omega =Pdx+Qdy,$ and is determined by two continuous real-valued functions $P$ and $Q$ on $U.$ We say that $\omega$ is bounded or uniformly continuous if the functions $P$ and $Q$ are bounded or uniformly continuous functions on $U.$ We say that the differential form $\omega$ is smooth of order $k$ if the set $U$ is open, and the functions $P$ and $Q$ have continuous mixed partial derivatives of order $k.$

If $\omega =Pdx+Qdy$ is a differential form on a set $U,$ and if $C$ is any piecewise smooth curve of finite length contained in $U,$ then we define the line integral ${\int }_{C}\omega$ of $\omega$ over $C$ by

${\int }_{C}\omega ={\int }_{C}P\phantom{\rule{0.166667em}{0ex}}dx+Q\phantom{\rule{0.166667em}{0ex}}dy={\int }_{0}^{L}P\left(\gamma \left(t\right)\right){x}^{\text{'}}\left(t\right)+Q\left(\gamma \left(t\right)\right){y}^{\text{'}}\left(t\right)\phantom{\rule{0.166667em}{0ex}}dt,$

where $\gamma \left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$ is a parameterization of $C$ by arc length.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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not now but maybe in future only AgNP maybe any other nanomaterials
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