Path integral examples

Example 1

Let

c(t) = (3t - 2,t + 1), 1 ≤ t ≤ 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 ∫ cfds. Since,

c'(t) = (3, 1),
||c'(t)|| = √ ------- 32 + 12 = √ --- 10,
f(c(t)) = (3t - 2) + (t + 1) = 4t - 1,
the integral is
∫ cfds = ∫ abf(c(t))||c'(t)||dt
= ∫ 12(4t - 1)√10--dt
= (2t2 - t)√ --||2 10| 1
= (8 - 2 - (2 - 1)) √ --- 10 = 5√ --- 10

So, if f were given in grams/cm and c(t) were given in cm, then the mass of the wire would 5√ --- 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 ≤ t ≤ 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)|| = √ -2----2 9 + 3 = √ --- 90 = 3√ --- 10,
f(p(t)) = (9t - 2) + (3t + 1) = 12t - 1,
the integral is
∫ pfds = ∫ 1∕32∕3f(p(t))||p'(t)||dt
= ∫ 1∕32∕3(12t - 1)3√ --- 10dt
= (6t2 - t) | √ ---|2∕3 3 10 |1∕3
= [ ( )2 ( ( )2 ) ] 2- 2- 1- 1- 6 3 - 3 - 6 3 - 3 3√ --- 10
= [ ] 24- 2- 6- 1- 9 - 3 - 9 + 3 3√ --- 10 = 5- 33√ --- 10 = 5√--- 10,
which matches Example 1.

Note that the “speed” of the parametrization ||p'(t)|| = 3√ --- 10 was three times greater than the speed ||c'(t)|| = √ --- 10 of Example 1. But the range of integration 1∕3 ≤ t ≤ 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 ≤ t ≤√ -- 2.
Still using the density f(x,y) = x + y, calculate the mass of the wire.

Solution: Since

q'(t) = (6t, 2t),
||q'(t)|| = ------------ ∘ (6t)2 + (2t)2 = √ ---- 40t2 = 2t --- √ 10,
f(q(t)) = (3t2 - 2) + (t2 + 1) = 4t2 - 1.
the integral is
∫ qfds = ∫ 1√- 2f(q(t))||q'(t)||dt
= ∫ 1√- 2(4t2 - 1)2t√--- 10dt
= ∫ 1√2(8t3 - 2t)√ --- 10dt
= (2t4 - t2)√ --|√2 10|| 1
= [ ] 2(√2-)4 - (√2-)2 - 2 + 1√10-- = 5√10-,
which matches Examples 1 and 2.