Laws of area and conservation of momentum

[almo]

i am having problem getting the dθ in the laws of area.

what the textbook states is,
dθ=

xdy-ydx
x²+y²

I can't get the equation..
the only thing i could think of is trigo functions, but i got stuck...
any help?

 

[Ratch]

almo,

laws of area and conservation of momentum

--------------------------------------------------------------------------------
i am having problem getting the dθ in the laws of area.

what the textbook states is,
dθ=

xdy-ydx
x2+y2

I can't get the equation..
the only thing i could think of is trigo functions, but i got stuck...
any help?

What are you trying to do? What are the "laws of area". What equation are you trying to find? How does the conservation of momentum fit into this?

Ratch

 

[Vargo]

"Laws of Area" is vague, but the formula for dtheta is straightforward.

[itex] \theta = \arctan (y/x)+C[/itex]

Here C is an additive constant that depends on 1. Which region of the plane you are in, and 2. How you assign angles to points in the plane. For example, if you define theta to be between 0 and 2pi, then C=0 for (x,y) in the first quadrant, C=pi in the second and third quadrants, and C=2pi in the fourth quadrant. If you define theta to lie between -pi and +pi, then C works differently. But all that is irrelevant, because even though theta as a function of x and y has some ambiguity, no matter what your convention is for assigning angles to points, dtheta is well defined by the formula:

[itex] d\theta = \frac{\partial \theta}{\partial x} dx +\frac{\partial \theta}{\partial y}dy[/itex]

There are ways to derive the formula for dtheta without relying on a formula for theta as a function of x and y. For example, no matter how you assign angles, you have:

[itex] x=r\cos(\theta), \hspace{1cm} y=r\sin(\theta).[/itex]

Use these formulas to calculate dx and dy as linear functions of dr and dtheta (with a basepoint understood to be fixed). You get:

[itex] dx = (x/r)dr -yd\theta,\hspace{1cm} dy= (y/r)dr+xd\theta[/itex].

Multiply the first equation by y, the second equation by x and subtract to obtain:
[itex] ydx - xdy = -(y^2+x^2)d\theta =-r^2d\theta[/itex]
Dividing out the - (r squared) gives you the formula for dtheta.

 

[almo]

Thanks vargo!.. I figure out to just differentiate theta, which gives me the equation. I am too use to differentiate something against something else, that I forget about some basic stuff.

The parametric method seems quite confusing when multiplication of x and y is involved though, but its brilliant.

 

[CellCoree]

The laws of area and conservation of momentum are fundamental principles in physics that help us understand the behavior of objects in motion. The equation you are struggling with, dθ= (xdy-ydx)/(x2+y2), is known as the differential form of the law of area. This equation represents the change in angle (dθ) of a moving object as it travels along a curved path.

To better understand this equation, let's break it down into its components. The numerator, xdy-ydx, represents the difference in the x and y components of the object's displacement. This difference is multiplied by the inverse of the distance from the origin, (x2+y2), which represents the magnitude of the object's displacement.

In simpler terms, this equation tells us that the change in angle of a moving object is directly proportional to the difference in its horizontal and vertical displacements, and inversely proportional to the distance it travels from the origin. This is important because it helps us understand how objects move along curved paths and how their velocities change.

As for the use of trigonometric functions, they can be helpful in visualizing the relationship between the different components of the equation. For example, we can use the sine and cosine functions to represent the x and y components of the object's displacement, respectively. However, it is not necessary to use these functions to solve the equation.

I hope this explanation helps clarify the meaning of the equation and its relevance in the laws of area and conservation of momentum. Keep practicing and seeking guidance if you are still struggling to understand it. As a scientist, it is important to persist in the face of challenges and continue to seek understanding. Good luck!