Simple Differential Equation Question

In summary, we use field lines to represent vector functions and a field line y=y(x) is a solution of the differential equation dy/dx = F_y(x,y)/F_x(x,y) for a vector function F(x,y).b) The field lines of the function v(x,y) = iy + jx have no unique slope in 3D, but can be represented by a unit vector with direction cosines cos(theta) and cos(phi), where theta is the angle with the x-axis and phi is the angle with the y-axis.
  • #1
Saladsamurai
3,020
7
So I think that this question is easier than I am making it, but it's been awhile since I have had to solve a Diff EQ. And I am not even sure that the question even requires me to solve one:

Problem

Instead of using arrows to represent vector functions, we sometimes use families of curves called field lines. A curve y = y(x) is a field line of the vector function F(x,y) if at each point (xo, yo) on the curve, F(xo, yo) is tangent to the curve.(a) Show that the field lines y = y(x) of a vector function F(x,y) = iFx(x,y) + jFy(x,y) are solutions of the differential equation

[tex]\frac{dy}{dx}=\frac{F_y(x,y)}{F_x(x,y)}[/tex]

--------------------------------------------------------------

(b) Determine the field lines of the function v(x,y) = iy + jxSolution

I know that the definition of field lines is very similar, if not identical, to that of the curves in an integral field.

But I am really not sure, for part (a), how I am supposed to start. I am trying to show that y = y(x) is a solution to the diff eq, but I am only given the 'properties' of y(x). So I am a little confused as to how to start the math.

Any hints?

Casey
 
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  • #2
Right. You don't have to solve the differential equation. The slope of the tangent to the curve y=y(x) is dy/dx, yes? What's the slope of the vector F(x,y)?
 
  • #3
Dick said:
Right. You don't have to solve the differential equation. The slope of the tangent to the curve y=y(x) is dy/dx, yes? What's the slope of the vector F(x,y)?

Sorry Dick; I think it is going over my head :redface: When you ask what the slope of F(x,y) is, it is a little ambiguous to me. Do you mean what is the gradient?

grad( F(x,y)) = grad(iFx(x,y) + jFy(x,y) )

Wow. This brings to surface another of my mathematical deficiencies (what else is new); I have to go look up how to take the gradient of a vector function now. For some reason the fact that it is not a scalar function is throwing me for a loop.

But, that aside, is that what you are referring to? Its gradient?
 
  • #4
No, you don't have to take the gradient. The slope of a vector is just the y component over the x component, isn't it?
 
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  • #5
Dick said:
No, you don't have to take the gradient. The slope of a vector is just the y component over the x component, isn't it?

Oh. I guess it is :redface:

But since that I have brought it up and I am supposed to be working right now instead of being on PF, I cannot think of a better time to ask another question:

1.) I am having a hard time wrapping my head around what the gradient of a vector function is. I am assuming it should be analogous to that of a scalar function. Would it just be the direction in the vector field that 'sees' sharpest change in vector magnitude?


2.) How would you find the slope of a vector that is 3D? Or is that the ambiguous question? It has more than 1? Or is it a meaningless question?


thanks!
 
  • #6
The 'gradient' of a vector is a tensor. It's basically a collection of vectors, one for each scalar component of the vector. The thing that would see the sharpest change in the vector magnitude isn't that, it's just grad(|F|) - a different animal. And, yes, a vector in 3d doesn't have a unique 'slope'. You have to define the x and y axes of a plane containing it to measure a 'slope'.
 
  • #7
Saladsamurai said:
Oh. I guess it is :redface:

But since that I have brought it up and I am supposed to be working right now instead of being on PF, I cannot think of a better time to ask another question:

1.) I am having a hard time wrapping my head around what the gradient of a vector function is. I am assuming it should be analogous to that of a scalar function. Would it just be the direction in the vector field that 'sees' sharpest change in vector magnitude?
The "gradient of a vector function" is not defined (unless you want to think of it as the linear transformation represented by the matrix having each row the gradient a component of the vector and I don't think that's what you are talking about). Dick was just saying that we can think of a vector (in two dimensions) as the direction of a line through the origin and he was asking about the slope of that line.


2.) How would you find the slope of a vector that is 3D? Or is that the ambiguous question? It has more than 1? Or is it a meaningless question?
The problem is that while direction of a line (perhaps determined by a vector) in 2 dimensions can be given as a single angle, the direction of a line in 3 dimensions cannot- although you could do it with two angles. The "direction cosines" the cosines of the angles the line makes with lines parallel to the axes. One can show that, if we call those angles, the angles a line makes with lines parallel to the x, y, and z axes, [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\psi[/itex] respectively, then \(\displaystyle cos^2(\theta)+ cos^2(\phi)+ cos^2(\psi)= 1[/itex] so that, knowing two of the angles, you can calculate the third. In fact, if you represent the line by the vector having those direction cosines as components, [itex]cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}[/itex] you just have a unit vector pointing in the same direction as the line.

The corresponding thing in two dimensions would be [itex]cos(\theta)[/itex] and [itex]cos(\phi)[/itex], the cosines of the angles the line makes with the x and y axes respectively. But, of course, looking at the right triangle formed by that line and the two axes, those two angles are "complimentary": [itex]\phi= \pi/2- \theta[/itex] and so [itex]cos(\phi)= sin(\theta)[/itex]. You only need the single angle [itex]\theta[/itex].
thanks![/QUOTE]\)
 

1. What is a simple differential equation?

A simple differential equation is a mathematical equation that relates the rate of change of a function to its current value. It is often used to model real-world systems that involve changing quantities, such as population growth or chemical reactions.

2. How do you solve a simple differential equation?

The method for solving a simple differential equation depends on the specific equation and its form. In general, you can start by separating the variables and then integrating both sides. Other methods include using substitution, integrating factors, and using an integrating factor.

3. What is the difference between a simple differential equation and a partial differential equation?

A simple differential equation involves only one independent variable, while a partial differential equation involves multiple independent variables. Simple differential equations are often used to describe systems that change over time, while partial differential equations are used to describe systems that change over space and time.

4. What are the applications of simple differential equations?

Simple differential equations have numerous applications in various fields, including physics, chemistry, biology, economics, and engineering. They are used to model and predict the behavior of systems that involve changing quantities, such as population growth, radioactive decay, and heat transfer.

5. Can you provide an example of a simple differential equation?

One example of a simple differential equation is the equation for exponential growth: dy/dt = ky, where y represents the quantity, t represents time, and k is a constant. This equation describes the rate of change of a quantity as proportional to its current value. Another example is the equation for a simple harmonic oscillator: d^2x/dt^2 + kx = 0, where x represents displacement, t represents time, and k is a constant. This equation describes the motion of a mass attached to a spring.

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