A couple of questions regarding to my calculus work.

In summary, vector fields are not conservative for certain values if their integral from one point in a region to another is independent of the path taken.
  • #1
JJRKnights
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1.If you have a vector field F = <-y,x>/(x^2 + y^2)

is that the same thing as saying: F = < -y/(x^2+y^2), x/(x^2 + y^2)>?
That's basically saying that the function 1/(x^2 + y^2) is a scalar.




2.How do you show that vector fields are not conservative for certain values?

Ex: F = <-y,x>/|r|^p show it's not conservative for p = 2.




3. How do you find normal vectors to rounded surfaces given a vector field and the function of the rounded surface? (paraboloids, spheres)
 
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  • #2
Can anybody say anything on this?

They should be simple questions to people more skilled than I am.
 
  • #3
JJRKnights said:
1.If you have a vector field F = <-y,x>/(x^2 + y^2)

is that the same thing as saying: F = < -y/(x^2+y^2), x/(x^2 + y^2)>?
That's basically saying that the function 1/(x^2 + y^2) is a scalar.
Yes, that is correct.




2.How do you show that vector fields are not conservative for certain values?
A vector field, F, is "conservative" on a region if its integral from one point in that region to another is independent of the path. That leads to the fact that the integral around a closed path must be 0. It can be shown that that is the same as saying it has a "potential function", a scalar function, [itex]\phi[/itex] such that [itex]\nabla \phi= F[/itex]. In that case, if F= <f(x,y), g(x,y)>, we must have [itex]\partial \phi/\partial x= f(x, y)[/itex] and [itex]\partial \phi/\partial y= g(x,y)[/itex]. And now the "mixed derivatives", [itex]\partial^2\phi/\partial x\partial y= f_y(x,y)[/itex] and [itex]\partial^2\phi/\partial x\partial y= g_x[/itex], must be equal so we must have [itex]f_y= g_x[/itex]. This last is a necessary condition for a "conservative" vector field but is a "sufficient" condition only if the function is continuous in a "simply connected" region. I mention that because in your next example the function is not continuous at (0, 0) and so not in any region containing that point.

Ex: F = <-y,x>/|r|^p show it's not conservative for p = 2.
Integrate around the unit circle: [itex]x= cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex] so that [itex]d\vec{r}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta[/itex] and the function is [itex]-y\vec{i}+ x\vec{j}= -sin(\theta)\vec{i}+ cos(\theta)\vec{j}[/itex] and the integral becomes
[tex]\int_0^{2\pi} (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})(-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta= \int_0^{2\pi} (1)d\theta= 2\pi[/tex]
Since that is not 0, the function is not conservative.

3. How do you find normal vectors to rounded surfaces given a vector field and the function of the rounded surface? (paraboloids, spheres)
A surface is two dimensional and so can be written in terms of two parameters: [itex]\vec{r}= f(u, v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex]. The two derivative vectors, [itex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/itex] and [itex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/itex] lie in the tangent plane at each point. their cross product is a normal vector to the surface and, in fact, its length gives the "differential of surface area".

For example, the sphere of radius R can be written using spherical coordinates with [itex]\rho= R[/itex], a constant. Then
[tex]\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}[/tex]

The two derivative vectors,
[tex]\vec{r}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
and
[tex]\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}[/tex]
have cross product
[tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ R cos(\theta)cos(\phi) & Rsin(\theta)cos(\phi) & -Rsin(\phi) \\ -Rsin(\theta)sin(\phi) & Rcos(\theta)sin(\phi) & 0 \end{array}\right|= R^2cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2sin(\phi)cos(\phi)\vec{k}[/tex]
(This is called the "fundamental vector product" for this surface.)

Note that the length of that vector is [itex]R^2sin(\phi)[/itex] and that [itex]R^2sin(\phi)d\theta d\phi[/itex] is the differential of surface area for a sphere.

To determing the flux of a vector function, [itex]F= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ g(x,y,z)\vec{k}[/itex] through that surface, convert x, y, and z to [itex]\theta[/itex] and [itex]\phi[/itex] , take the dot product of F with that "fundamental vector product" and integrate with respect to [itex]\theta[/itex] and [itex]\phi[/itex].
 
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Related to A couple of questions regarding to my calculus work.

1. What is the purpose of calculus?

The purpose of calculus is to study and analyze quantities that change continuously. It provides a framework for understanding and solving problems related to rates of change, accumulation, and optimization.

2. What are the two main branches of calculus?

The two main branches of calculus are differential calculus and integral calculus. Differential calculus deals with the study of rates of change and slopes of curves, while integral calculus focuses on the accumulation of quantities and the area under curves.

3. How is calculus used in real life?

Calculus has numerous applications in various fields such as physics, engineering, economics, and statistics. It is used to model and solve problems related to motion, optimization, and growth.

4. What are the basic concepts of calculus?

The basic concepts of calculus include limits, derivatives, and integrals. Limits are used to describe the behavior of a function near a particular point, derivatives give the rate of change of a function at a specific point, and integrals represent the accumulation of a quantity over a given interval.

5. Is calculus difficult to learn?

Calculus can be challenging for some people, but with practice and a good understanding of the fundamental concepts, it can be mastered. It is important to have a strong foundation in algebra and trigonometry before diving into calculus.

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