# Finding integral curves of a vector field

• CptXray
In summary, the vector field X satisfies the following differential equation:x^2 - y^2 = k, where k is a constant. X is tangent to the family of curves x^2 - y^2 = k if and only if k = 1.
CptXray

## Homework Statement

For a vector field $$X:=y\frac{\partial{}}{\partial{x}} + x\frac{\partial{}}{\partial{y}}$$
Find it's integral curves and the curve that intersects point $$p = \left(1, 0 \right).$$
Show that $$X(x,y)$$ is tangent to the family of curves: $$x^2 - y^2 = k,k∈ℝ$$

## The Attempt at a Solution

I know that a integral curve here is:
$$\begin{bmatrix} \dot{x} \\ \dot{y}\\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0\\ \end{bmatrix} \begin{bmatrix} x(t)\\ y(t)\\ \end{bmatrix}$$
Solving these gives me:
$$\begin{cases} x(t) = yt + x_{0} & \\ y(t) = xt + y{0} \end{cases}$$
For point (1, 0):
$$\begin{cases} x(0) = 0 \rightarrow x_{0} = 1 & \\ y(0) = 0 \rightarrow y_{0} = 0 \end{cases}$$
I guess that's what I was supposed to do here but i can't find a way to prove that $$x^2 - y^2 = k$$
I'd be glad for help because I couldn't find anything helpful in my textbooks.

P.S.
Hello people, I'm new and happy to find this place :)

CptXray said:
Solving these gives me:
$$\begin{cases} x(t) = yt + x_{0} & \\ y(t) = xt + y{0} \end{cases}$$
I don't think that can be correct because it leads to a parametrisation ##x=\frac1{1-t^2},y= \frac t{1-t^2}##, and that doesn't satisfy the hyperbola equation ##x^2-y^2=1##.

I'm not very good at integration but fortunately we don't need to integrate. They've told us that the result satisfies ##x^2-y^2=k## and that it goes through (1,0), from which we can infer that ##k=1##. So now we just need to find a parametrisation of that hyperbola and check that it satisfies the original DEs.

There are enough search terms in the above that an answer could be found by internet search. But more challenging and instructive is to note that the equation ##x^2-y^2=1## looks like a trig identity where ##x## and ##y## are trig functions of some parameter ##t##. Can you think of a trig identity that has that general form?

If you can follow that path to get a parametrisation, you then just need to check it satisfies the given DEs.

CptXray said:

## Homework Statement

For a vector field $$X:=y\frac{\partial{}}{\partial{x}} + x\frac{\partial{}}{\partial{y}}$$
Find it's integral curves and the curve that intersects point $$p = \left(1, 0 \right).$$
Show that $$X(x,y)$$ is tangent to the family of curves: $$x^2 - y^2 = k,k∈ℝ$$

## The Attempt at a Solution

I know that a integral curve here is:
$$\begin{bmatrix} \dot{x} \\ \dot{y}\\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0\\ \end{bmatrix} \begin{bmatrix} x(t)\\ y(t)\\ \end{bmatrix}$$
Solving these gives me:
$$\begin{cases} x(t) = yt + x_{0} & \\ y(t) = xt + y{0} \end{cases}$$
Hello people, I'm new and happy to find this place :)
No, you cannot solve the coupled DEs the way you did. The DEs read as
$$\begin{array}{rcl} \dot{x}(t) &=& y(t) \\ \dot{y}(t) & =& x(t) \end{array}$$ You cannot just erase the "##(t)##" part of ##y(t)## and then declare that ##x(t) = x_0 + y t.## The actual solution ##(x(t),y(t))## is not given by a pair of linear functions of ##t##.

I think I've found general solution:
$$\dot{\gamma}(t) = \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{t} + \beta \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t},$$
where
$$\alpha,\beta = const.$$
If anyone doesn't mind I'll upload detailed solution step by step later because it's really late here in my time zone.

## 1. What are integral curves of a vector field?

Integral curves are smooth curves that are tangent to a vector field at each point. This means that at every point on the curve, the direction of the curve matches the direction of the vector field at that point.

## 2. How do you find integral curves of a vector field?

The process of finding integral curves involves solving a set of differential equations, known as the "integral curve equations", that relate the vector field to the curve. These equations can be solved using various techniques, such as separation of variables or using integrating factors.

## 3. What is the significance of integral curves in vector field analysis?

Integral curves are important because they provide a way to visualize and understand the behavior of a vector field. They can also be used to solve various problems in physics and engineering, such as predicting the path of a particle in a fluid flow.

## 4. Can integral curves intersect or cross each other?

No, integral curves of a vector field cannot intersect or cross each other. This is because at any point of intersection, the vector field would have two different directions, which is not possible. However, integral curves can be tangent to each other at a point.

## 5. Are there any real-world applications of finding integral curves of a vector field?

Yes, there are many real-world applications of finding integral curves. For example, in fluid mechanics, integral curves can be used to model the path of a fluid particle in a flow field. In electromagnetism, integral curves can be used to represent the path of a charged particle in an electric or magnetic field. They are also used in image processing and computer graphics to create smooth curves and surfaces.

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