Question
particle's position given by x=cos(2t) and y=sin(t)
Answers
GPT-4o mini
The equations given for the particle's position in terms of time \( t \) are:
\[
x(t) = \cos(2t)
\]
\[
y(t) = \sin(t)
\]
These equations define the trajectory of the particle in a two-dimensional plane, where \( x \) and \( y \) are functions of time \( t \).
### Analyzing the Motion:
1. **Type of Motion**:
- The \( x \)-coordinate \( \cos(2t) \) suggests that the motion in the \( x \)-direction is periodic with a period of \( \pi \) (since \( \cos(2t) \) completes a full cycle as \( t \) goes from \( 0 \) to \( \frac{\pi}{2} \)).
- The \( y \)-coordinate \( \sin(t) \) has a period of \( 2\pi \).
2. **Parameterization**:
- This representation relates \( x \) and \( y \) through time \( t \).
3. **Eliminating the Parameter**:
To understand the relationship between \( x \) and \( y \), we can eliminate the parameter \( t \). We know:
- From \( y(t) = \sin(t) \), we have \( t = \arcsin(y) \) for \( t \) in the range where \( \sin(t) \) is defined.
- Substituting \( t \) into the \( x \) equation, we get:
\[
x = \cos(2t) = \cos(2\arcsin(y))
\]
- Using the double angle formula:
\(\cos(2\theta) = 1 - 2\sin^2(\theta)\), we substitute \(\theta = \arcsin(y)\):
\[
x = 1 - 2y^2
\]
### Conclusion:
The trajectory of the particle in the \( x \)-\( y \) plane can be described by the equation:
\[
x = 1 - 2y^2
\]
This is a downward-opening parabola that intersects the \( x \)-axis at \( (1, 0) \) and has its vertex at \( (1, 0) \). The values for \( y \) are constrained by the range of the sine function, \( -1 \leq y \leq 1 \), which gives corresponding values for \( x \).
\[
x(t) = \cos(2t)
\]
\[
y(t) = \sin(t)
\]
These equations define the trajectory of the particle in a two-dimensional plane, where \( x \) and \( y \) are functions of time \( t \).
### Analyzing the Motion:
1. **Type of Motion**:
- The \( x \)-coordinate \( \cos(2t) \) suggests that the motion in the \( x \)-direction is periodic with a period of \( \pi \) (since \( \cos(2t) \) completes a full cycle as \( t \) goes from \( 0 \) to \( \frac{\pi}{2} \)).
- The \( y \)-coordinate \( \sin(t) \) has a period of \( 2\pi \).
2. **Parameterization**:
- This representation relates \( x \) and \( y \) through time \( t \).
3. **Eliminating the Parameter**:
To understand the relationship between \( x \) and \( y \), we can eliminate the parameter \( t \). We know:
- From \( y(t) = \sin(t) \), we have \( t = \arcsin(y) \) for \( t \) in the range where \( \sin(t) \) is defined.
- Substituting \( t \) into the \( x \) equation, we get:
\[
x = \cos(2t) = \cos(2\arcsin(y))
\]
- Using the double angle formula:
\(\cos(2\theta) = 1 - 2\sin^2(\theta)\), we substitute \(\theta = \arcsin(y)\):
\[
x = 1 - 2y^2
\]
### Conclusion:
The trajectory of the particle in the \( x \)-\( y \) plane can be described by the equation:
\[
x = 1 - 2y^2
\]
This is a downward-opening parabola that intersects the \( x \)-axis at \( (1, 0) \) and has its vertex at \( (1, 0) \). The values for \( y \) are constrained by the range of the sine function, \( -1 \leq y \leq 1 \), which gives corresponding values for \( x \).