MATH 251 Lecture 29

« previous | Monday, April 9, 2012 | next »

Fundamental Theorem of Line Integrals

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

Suppose ${\vec {F}}={\vec {\nabla }}f=f_{x}{\hat {\imath }}+f_{y}{\hat {\jmath }}+f_{z}{\hat {k}}$.

$\int _{C}{\vec {F}}\cdot \mathrm {d} {\vec {r}}=\int _{C}f_{x}\,\mathrm {d} x+f_{y}\,\mathrm {d} y+f_{z}\,\mathrm {d} z=\int _{C}\mathrm {d} f$.

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

${\begin{aligned}{\vec {F}}&=\sum \mathrm {d} f({\vec {v}})\\\sum \mathrm {D} _{\vec {v}}f&=\sum f(P_{i})-f(P_{i-1})\\&=\sum \Delta f_{i}\\&=f(B)-f(A)\end{aligned}}$

Example

Gravitational Force:

${\vec {F}}=-{\frac {GMm}{\left\|{\vec {r}}\right\|^{2}}}{\hat {r}}$

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

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 $GMm\left({\frac {1}{13}}-1\right)$.

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):

${\begin{aligned}U(t)&=-f(x(t),y(t),z(t))=-f(\alpha (t))\\K(t)&={\frac {1}{2}}m\alpha '(t)\cdot \alpha '(t)\end{aligned}}$

If $E(t)=U(t)+K(t)$ is a constant, then $E'(t)=0$:

${\begin{aligned}E(t)&={\frac {1}{2}}m\alpha '(t)\cdot \alpha '(t)-f(\alpha (t))\\E'(t)&=\left[m\alpha ''(t)-{\vec {F}}(\alpha (t))\right]\cdot \alpha '(t)\end{aligned}}$