# Solve for Path of Particle: x - 2cos(arcsin(y/2)) = 0

• Kaura
In summary, squaring the variables x and y in the given parametrization r(t) = 2cos(t)i + 2sin(t)j and setting them equal to 0 results in the equation x^2 + y^2 - 4 = 0, which is the desired equation whose solutions consist of the path of the particle. This method can be useful when dealing with equations involving both sine and cosine functions.
Kaura

## Homework Statement

Suppose that a particle follows the path
r(t) = 2cos(t)i + 2sin(t)j
Give an equation (in the form of a formula involving x and y set equal to 0 ) whose whose solutions consist of the path of the particle.

## Homework Equations

None that come to mind

## The Attempt at a Solution

I set x = 2cos(t) and y = 2sin(t)
thus t = arcsin(y/2)
then x = 2cos(arcsin(y/2))
then x - 2cos(arcsin(y/2)) = 0

It says that this is wrong
I am not all to familiar with doing this type of problem though I suspect that the inverse trigonometric could be messing up the domain of the solution

This is the last problem I need for the homework so any help would be much appreciated

Kaura said:

## Homework Statement

Suppose that a particle follows the path
r(t) = 2cos(t)i + 2sin(t)j
Give an equation (in the form of a formula involving x and y set equal to 0 ) whose whose solutions consist of the path of the particle.

## Homework Equations

None that come to mind

## The Attempt at a Solution

I set x = 2cos(t) and y = 2sin(t)

This is the last problem I need for the homework so any help would be much appreciated
Try squaring ##x## and ##y## and see if anything comes to mind.

Can you draw a few points of the path ? Take easy values for ##t##, like ##{\pi\over 6},\ {\pi \over 4},\ {\pi\over 3}\ ## etc.

Alright I finally got it
r(t) = 2cos(t)i + 2sin(t)j
x = 2cos(t)
y = 2sin(t)
x^2 = 4cos^2(t)
y^2 = 4sin^2(t)
x^2+y^2=4
x^2+y^2-4=0
This answer was accepted as correct
I did not even think about squaring the variables thanks

Kaura said:
Alright I finally got it
r(t) = 2cos(t)i + 2sin(t)j
x = 2cos(t)
y = 2sin(t)
x^2 = 4cos^2(t)
y^2 = 4sin^2(t)
x^2+y^2=4
x^2+y^2-4=0
This answer was accepted as correct
I did not even think about squaring the variables thanks

Whenever you see ##\sin(at)## and ##\cos(at)## appearing together in some equation or parametrization, you should always look at what happens when you square them. Sometimes squaring will not work, but sometimes it solve a problem very easily---all you can do is try it and see.

## What is the equation "x - 2cos(arcsin(y/2)) = 0" used for?

The equation is used to determine the path of a particle in a two-dimensional plane.

## What does the variable x represent in the equation?

The variable x represents the horizontal position of the particle.

## What does the variable y represent in the equation?

The variable y represents the vertical position of the particle.

## What is the significance of the arcsin function in the equation?

The arcsin function is used to calculate the angle of the particle's path in relation to the y-axis.

## How can this equation be solved to determine the path of the particle?

To solve for the path of the particle, the equation can be rewritten as y = 2sin(x/2) and graphed to visualize the path. Alternatively, the equation can be solved algebraically by isolating either x or y and substituting values for the other variable.

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