MATH 251 Lecture 29

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Fundamental Theorem of Line Integrals

(See Vector Analysis Theorems#Fundamental Theorem of Calculus for Curves→)

Suppose $\overrightarrow{F}=\overrightarrow{\mathrm{\nabla }}f=f_x\hat{ı}+f_y\hat{ȷ}+f_z\hat{k}$.

$\underset{C}{\int }\overrightarrow{F}\cdot \mathrm{d}\overrightarrow{r}=\underset{C}{\int }{f}_x\mathrm{d}x+f_y\mathrm{d}y+f_z\mathrm{d}z=\underset{C}{\int }\mathrm{d}f$.

$F$ if conservative since it is a vector field that represents the gradient of a function $f$. Therefore,

$\begin{array}{rl}\overrightarrow{F}& =\sum \mathrm{d}f(\overrightarrow{v})\\ \sum {\mathrm{D}}_{\overrightarrow{v}}f& =\sum f(P_i)-f(P_{i-1})\\ & =\sum \mathrm{\Delta }{f}_i\\ & =f(B)-f(A)\end{array}$

Example

Gravitational Force:

$\overrightarrow{F}=-\frac{GMm}{{\Vert \overrightarrow{r}\Vert }^2}\hat{r}$

Find the potential function $f$ for $\overrightarrow{F}$. From physics, we already know this is $\frac{GMm}{\Vert r\Vert }$. This is correct since $\overrightarrow{\mathrm{\nabla }}f=\overrightarrow{F}$.

Therefore, gravity is a conservative force, and the work done by gravity to move a particle $m$ from (1,0,0) to (0,5,12), where $M$ is at the origin, is $GMm(\frac{1}{13}-1)$.

Since gravity is conservative, the total energy should be conserved. Total energy is the sum of potential and kinetic energy (assume that $\alpha (t)$ gives the position):

$\begin{array}{rl}U(t)& =-f(x(t),y(t),z(t))=-f(\alpha (t))\\ K(t)& =\frac{1}{2}m{\alpha }^{\prime }(t)\cdot {\alpha }^{\prime }(t)\end{array}$

If $E(t)=U(t)+K(t)$ is a constant, then $E^{\prime }(t)=0$:

$\begin{array}{rl}E(t)& =\frac{1}{2}m{\alpha }^{\prime }(t)\cdot {\alpha }^{\prime }(t)-f(\alpha (t))\\ E^{\prime }(t)& =[m{\alpha }^{″}(t)-\overrightarrow{F}(\alpha (t))]\cdot {\alpha }^{\prime }(t)\end{array}$