Navigating a Complex Plane Curve: A Homework Guide

The difference is that the first integral is over x while the second is over t. As far as the integral is concerned, as long as the limits of integration are the same, it doesn't matter what you call the variable of integration. What is important is that the limits of integration are the same as the limits of the parametric equation. Notice that the "t" in the first integral is not the same as the "t" in the second because the first is over x while the second is over t. As long as the limits of integration are the same, you can use any symbol you like, it doesn't have to be the same as what you used in the parametric equation.
  • #1
dwn
165
2

Homework Statement



Attached Image

Homework Equations


this is not a simple plane curve or a close plane curve so I use the formula:
∫ F ⋅ dr/dt dt

The Attempt at a Solution


From the point (0,0) to (2,4)
Direction Vector v(t) = <2-0, 4-0>
Parametric Equation:
r(t) = (2t + 0) i + (4t + 0) j
r'(t) = 2i + 4j

∴ F(x(t),y(t)) = (4t)2i + 2(2t)(4t)j = (16t2)i + (16t2)j
F(x(t),y(t)) ⋅ r'(t) =( (16t2)i + (16t2)j ) ⋅ (2i + 4j) = 32t2 + 64t2

I feel my math is correct which leads me to believe my approach is incorrect. The final answer is supposed to be 32 evaluated from 0 to 2.

But if I finished my integral : ∫ 96t2dt = 32t3

Can someone please help me.
 

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  • #2
dwn said:

Homework Statement



Attached Image

Homework Equations



∫ F ⋅ dr/dt dt

The Attempt at a Solution


From the point (0,0) to (2,4)
Direction Vector v(t) = <2-0, 4-0>
Parametric Equation:
r(t) = (2t + 0) i + (4t + 0) j
r'(t) = 2i + 4j

∴ F(x(t),y(t)) = (4t)2i + 2(2t)(4t)j = (16t2)i + (16t2)j
F(x(t),y(t)) ⋅ r'(t) =( (16t2)i + (16t2)j ) ⋅ (2i + 4j) = 32t2 + 64t2

I feel my math is correct which leads me to believe my approach is incorrect. The final answer is supposed to be 32 evaluated from 0 to 2.

But if I finished my integral : ∫ 96t2dt = 32t3

Can someone please help me.

The parameter t should run from 0 to 1, rather than 0 to 2. Plug these limits into your flux integral and everything should work out.

Try the next path.
 
  • #3
SteamKing said:
The parameter t should run from 0 to 1, rather than 0 to 2. Plug these limits into your flux integral and everything should work out.

Try the next path.
Thank you for your help. I had considered that but I'm still not confident on how interval is defined. Would you mind explaining that to me? All of the examples I see in the book simply set the interval from 0 to 1. Is this always the case?
 
  • #4
dwn said:
Thank you for your help. I had considered that but I'm still not confident on how interval is defined. Would you mind explaining that to me? All of the examples I see in the book simply set the interval from 0 to 1. Is this always the case?

It doesn't have to be, but the interval from 0 to 1 is usually the simplest to evaluate.

The path over which you were evaluating the flux integral was y = 2x, where 0 ≤ x ≤ 2. By choosing the parameterization r(t) = 2t i + 4t j for this path, it is important to check and make sure that by plugging in the limits of t into the parametric equation, you obtain the same values of x and y for the limits of the actual path expression, hence the interval 0 ≤ t ≤ 1 when plugged into r(t) gives (0, 0) and (2, 4) for the end points of the path, which are identical to y = 2x for 0 ≤ x ≤ 2.

You could have specified r(t) = t i + 2t j, for 0 ≤ t ≤ 2, and obtained the same evaluation for the flux integral after making the appropriate integration.
 
  • #5
There are an infinite number of ways to set up parametric equations for a curve. Myself, since in both of these y is a function of x, I would have just used x itself as parameter. In the first, y= 2x so that [itex]F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (2x)^2\vec{i}+ 2x(2x)\vec{j}=4x^2\vec{i}+ 4x^2\vec{j}[/itex]. Since y= 2x, dy= 2dx and [itex]d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2dx\vec{j}= (\vec{i}+ 2\vec{j})dx[/itex]. The integral becomes
[itex]\int_0^2 (4x^2\vec{i}+ 4x^2\vec{j})\cdot(\vec{i}+ 2\vec{j})dx= \int_0^2 (4x^2+ 8x^2) dx= 12\int_0^2 x^2 dx[/itex].
The limits of integration are 0 and 2, of course, because the variable of integration is x and x goes from 2 to 4.

Similarly, the second integral is over the curve [itex]y= x^2[/itex] so that [itex]F(x, y)= y^2\vec{i}+ 2xy\vec{j}= (x^2)^2\vec{i}+ 2x(x^2)\vec{j}= x^4\vec{i}+ 2x^3\vec{j}[/itex]. Further, since [itex]y= x^2[/itex], [itex]dy= 2xdx[/itex] so that [itex]d\vec{r}= dx\vec{i}+ dy\vec{j}= dx\vec{i}+ 2xdx\vec{j}= (\vec{i}+ 2x\vec{j})dx[/itex]. The integral becomes
[itex]\int_0^2 (x^4\vec{i}+ 4x^2\vec{j})\cdot (\vec{i}+ 2x\vec{j})dx= \int_0^2( x^4+ 8x^3) dx[/itex].

Again, the limits of integration are 0 to 2 because x is the variable of integration and x goes from 0 to 2.
 

Related to Navigating a Complex Plane Curve: A Homework Guide

1. What is a complex plane curve?

A complex plane curve is a mathematical concept that describes a curve in the complex plane, which is a two-dimensional space where each point has two coordinates, a real and an imaginary value. This type of curve is often represented by a function that relates the real and imaginary parts of a complex number.

2. How do I navigate a complex plane curve?

Navigating a complex plane curve involves understanding the properties of the curve and using mathematical tools such as calculus and geometry. This includes finding the slope, curvature, and arc length of the curve, as well as identifying singularities and points of intersection.

3. What is a singularity in a complex plane curve?

A singularity in a complex plane curve is a point where the curve is not smooth or continuous. This can occur when the curve has a cusp, corner, or self-intersection. Singularities can be important in understanding the behavior of the curve and can help identify critical points.

4. What is the importance of understanding a complex plane curve?

Understanding a complex plane curve is important in various fields of science and engineering, including physics, chemistry, and computer graphics. It allows for the analysis and prediction of complex systems and phenomena, and can also provide insights into the behavior of simpler curves.

5. Are there any real-world applications of complex plane curves?

Yes, there are many real-world applications of complex plane curves. For example, they are used in physics to describe the motion of particles and waves, in engineering to design structures and optimize processes, and in computer graphics to create 3D images and animations. They are also used in economics, biology, and other fields to model and analyze complex systems.

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