Question

Line integrals. I have to evaluate a line integral along the curve of a parabole x=t, y=t²+1. I'm not sure how to parametrize this.

Answer 1

The integral given is
\[\int\limits_{(01)}^{(1,2)}[(x^2-y)dx+(y^2+x)dy]\]

Answer 2

Do I just plug in the t values for the x and y as well as the derivatives?

Answer 3

yes,

x = t
dx = ?

y = t^2+1
dy = ?

Answer 4

sorry that lower bound ought to be (0,1)
dx=1
dy=2t

Answer 5

dx=1 dt
dy=2t dt

Answer 6

but don't I have to parametrize the bounds?

Answer 7

\[\large \int\limits_{(0,1)}^{(1,2)}[(x^2-y)dx+(y^2+x)dy]\]

after parameterization :

\[\large \int\limits_{0}^{1}[(t^2-(t^2+1))dt+((t^2+1)^2+t)2tdt]\]

Answer 8

how come it's

\[0 \le t \le 1\]

Answer 9

Because your vector is really: \[\LARGE \bar v(t)=<x(t), y(t)>=<t, t^2+1>\] So now let's evaluate v(t) at the end points what do we get? \[\LARGE \bar v(0)=<0, 1> \\ \LARGE v(1)=<1,2>\] Similarly if we just graphed the non parameterized parabola y=x^2+1 we would get the same ordered pairs! Any time you want to parameterize a function you can simply just put in x=t and replace all the x's in the function y(x) with t's to get y(t). Play around with this a little bit and I think it will become clear how you could turn any function into a function of a single parameter. From a vector point of view, all along you were really just looking at a function where x only scaled by 1 every time, hence x(t)=t!

Answer 10

Thanx, @kainui, and @rational. I found 35/3.