Path integral examples

Example 1

Let

c(t) = (3t - 2, t + 1),        1$\displaystyle \le$t$\displaystyle \le$2    

be a parameterization of a wire. Let the density of the wire at point (x, y) be given by

f (x, y) = x + y.    

Compute the mass of the wire.

Solution: Mass is the integral of the density along the wire. So we must compute $ \int_{{\mathbf{c}}}^{}$f ds. Since,

c'(t) = (3, 1),    
|c'(t)| = $\displaystyle \sqrt{{3^2+1^2}}$ = $\displaystyle \sqrt{{10}}$,    
f (c(t)) = (3t - 2) + (t + 1) = 4t - 1,    

the integral is

$\displaystyle \int_{{\mathbf{c}}}^{}$f ds = $\displaystyle \int_{a}^{b}$f (c(t))|c'(t)| dt    
  = $\displaystyle \int_{{1}}^{2}$(4t - 1)$\displaystyle \sqrt{{10}}$dt    
  = (2t2 - t)$\displaystyle \sqrt{{10}}$$\displaystyle \Big\vert _{{1}}^{2}$    
  = $\displaystyle \left(\vphantom{8-2-(2-1)}\right.$8 - 2 - (2 - 1)$\displaystyle \left.\vphantom{8-2-(2-1)}\right)$$\displaystyle \sqrt{{10}}$ = 5$\displaystyle \sqrt{{10}}$    

So, if f were given in grams/cm and c(t) were given in cm, then the mass of the wire would 5$ \sqrt{{10}}$ grams.

Example 2

Neither the length of the wire nor its mass can depend on parametrization. Check that you get the same answer with the parametrization

p(t) = (9t - 2, 3t + 1),        1/3$\displaystyle \le$t$\displaystyle \le$2/3.    

Not that this is the same wire as in Example 1, which is a straight line from the point (1, 2) to (4, 3).

Solution: We repeat the same calculations as in Example 1. (Recall that f (x, y) = x + y.) Since

p'(t) = (9, 3),    
|p'(t)| = $\displaystyle \sqrt{{9^2+3^2}}$ = $\displaystyle \sqrt{{90}}$ = 3$\displaystyle \sqrt{{10}}$,    
f (p(t)) = (9t - 2) + (3t + 1) = 12t - 1,    

the integral is

$\displaystyle \int_{{\mathbf{p}}}^{}$f ds = $\displaystyle \int_{{1/3}}^{{2/3}}$f (p(t))|p'(t)| dt    
  = $\displaystyle \int_{{1/3}}^{{2/3}}$(12t - 1)3$\displaystyle \sqrt{{10}}$dt    
  = (6t2 - t)3$\displaystyle \sqrt{{10}}$$\displaystyle \Big\vert _{{1/3}}^{{2/3}}$    
  = $\displaystyle \left[\vphantom{6\left(\frac{2}{3}\right)^2-\frac{2}{3} -\left(6\left(\frac{1}{3}\right)^2 -\frac{1}{3}\right)}\right.$6$\displaystyle \left(\vphantom{\frac{2}{3}}\right.$$\displaystyle {\frac{{2}}{{3}}}$$\displaystyle \left.\vphantom{\frac{2}{3}}\right)^{2}_{}$ - $\displaystyle {\frac{{2}}{{3}}}$ - $\displaystyle \left(\vphantom{6\left(\frac{1}{3}\right)^2 -\frac{1}{3}}\right.$6$\displaystyle \left(\vphantom{\frac{1}{3}}\right.$$\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left.\vphantom{\frac{1}{3}}\right)^{2}_{}$ - $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left.\vphantom{6\left(\frac{1}{3}\right)^2 -\frac{1}{3}}\right)$$\displaystyle \left.\vphantom{6\left(\frac{2}{3}\right)^2-\frac{2}{3} -\left(6\left(\frac{1}{3}\right)^2 -\frac{1}{3}\right)}\right]$3$\displaystyle \sqrt{{10}}$    
  = $\displaystyle \left[\vphantom{\frac{24}{9} - \frac{2}{3} - \frac{6}{9} + \frac{1}{3}}\right.$$\displaystyle {\frac{{24}}{{9}}}$ - $\displaystyle {\frac{{2}}{{3}}}$ - $\displaystyle {\frac{{6}}{{9}}}$ + $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left.\vphantom{\frac{24}{9} - \frac{2}{3} - \frac{6}{9} + \frac{1}{3}}\right]$3$\displaystyle \sqrt{{10}}$ = $\displaystyle {\frac{{5}}{{3}}}$3$\displaystyle \sqrt{{10}}$ = 5$\displaystyle \sqrt{{10}}$,    

which matches Example 1.

Note that the "speed" of the parametrization |p'(t)| = 3$ \sqrt{{10}}$ was three times greater than the speed |c'(t)| = $ \sqrt{{10}}$ of Example 1. But the range of integration 1/3$ \le$t$ \le$2/3 was one third the range of integration from Example 1.

Example 3

Here's another parametrization of the same straight-line wire from (1, 2) to (4, 3). This time, the "speed" of the parametrization is not constant, but depends on t:

q(t) = (3t2 -2, t2 +1),        1$\displaystyle \le$t$\displaystyle \le$$\displaystyle \sqrt{{2}}$.    

Still using the density f (x, y) = x + y, calculate the mass of the wire.

Solution: Since

q'(t) = (6t, 2t),    
|q'(t)| = $\displaystyle \sqrt{{(6t)^2+(2t)^2}}$ = $\displaystyle \sqrt{{40t^2}}$ = 2t$\displaystyle \sqrt{{10}}$,    
f (q(t)) = (3t2 -2) + (t2 +1) = 4t2 - 1.    

the integral is

$\displaystyle \int_{{\mathbf{q}}}^{}$f ds = $\displaystyle \int_{{1}}^{{\sqrt{2}}}$f (q(t))|q'(t)| dt    
  = $\displaystyle \int_{{1}}^{{\sqrt{2}}}$(4t2 -1)2t$\displaystyle \sqrt{{10}}$dt    
  = $\displaystyle \int_{{1}}^{{\sqrt{2}}}$(8t3 -2t)$\displaystyle \sqrt{{10}}$dt    
  = (2t4 - t2)$\displaystyle \sqrt{{10}}$$\displaystyle \Big\vert _{{1}}^{{\sqrt{2}}}$    
  = $\displaystyle \left[\vphantom{2\bigl(\sqrt{2}\bigr)^4 - \bigl(\sqrt{2}\bigr)^2 -2 +1}\right.$2$\displaystyle \bigl($$\displaystyle \sqrt{{2}}$$\displaystyle \bigr)^{4}_{}$ - $\displaystyle \bigl($$\displaystyle \sqrt{{2}}$$\displaystyle \bigr)^{2}_{}$ - 2 + 1$\displaystyle \left.\vphantom{2\bigl(\sqrt{2}\bigr)^4 - \bigl(\sqrt{2}\bigr)^2 -2 +1}\right]$$\displaystyle \sqrt{{10}}$ = 5$\displaystyle \sqrt{{10}}$,    

which matches Examples 1 and 2.


Duane Nykamp
nykamp@math.umn.edu
2005-10-14