- #1

dwn

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For example: ∫ ydx + x^2dy (0,-1) to (4,-1) to (4,3)

How do you determine the parametric equations for this problem?

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- Thread starter dwn
- Start date

In summary, I am having difficulty finding the parametric equations x = x(t) and y = y(t) for line integrals. I know how to find them when dealing with circles, but when it comes to finding them for anything else, I don't see the method...it all seems very random.

- #1

dwn

- 165

- 2

For example: ∫ ydx + x^2dy (0,-1) to (4,-1) to (4,3)

How do you determine the parametric equations for this problem?

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- #2

CAF123

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You first have to specify the path with which you want to integrate your function over. The given points is suggestive of a path consisting of two straight lines, one from (0,-1) to (4,-1) and an another from (4,-1) to (4,3).dwn said:How do you determine the parametric equations for this problem?

So, in this case, you have to parametrise a straight line since this is the path you are integrating over.

- #3

dwn

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Which interval is integrated?

thanks

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- #5

jackarms

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One fairly simple way of determining an equation is to first find a function in terms of x and y -- for two points, you can find the slope and then use point-slope form to find the equation. Then let x equal t, and then solve for y in terms of t. The interval of t in this case would be the interval in x.

Another, more creative way of finding them is to first start with your starting points -- for the first interval, [itex](0, -1)[/itex] -- then think about how you get to your end point. To go from 0 to 4, you need an increase of 4, and you don't need to change -1 at all. If you think about t being a step, this means x has to go from 0 to 4 in t steps. So maybe 4 in 1 step, 2 in 2 steps, 3 in 4/3 steps, etc. This is where the infinitely many paths comes in. The simplest is usually 1 step -- that is, [itex]0 \leq t \leq 1[/itex]. This would give you the parametric equations:

[itex]x(t) = 0 + 4t, y(t) = -1, 0\leq t \leq 1[/itex]. You can check this by plugging in 0 and 1 for t and seeing that you get the points corresponding to your endpoints.

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jackarms said:[itex]x(t) = 0 + 4t, y(t) = -1, 0\leq t \leq 1[/itex]. You can check this by plugging in 0 and 1 for t and seeing that you get the points corresponding to your endpoints.

What is the purpose of that parameterisation? Why not use x as the parameter?

- #7

jackarms

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PeroK said:What is the purpose of that parameterisation? Why not use x as the parameter?

Yes, I know using t in this case is a bit overkill, but it's important to know how to parameterize things since you don't always have just one variable changing.

- #8

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dwn said:

Which interval is integrated?

thanks

Okay, so you've got two ways to do this. Maybe try it both ways and make sure you get the same result:

a) Use an additional parameter t (going from 0 to 1) in each case.

b) Use x and y as your parameters for the first and second parts of your curve.

- #9

dwn

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PeroK said:

What about the point (4,3)? Where does this come into play?

Is it just common practice to setup the interval 0 to 1 for t, when you parameterize in terms of t?

Thanks for all the help, I'm taking this class online and this has me stumped.

- #10

jackarms

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- #11

dwn

- 165

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How I understand it based on your responses...is this correct?

Δx = 4-0, therefore x = 0 +4t → x = 4t

Why is it -1 and not Δy = -1 - (-1) = 0 ?

Code:

```
∫ ydx + x^2dy
y = -1 x = 4t
dy = 0 dx = 4
0 to 1 interval for t
∫ -1(4)dt + 4t^2(0)dt
-4t |[SUP]1[/SUP] = -4
```

Δx = 4-0, therefore x = 0 +4t → x = 4t

Why is it -1 and not Δy = -1 - (-1) = 0 ?

Last edited:

- #12

CAF123

Gold Member

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It is correct, yes, so now do the same again for the other straight line segment comprising your path. Note that in some cases, the conventional parametrisation for t in the interval [0,1] will not allow you to actually compute the integral. Here though, it is ok.dwn said:How I understand it based on your responses...is this correct?

I am not quite sure I understand the question. On segment 1, Δy = 0 and only on segment 2, which is treated separately, does y change.Why is it -1 and not Δy = -1 - (-1) = 0 ?

A line integral is a type of integral that is used to calculate the total value of a function along a given curve or path. It takes into account the direction and length of the curve, as well as the function being integrated.

Unlike a regular integral, which calculates the area under a curve, a line integral calculates the value of a function along a specific curve or path. It takes into account the direction and length of the curve, as well as the function being integrated.

Line integrals have various applications in physics and engineering, such as calculating work done by a force, finding the mass of a wire, and calculating the flow of a fluid in a vector field.

To find the parametric equations of a curve, you first need to determine the parametric variables (usually denoted by t and/or u). Then, you can express the x and y coordinates of the curve in terms of these variables. The resulting equations are the parametric equations of the curve.

Parametric equations are useful in line integrals because they allow us to express a curve in terms of its parametric variables, making it easier to calculate the line integral along that curve. They also allow us to represent more complex curves, such as circles and spirals, in a simpler form.

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