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

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.
  • #1
Kaura
122
22

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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
193
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
571
Back
Top