Significance of parabola in a line integral?

In summary, when evaluating a path integral, if $${\tmmathbf{r}}$$ and $${\tmmathbf{s}}$$ are equivalent paths that trace out a curve $${C}$$ in either the same direction or the opposite direction, then the path integral along $${C}$$ can be found by taking the dot product of $${\tmmathbf{F}}$$ and $${\tmmathbf{r}}'(t)$$ or $${\tmmathbf{s}}'(t)$$ and integrating it with respect to time $${t}$$. The significance of mentioning that $${C}$$ is a parabola $${y = x^2}$$
  • #1
hivesaeed4
217
0
If $${\tmmathbf{r}}$$ and $${\tmmathbf{s}}$$ are piecewise smooth paths, which have the same graph, then they are said to be equivalent paths.

They either trace out a set of points in the same direction, or in the opposite direction.

If they trace out a curve $${C}$$ in the same direction

$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$ $${{\int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

If they trace out a curve $${C}$$ in the opposite direction

$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$$${{- \int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

Evaluate the path integral $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t}$$ where

$${\tmmathbf{F}=\tmmathbf{i}}$$

and

$${C}$$ is the parabola $${y = x^2}$$ traced out by $${\tmmathbf{r}}$$ so that the path goes from (0,0) to (1,1).

What's the significance of mentioning that C is a parabola y=x^2? I mean can't we do without it?
 
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  • #2
hivesaeed4 said:
If $${\tmmathbf{r}}$$ and $${\tmmathbf{s}}$$ are piecewise smooth paths, which have the same graph, then they are said to be equivalent paths.

They either trace out a set of points in the same direction, or in the opposite direction.

If they trace out a curve $${C}$$ in the same direction

$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$ $${{\int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

If they trace out a curve $${C}$$ in the opposite direction

$${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$$${{- \int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

Evaluate the path integral $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t}$$ where

$${\tmmathbf{F}=\tmmathbf{i}}$$

and

$${C}$$ is the parabola $${y = x^2}$$ traced out by $${\tmmathbf{r}}$$ so that the path goes from (0,0) to (1,1).

What's the significance of mentioning that C is a parabola y=x^2? I mean can't we do without it?



You should "preview post" before "submit post": something's wrong with your writing.

DonAntonio
 
  • #3
I agree. So here I've rewritten it:
Evaluate the path integral
∫F.dr (restricted to path C)

where F=i
and C is the parabola y=x^2 traced out by r.

so that the path goes from (0,0) to (1,1).

What's the significance of mentioning that C is a parabola y=x^2? I mean if we forget that C is the parabola y=x^2 and just take y=x^2 can't we do the question.
 
  • #4
hivesaeed4 said:
I agree. So here I've rewritten it:
Evaluate the path integral
∫F.dr (restricted to path C)

where F=i
and C is the parabola y=x^2 traced out by r.

so that the path goes from (0,0) to (1,1).

What's the significance of mentioning that C is a parabola y=x^2? I mean if we forget that C is the parabola y=x^2 and just take y=x^2 can't we do the question.



I'm not sure I get your point: are asking why didn't they just wrote "..and C is y=x^2..", instead of "...and C is the parabola y = x^2..."??

Well, who cares? As it happens, y = x^2 is a parabola, so they just point out his trivial fact.

DonAntonio
 
  • #5
Thanks for clearing that part of the question.

Now the curve is y=x^2. So what would r be? The same as the path or would it be ti +tj since we're going from (0,0) to (1,1).

And now the last part (promise) I'm confused about.

When we do ∫ on curve C of F.dr we only are going to do as follows:

Define the terms of F in terms of r. (i.e. Suppose r=10ti+5tj and F= (2y+1)i +10xyj; F would be (10t+1)i+500t^2j.)

Take dot product of F and dr ( dr=in this case would be 10dti+5dtj;)

Integrate the result of the above dot product.

That's all there is to it right. And that is this integrating over a curve?
 
  • #6
hivesaeed4 said:
Thanks for clearing that part of the question.

Now the curve is y=x^2. So what would r be? The same as the path or would it be ti +tj since we're going from (0,0) to (1,1).


*** You have to fo from (0,0) to (1,1) along [itex]\,\,y=x^2\,\,[/itex] , or in other words: if you write parametrically

this parabola, you get [itex]\,\,\{(t, t^2)\,;\,\,t\in\mathbb{R}\}\,\,[/itex] . Well, what is the range of values

t has to obtain for the above parametric curve to go from (0,0) to (1,1)?



And now the last part (promise) I'm confused about.

When we do ∫ on curve C of F.dr we only are going to do as follows:

Define the terms of F in terms of r. (i.e. Suppose r=10ti+5tj and F= (2y+1)i +10xyj; F would be (10t+1)i+500t^2j.)

Take dot product of F and dr ( dr=in this case would be 10dti+5dtj;)

Integrate the result of the above dot product.

That's all there is to it right. And that is this integrating over a curve?


Yes, that's all...and the important thing here is "integrating over (or better, along) the given curve, not ANY curve...

DonAntonio
 
  • #7
Okay I get why the curve part is necessary. So in your e.g.( {(t,t^2 );t∈R} ) r would be ti +t^2j.

And then using this value of r we would do the whole procedure of finding the integral of the dot product of F and dr. And this will integrating along the given curve since the r we are using is along the given curve. Right?
 

What is a line integral?

A line integral is a mathematical concept that involves calculating the area under a curve in a two-dimensional space. It is typically used to determine the work done by a force along a particular path.

What is the significance of parabola in a line integral?

The parabola is a specific type of curve that can be represented by a quadratic equation. In a line integral, the use of a parabola can help to simplify the calculations and make them more manageable.

How is a parabola used in a line integral?

A parabola is used in a line integral by representing the curve in terms of its parametric equations. These equations can then be used to calculate the area under the parabola, which is the line integral.

What are some real-world applications of the significance of parabola in a line integral?

The significance of parabola in a line integral can be seen in various real-world applications, such as calculating the work done by a force on a moving object, finding the electric potential of a charged particle in an electric field, and determining the center of mass of a curved object.

Are there any limitations to the use of parabola in a line integral?

While the parabola is a useful curve for simplifying line integrals, it is not always applicable in every situation. In some cases, other types of curves may be more appropriate for representing the path and calculating the line integral.

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