Conservative Vector Fields

Conservative Vector Fields and Independence of Path


Master of Science in Mathematics
Lecture Notes. Accessed on: 2019-10-21 17:50:03

Independence of Path

Definition. The line integral is called independent of path if in a region $D$, if for any two points $P$ and $Q$ in $D$ then the line integral along every piecewise smooth curve in $D$ from $P$ to $Q$ has the same value.

Theorem. (Independence of Path) If $V$ is a continuous vector field on the open connected set $D,$ then the following three conditions are either all true or all false:

$(1) \quad$ $V$ is conservative on $D$

$(2) \quad$ $\int_C c{V}\cdot d{R}=0$ for every piecewise smooth closed curve $C$ in $D.$

$(3) \quad$ $\int_C {V}\cdot d{R}$ is independent of path within $D.$

Example. Let $F=\langle y,-x\rangle$ and let $C_1$ and $C_2$ be the following two paths joining $(0,0)$ to $(1,1)$; $C_1$: $y=x$ for $0\leq x\leq 1$ and $C_2$: $y=x^2$ for $0\leq x\leq 1.$ Show that $$ \int_{C_1}{F}\cdot d {R}\neq \int_{C_2}{F}\cdot d {R}. $$ Explain what this means? On $C_1:$ $x=t,$ $y=t$ with $0\leq t\leq 1,$ $$ \int_{C_1} y \, dx-x \, dy=\int_0^1 (t-t) \, dt=0. $$ On $C_2:$ $x=t,$ $y=t^2,$ with $0\leq t\leq 1,$ $$ \int_{C_2}y \,dx-x\,dy=\int_0^1 \left(t^2-t(2t)\right) \, dt=-\frac{1}{3}. $$

Solution. Since these two line integrals do not have the same value we see that not all line integrals are independent of path.

Fundamental Theorem of Line Integrals

Now we are ready for a generalization of the Fundamental Theorem of Calculus.

Theorem. (Fundamental Theorem of Line Integrals) Let $C$ be a piecewise smooth curve that is parametrized by the vector function ${R}(t)$ for ${a\leq t\leq b}$ and let ${V}$ be a vector field that is continuous on $C.$ If $f$ is a scalar function such that ${V}=\nabla f,$ then $$ \int_C {V}\cdot d{R}=f(Q)-f(P)$$ where $Q=R(b)$ and $P=R(a)$ are the endpoints of $C.$

Proof. Suppose ${R}(t)=x(t) {i}+y(t) {j}+z(t) {k}$ and let $G$ be the composite function $G(t)=f(x(t),y(t),z(t)).$ We have \begin{align} \int_C {F}\cdot d{R} & =\int_C\nabla f\cdot d{R} =\int_C \left[\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\right] \\ & =\int_a^b \left[\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}\right] \, dt \\ & =\int_a^b \frac{d G}{d t} \, dt =G(b)-G(a) =f(Q)-f(P). \end{align}

Example. Show that the vector field $$ {F}(x,y,z)=\left\langle \frac{y}{1+x^2}+\tan ^{-1}z,\tan ^{-1}x,\frac{x}{1+z^2}\right\rangle, $$ is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R}. $$ where $C$ is any piecewise smooth path connecting $A(1,0,-1)$ to $B(0,-1,1).$

Solution. Since $\mathop{curl} {V}={0},$ ${V}$ is conservative. Now we set out to find $f.$ Since $$ \frac{\partial f}{\partial x}=\frac{y}{1+x^2}+\tan ^{-1}z $$ we set $$ f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c(y,z). $$ Since $$ \frac{\partial f}{\partial y}=\tan ^{-1}z=\frac{\partial }{\partial y}\left(y \tan ^{-1}x+x \tan ^{-1}z+c\right)=\tan ^{-1}x+\frac{\partial c}{\partial y}, $$ so $\frac{\partial c}{\partial y}=0$ and $c=c_1(z)$ and so we set $f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c_1(z).$ Since $$ \frac{\partial f}{\partial z}=\frac{x}{1+z^2}=\frac{\partial }{\partial z}\left[y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\right]=\frac{x}{1+z^2}+c’_1(z),$$ so $c_1′(z)=0,$ $ c_1=0$ and so we set $ f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z.$ \begin{equation} \int_C{F}\cdot d {R}=f(1,0,-1)-f(0,-1,1) =\frac{3\pi }{4}+\frac{\pi }{4} =\pi . \end{equation}

Example. Show that the vector field $$ {V}(x,y,z)=\left(x y^2+y z\right){i}+\left(x^2y+x z+3y^2z\right){j}+\left(x y+y^3\right){k} $$ is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R}. $$ where $C$ is any piecewise smooth path joining $A(8,6,1)$ to $B(1,3,5).$

Solution. First we determine $\mathop{curl} {V}.$ We find \begin{align} \mathop{curl} {V} & =\left| \begin{array}{cc} \begin{array}{c} {i} \\ \begin{array}{c} \partial / \partial x \\ x y^2+y z \end{array} \end{array} & \begin{array}{cc} {j} & {k} \\ \begin{array}{c} \partial / \partial y \\ x^2y+x z+3y^2z \end{array} & \begin{array}{c} \partial / \partial z \\ x y+y^3 \end{array} \end{array} \end{array} \right| \\ & =\left(x+3y^2-x-3y^2\right){i}-(y-y){j}+(2x y+z-2x y-z){k} ={0}\end{align} and so the vector field is conservative and we can find the scalar potential function. Now we set out to find $f$ with $\nabla f={V}.$ Since $f_x=x y^2+y z$ we know that $$ f(x,y,z)=\frac{x^2}{2}y^2+x y z+C_1(y,z). $$ Then we find $$ f_y=x^2y+x z+\frac{ \partial C_1}{\partial y}; $$ and so comparing this with the given $x^2y+x z+3y^2z$ we determine that $$ \frac{ \partial C_1}{\partial y}=3y^2z. $$ So $C_1=y^3z+C_2(z).$ So far we have $$ f=\frac{x^2}{2}y^2+x y z+y^3z+C_2(z). $$ Also since $f_z=x y+y^3+\frac{dC_2}{d z}$ and comparing this to the given $x y+y^3$ we determine $\frac{dC_2}{dz}=0.$ So $C_2$ is a constant with respect to $x,$ $y,$ and $z.$ Therefore a scalar potential function is $$ f(x,y,z)=\frac{x^2}{2}y^2+x y z+y^3z$$ (taking the constant to be zero). Finally \begin{equation} \int_C {V}\cdot d{R}=f(8,6,1)-f(1,3,5)=\frac{2523}{2}. \end{equation}

Example. Show that the vector field is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R} = {F}(x,y)=\frac{(y+1){i}-x {j}}{(y+1)^2} $$ where $C$ is any smooth path connecting $A(0,0)$ to $B(1,1).$

Solution. Let ${F}(x,y)=u(x,y){i}+v(x,y){j}$, then since $$ \frac{\partial u}{\partial y}=\frac{-1}{(y+1)^2}=\frac{\partial v}{\partial x}, $$ we know that $F$ is conservative. By definition, we know $$ f_x(x,y)=\frac{1}{y+1} \qquad \text{and}\qquad f_y(x,y)=\frac{-x}{(y+1)^2}. $$ Integrating with respect to $x,$ $f(x,y)=\frac{x}{y+1}+c(y).$ Since, $$ f_y(x,y)=-\frac{x}{(y+1)^2}+c'(y)=-\frac{x}{(y+1)^2}, $$ so $c'(y)=0.$ For $c(y)=0;$ and then $f(x,y)=\frac{x}{y+1}$ is a scalar potential function for ${F}.$ By the fundamental theorem of line integrals, \begin{equation} \int_C {F}\cdot d{R}=f(1,1)-f(0,0)=\frac{1}{2}-0=\frac{1}{2}. \end{equation}

Work in a Conservative Vector Field

Example. Find the work done an object moves in the force field $$ {F}(x,y,z)=x {i}+y {j}+(x z-y){k} $$ along the curve $C$ defined parametrically by ${R}(t)=t^2{i}+2t {j}+4t^3 {k}$ for $0\leq t\leq 1.$

Solution. We determine \begin{align} & d{R}=\left(2t {i}+2{j}+12t^2{k}\right)dt, \qquad {F}(t)=t^2{i}+2t{j}+\left(4t^5-2t\right){k}, \end{align}
(from $x(t)=t^2, y(t)=2t,$ and $z=4t^3$) and $$ {F}\cdot d{R}=\left(2t^3+4t+48t^7-24t^3\right)dt. $$ Thus, \begin{equation} W=\int_C {F}\cdot d{R}=\int_0^1 \left(2t^3+4t+48t^7-24t^3\right) \, dt=\frac{5}{2}. \end{equation}

Example. Find the work done when an object moves along a closed path in a connected domain where the force field is conservative.

Solution. In such a force field ${F},$ where $f$ is a scalar potential of ${F}$ and because the path of motion is closed, it begins and ends at the same point $P.$ Thus, the work is given by $$ W=\oint _C {F}\cdot d{R}=f(P)-f(P)=0. $$

Example. Let $$ {F}(x,y)=\frac{-y {i}+x {j}}{x^2+y^2}. $$ Evaluate the line integral $$ \int_{C_1}{F}\cdot d {R} $$ where $C_1$ is the upper semicircle $y=\sqrt{1-x^2}$ transversed counterclockwise. What is the value of $$ \int_{C_2}{F}\cdot d {R} $$ if $C_2$ is the lower semicircle $y=-\sqrt{1-x^2}$ also transversed counterclockwise?

Solution. On the upper semi-circle $C_1$, we let $x=\cos (t)$ and $y=\sin (t)$ for $0\leq t\leq \pi ,$ and then $$ \int_{C_1} \frac{-y \, dx+x \, dy}{x^2+y^2}=\int_0^{\pi } \frac{-\sin (t)(-\sin (t))+\cos (t)(\cos (t))}{\sin ^2(t)+\cos ^2(t)} \, dt=\int_0^{\pi }dt=\pi . $$ For $\pi \leq t\leq 2\pi ,$ we have the lower semi-circle $C_2$: \begin{equation} \int{C_2}\frac{-y\, dx+x\, dy}{x^2+y^2}=\int_{\pi }^{2\pi }dt=\pi . \end{equation}

Finding Area with Line Integral

Theorem. If $R$ is a region bounded by a piecewise smooth simple closed curve $C$ oriented counterclockwise, then the area of $R$ is given by $$ A=\oint _C x dy=-\oint _C y dx=\frac{1}{2}\oint _C x dy-y dx. $$

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