Understanding Vector Fields and Their Relationships to Circles

In summary: I had the same experiences studying physics for the first time and I know how it feels like.It's okay to feel confused, but it's important to take the time to really understand the basics before moving on to more complex ideas. I would suggest reviewing vector operations and notations, as well as the concept of level surfaces and gradients, before continuing with this problem. Once you have a strong foundation, you will be able to understand this problem and others like it more easily.
  • #1
Calpalned
297
6
Oops. I just realized that this is the physics homework forum... This is actually calculus homework...

1. Homework Statement


b.png
a.png


Homework Equations


n/a

The Attempt at a Solution


I read, analyzed and reread the text, but I am still confused.

1) How was the position vector ##<x,y>## determined?

2) How is it related to the circle with centered at the origin?

3) We are trying to prove that ##F(x,y)## is tangent to a circle with center at the origin. Why then, are we taking a dot product? The dot product shows that ##F(x.y)## is perpendicular to the position vector ##<x,y>##. But isn't being perpendicular the opposite of being tangent?

4) The magnitude of the vector ##F(x,y)## is equal to the radius of the circle. If the circle wasn't centered at the origin, I believe that it will still be true. What is the significance of the magnitude in other vector fields, ones that don"t involve circles?

Thank you all so much!
 
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  • #2
The position vector is simply the vector from the origin to a given point. A circle centered on the origin fulfills ##F(\vec x) = \vec x^2 = R^2## and is therefore a level surface of ##F##. The normal of a level surface is the gradient of the function and in this case it is proportional to ##\vec x##.

A tangent to a surface is orthogonal to its normal, so a tangent to the circle is necessarily also orthogonal to ##\vec x##.
 
  • #3
Orodruin said:
The position vector is simply the vector from the origin to a given point. A circle centered on the origin fulfills ##F(\vec x) = \vec x^2 = R^2## and is therefore a level surface of ##F##. The normal of a level surface is the gradient of the function and in this case it is proportional to ##\vec x##.

A tangent to a surface is orthogonal to its normal, so a tangent to the circle is necessarily also orthogonal to ##\vec x##.

While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was ##F(\vec x) = \vec x^2 = R^2## determined? Does ##R^2## mean two dimensional? A level surface is a projection (or shadow) of a 3D object onto the xy plane. The circle is not 3D. I agree that the normal to a level surface is the gradient, but I don't see how it's proportional to ##\vec x ##

Thank you Orodriun.

 
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  • #4
Calpalned said:
While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was ##F(\vec x) = \vec x^2 = R^2## determined? Does ##R^2## mean two dimensional?

No, ##R## is the radius of the circle.

Calpalned said:
A level surface is a projection (or shadow) of a 3D object onto the xy plane.

No, a level surface is the set of points where a function has a given value. It has a dimension of one less than the full space. For this reason, a level surface in two dimensions is a one dimensional curve.

Calpalned said:
agree that the normal to a level surface is the gradient, but I don't see how it's proportional to ##\vec x ##

Did you try taking the gradient of the function ##F(\vec x) = \vec x^2##?
 
  • #5
Orodruin said:
The position vector is simply the vector from the origin to a given point. A circle centered on the origin fulfills ##F(\vec x) = \vec x^2 = R^2## and is therefore a level surface of ##F##. The normal of a level surface is the gradient of the function and in this case it is proportional to ##\vec x##.

A tangent to a surface is orthogonal to its normal, so a tangent to the circle is necessarily also orthogonal to ##\vec x##.

Calpalned said:
While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was ##F(\vec x) = \vec x^2 = R^2## determined? Does ##R^2## mean two dimensional? A level surface is a projection (or shadow) of a 3D object onto the xy plane. The circle is not 3D. I agree that the normal to a level surface is the gradient, but I don't see how it's proportional to ##\vec x ##

Thank you Orodriun.
Calpalned said:
While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was ##F(\vec x) = \vec x^2 = R^2## determined? Does ##R^2## mean two dimensional? A level surface is a projection (or shadow) of a 3D object onto the xy plane. The circle is not 3D. I agree that the normal to a level surface is the gradient, but I don't see how it's proportional to ##\vec x ##

Thank you Orodriun.
Is ##F(\vec x) = \vec x^2 \hat i = R^2 ## ?
 
  • #6
No, ##\vec x^2 = \vec x \cdot \vec x = x^2 + y^2##, which is a scalar quantity. A vector can never be equal to a scalar. This is just the equation for a circle, ##x^2 + y^2 = R^2##.
 
  • #7
Orodruin said:
No, ##\vec x^2 = \vec x \cdot \vec x = x^2 + y^2##, which is a scalar quantity. A vector can never be equal to a scalar. This is just the equation for a circle, ##x^2 + y^2 = R^2##.

Is the ##\vec x## the same as the ##x## two original equations? The one found in ##F(x,y) = -y \hat + x \hat j ## and/or ##x = x \hat i + y \hat j ##
 
  • #8
##\vec x## is the position vector ##x\hat i + y \hat j##. The vector arrow is important.
 
  • #9
Dear @Calpalned,
I would recommend you re-read everything you have on vectors to clear up some basic concepts before proceeding in your studies in Calculus III.
Otherwise I fear you will face a lot of confusion.
 

FAQ: Understanding Vector Fields and Their Relationships to Circles

1. What is a vector field?

A vector field is a mathematical concept used to represent a vector quantity, such as velocity or force, at different points in space. It is typically represented by arrows or lines, with the length and direction of the arrows indicating the magnitude and direction of the vector at a particular point.

2. How is a vector field analyzed?

A vector field is typically analyzed by looking at its properties, such as its direction, magnitude, and flow. This can be done visually by looking at the arrows or lines representing the vectors, or mathematically by using equations and formulas to determine the properties of the field at different points.

3. What is the significance of analyzing a vector field?

Analyzing a vector field can provide valuable insights into the behavior of a physical system. It can help predict the movement of objects, understand the forces at play, and identify patterns or relationships within the field.

4. What are some common applications of vector field analysis?

Vector field analysis has a wide range of practical applications, including in physics, engineering, meteorology, and computer graphics. It is used to study fluid dynamics, electric and magnetic fields, weather patterns, and many other phenomena.

5. What are some techniques for analyzing vector fields?

There are various techniques for analyzing vector fields, including graphical methods, numerical methods, and analytical methods. Graphical methods involve visualizing the field and looking for patterns and trends. Numerical methods involve using computers to calculate the properties of the field at different points. Analytical methods involve using mathematical equations and formulas to determine the properties of the field.

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