Question

Calculate the line integral of 4xy^2dx - 3x^4dy from (1,1) to (2,2) along y=(1/2)x^2

Answer 1

OK, so first you need to parameterise the curve. Any ideas?

Answer 2

r=<1,1> and r2=<2,2>

then r(t) = (1-t)<1,1> + t<2,2>

Answer 3

I think that's a line, not parabola.

Answer 4

That is where I'm getting confused, I don't understand how to parametize parabolas.

Answer 5

There's an issue with the problem. It asks you to calculate the line integral from (1,1) to (2,2), but (1,1) isn't even a solution to the curve?

Answer 6

No there is no solution at all.

Answer 7

Well, to parameterize along the curve... the two points you are integrating between need to be on the curve. So there is an issue with the problem. It can't be answered as it is. Typo?

Answer 8

No, there is no typo. That is what my book asks.

Answer 9

Then the question can't be answered. (1,1) doesn't lie on the curve:\[y=\frac{x^2}{2} \rightarrow 1\stackrel{?}{=}\frac{(1)^2}{2}\rightarrow 1 \ne\frac{1}{2}\]

Answer 10

I appreciate your time and effort.

Answer 11

Of course. If you have other problems, feel free to ask.

Answer 12

I could still help you parameterise the parabola, even though the problem can't be solved. If you're interested.

Answer 13

Yes, every little bit helps.

Answer 14

OK. There are two types of parameterising. This one is the simpler type.

The easier case is when the curve is a function in any of the variables. In this case, it's a function in x (it's one-to-one).

When this is the case, set t equal to the variable that the curve is a function of:
\[x=t\]Then, solving for y is easy. You just plug in t for x in the equation:\[y=\frac{1}{2}x^2=\frac{1}{2}t^2\]And that's it. You've defined both x and y in terms of t: parameterised.

Answer 15

Thanks. No I believe we can integrate from 2 to 1

Answer 16

*Now

Answer 17

No. The starting point and end point really do need to both lie on the curve.

The entire integration occurs ALONG the curve, so you have to start and end on the curve. I'm confident it's a typo.