Determine the parametric equations of the path of a particle that travels the circle: (x-5)^2 +...

Question:

Determine the parametric equations of the path of a particle that travels the circle: {eq}(x-5)^{2} + (y-3)^{2} = 81 {/eq} on a time interval of 0 t 2:

A) if the particle makes one full circle starting at the point (14 3) traveling counterclockwise x(t) = ?

y(t) = ?

B) if the particle makes one full circle starting at the point (51 2) traveling clockwise

x(t) = ?

y(t) = ?

C) if the particle makes one half of a circle starting at the point (14 3) traveling clockwise

x(t) = ?

y(t) = ?

Parametrization of Circles


To parametrize a circle given in Cartesian coordinates, {eq}\displaystyle (x-a)^2+(y-b)^2=r^2 {/eq}

we will sue sine and cosine function, as follows

{eq}\displaystyle x-a=r\cos t, y-b=r\sin t, \text{ ( for counterclockwise direction) } 0\leq t\leq 2\pi \text{ or any other interval depending on where the curve starts}\\ \displaystyle \text{ or } x-a=r\sin t, y-b=r\cos t, \text{ ( for clockwise direction)}. {/eq}

Answer and Explanation:


A) To find a parametrization for the circle centered at {eq}\displaystyle (5,3) {/eq} and radius 9, {eq}\displaystyle (x-5)^{2} + (y-3)^{2} = 81 {/eq}

starting at {eq}\displaystyle (14,3) {/eq} in counterclockwise direction, one full rotation,

we will choose cosine for the x variable and sine for y, like below.

{eq}\displaystyle x-5=9\cos t, y-3=9\sin t \iff x=5+9\cos t, y=3+9\sin t. {/eq}

In order to start at {eq}\displaystyle (14,3) {/eq} we need to start with {eq}\displaystyle t, {/eq} such that

{eq}\displaystyle 5+9\cos t=14 \text{ and }3+9\sin t = 3\implies t=0. {/eq}

So, a parametrization of the curve with the required direction is {eq}\displaystyle \boxed{x(t)=5+9\cos t, y(t)=3+9\sin t, 0\leq t\leq 2\pi}. {/eq}


B) To find a parametrization in clockwise direction, we will choose sine for the x variable and cosine function for y.

To start at {eq}\displaystyle (5,12), {/eq} we will chose {eq}\displaystyle t {/eq} to start such that

{eq}\displaystyle 5+9\sin t=5 \text{ and }3+9\cos t= 12 \implies t=0. {/eq}

So, a parametrization of the curve with the required direction is {eq}\displaystyle \boxed{x(t)=5+9\sin t, y(t)=3+9\cos t, 0\leq t\leq 2\pi}. {/eq}


C) To find a parametrization in the clockwise direction, starting at {eq}\displaystyle (14,3), {/eq} we will choose {eq}\displaystyle t {/eq} to start such that

{eq}\displaystyle 5+9\sin t=11 \text{ and }3+9\cos t= 3 \implies t=\frac{\pi}{2}, {/eq}

and for a half rotation, {eq}\displaystyle \pi, {/eq} we will end {eq}\displaystyle t \text{ at } \frac{3\pi}{2}. {/eq}

So, a parametrization of the curve with the required direction is {eq}\displaystyle \boxed{x(t)=5+9\sin t, y(t)=3+9\cos t, \frac{\pi}{2} \leq t\leq \frac{3\pi}{2}}. {/eq}


Learn more about this topic:

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Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5
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