How Do You Set Up r(t) for Line Integrals on Triangular Paths?

[makegooduseof]

Homework Statement

Evaluate a line integral where C is the triangle with the vertices (0,0), (1,0) and (1,2). [Green's Theorem not allowed]

I am omitting the integral expression as I am having trouble with setting up the r(t) expressions for each C.

Homework Equations

r(t) = (1-t)r0 + tr1 (o≤t≤1)

The Attempt at a Solution

So the way to tackle this problem would be to get the line segments C1, C2 and C3, C1 from (0,0) to (1,0), C2 from (1,0) to (1,2) and C3 from (1,2) to (0,0). To get the segments, the aforementioned equation would be used.

C1 was easy enough, with r0 being <0,0> and r1 being <1,0>. r(t) was found to be <t,0>.

C2 is where I began having problems. Calculating r(t) with r0 = <1,0> and r(1) = <1,2> gave the r(t) expression <1, 2t>. When I drew out the graph, I could see that it should be <0, t>. So my question is, what am I doing wrong? And if I see a similar problem where I can see a visible straight horizontal or vertical line, I should just go with a <0, t> or a <t, 0> expression?

 

[Simon Bridge]

If I follow you: just focus on where you are going and how you're are going to get there...

Going from (0,0) to (0,1) as t goes from 0 to a:
... you need x=0 all the time, and y=0 when t=0, and y=1 when t=a.
The easiest thing to do is fit that data into the equation for a line: y=mt+c.
- so y=t/a ... the line element here is dl = dy = dt/a

Going from (0,1) to (1,0) as t goes from a to b would be ...
... in this case, both of them are changing. note that dl/dt is determined by pythagoras.

Going from (1,0) to (0,0) as t goes from b to c would be ... similar to the first one so try this first.

You can make a,b,c anything you like.

 

[makegooduseof]

If I follow you: just focus on where you are going and how you're are going to get there...

Going from (0,0) to (0,1) as t goes from 0 to a:
... you need x=0 all the time, and y=0 when t=0, and y=1 when t=a.
The easiest thing to do is fit that data into the equation for a line: y=mt+c.
- so y=t/a ... the line element here is dl = dy = dt/a

Going from (0,1) to (1,0) as t goes from a to b would be ...
... in this case, both of them are changing. note that dl/dt is determined by pythagoras.

Going from (1,0) to (0,0) as t goes from b to c would be ... similar to the first one so try this first.

You can make a,b,c anything you like.

I'm sorry, I don't quite understand your explanation. Do you think you could frame this based on the problem I uploaded?

 

[Simon Bridge]

I believe I did relate it to you problem - that's what the examples were for.

I'd love to do more but that would involve doing your homework for you.
I'll try give you an example.

You are trying to get $\int_C f\;\text{d}l$ ...

You have f(x,y) and you need f(t) ... to turn this into an integral wrt t, you have to express x and y in terms of t and dl in terms of dt.

For instance, if the first leg of the path C is C1 goes along the $x$ axis from $x=0$ to $x=2$, then we pick a range of values of $t$ to represent this by ... 0-2 looks good, then $x(t)=t$ and $y(t)=0$. See how this works?

I am free to choose any range of t-values I like. eg. if I pick 0-1, then $x(t)=2t$, and if I picked 0-pi, then $x(t)=2t/\pi$ see the difference? So I pick the one that makes the math easiest. You get a feel for the good choices with practice so feel free to play around a bit.

Remains only to convert dl - which is the length along the path between t and t+dt
You can do this just by examination - but by algebra: dl=|PQ| where P is the position at time t, and Q is the position at time t+dt. In this case, $P=(t,0)$ and $Q=(t+dt,0)$ so $\text{d}l=|PQ|=dt$.

So the integral for the first leg is: $$\int_{C_1}f(x,y)\text{d}l = \int_0^2 f(t,0)dt$$

... clearer now?

It's a good idea to sketch the path out on graph paper.

 

[makegooduseof]

I believe I did relate it to you problem - that's what the examples were for.

I'd love to do more but that would involve doing your homework for you.
I'll try give you an example.

You are trying to get $\int_C f\;\text{d}l$ ...

You have f(x,y) and you need f(t) ... to turn this into an integral wrt t, you have to express x and y in terms of t and dl in terms of dt.

For instance, if the first leg of the path C is C1 goes along the $x$ axis from $x=0$ to $x=2$, then we pick a range of values of $t$ to represent this by ... 0-2 looks good, then $x(t)=t$ and $y(t)=0$. See how this works?

I am free to choose any range of t-values I like. eg. if I pick 0-1, then $x(t)=2t$, and if I picked 0-pi, then $x(t)=2t/\pi$ see the difference? So I pick the one that makes the math easiest. You get a feel for the good choices with practice so feel free to play around a bit.

Remains only to convert dl - which is the length along the path between t and t+dt
You can do this just by examination - but by algebra: dl=|PQ| where P is the position at time t, and Q is the position at time t+dt. In this case, $P=(t,0)$ and $Q=(t+dt,0)$ so $\text{d}l=|PQ|=dt$.

So the integral for the first leg is: $$\int_{C_1}f(x,y)\text{d}l = \int_0^2 f(t,0)dt$$

... clearer now?

It's a good idea to sketch the path out on graph paper.

Yes, much clearer, thank you.

So basically you're saying that when I set the interval for t, it doesn't always have to be from 0 to 1, and based on what I see on the graph, I don't always need to use the equation I mentioned earlier?

 

[Simon Bridge]

That's right - you can set the range from 0 to 1 if you want, it just changes some constants in the parameterization. You can try it out on a simple structure ... say $f(x,y)=x+y$