\[x^2+(y-1)^2=4\]
In the manner describe:
a) One around clockwise starting at $(2,1)$
b) Three times around counterclockwise starting at $(2,1)$
This question aims to understand the parametric equations and dependent and independent variables concepts.
A sort of equation that uses an independent variable named a parameter (t) and in which dependent variables are described as continuous functions of the parameter and are not dependent on another existent variable. When necessary More than one parameter can be used.
Expert Answer
Given that a particle moves around the circle having equation is $x^2+(y-1)^2=4$.
Part a:
$x^2+(y-1)^2=4$ is the path of the circle in which the particle moves in the manner once around clockwise, starting at $(2,1)$
\[x^2+(y-1)^2=4\]
\[\dfrac{x^2}{4}+\dfrac{(y-1)^2}{4}=1\]
\[\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1\]
$\cos^2t + \sin^2t =1$ is the parametric equation of the circle.
As the circle is revolving once in the clockwise direction then the limit $t$ is $0 \leq t \leq 2\pi$
By comparing the two equations $\left(\dfrac{x}{2}\right)^2 +\left(\dfrac{(y-1)}{2}\right)^2 =1$and$\cos^2t +\sin^2t=1$.
\[\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2}=\sin t\]
\[x=2\cos t\space\space and\space\space y-1=2\sin t\]
\[x=2\cos t \space\space and\space\space y=1+2\sin t \space\space \epsilon\space |0, 2\pi|\]
Part b:
$x^2+(y-1)^2 =4$ is the path of the circle in which the particle moves in the manner three times around counter-clockwise, starting at $(2,1)$
\[x^2+(y-1)^2=4\]
The circle has a radius of $2$ and the center is at $(0,1)$.
As the circle is revolving thrice, the $t$ is less than equal to $3(2\pi)$ that is, $0\leq t\leq 6\pi$
By comparing the two equations $\left(\dfrac{x}{2}\right)^2+\left(\dfrac{(y-1)}{2}\right)^2=1$ and $\cos^2t+\sin^2t=1$.
\[\dfrac{x}{2}=\cos t\space\space and \space\space\dfrac{y-1}{2} =\sin t\]
\[x =2\cos t\space\space and \space \space y-1= 2\sin t\]
\[x =2\cos t\space\space and \space \space y=1+2\sin t \space\space\epsilon\space |0, 6\pi| \]
Numerical Answer
Part a: $ x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |0, 2\pi| $
Part b: $ x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |0, 6\pi| $
Example
A particle moves along the circle. Find its parametric equation for the path in the manner halfway around counterclockwise starting at $(0,3)$.
$x^2 + (y-1)^2 =4$ is the path of the circle in which the particle moves in the manner halfway around counter-clockwise, starting at $(0,3)$.
\[x^2 + (y-1)^2 =4 \]
point $(0,3)$ lies on the y-axis.
\[\dfrac{x^2}{4} + \dfrac{(y-1)^2}{4} =1 \]
\[ \left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1 \]
$\cos^2t + \sin^2t =1$ is the parametric equation of the circle.
As the circle is revolving in halfway around the counterclockwise direction, the limit $t$ is $\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2} + \pi$
That is: $\dfrac{\pi}{2}\leq t \leq \dfrac{3\pi}{2}$
By comparing the two equations $\left( \dfrac{x}{2} \right)^2 + \left( \dfrac{(y-1)}{2} \right)^2 =1$ and $\cos^2t + \sin^2t =1$.
\[ \dfrac{x}{2} = \cos t \space \space and \space \space \dfrac{y-1}{2} = \sin t \]
\[ x = 2\cos t \space \space and \space \space y-1 = 2\sin t \]
\[ x = 2\cos t \space \space and \space \space y = 1+2\sin t \space \space \epsilon \space |\dfrac{\pi}{2}, \dfrac{3 \pi}{2}| \]