Q. 38 - Calculus

Question

 The curve is a circle centered at the point $(a, b)$. It is traced once, counterclockwise, starting at the point $(a+r, b)$ with $t\in [0,2\pi ]$

Step-by-Step Solution

Answer

The parametric equations are $x=a+r\cos \left(kt\right),y=b+r\sin \left(kt\right)$ .

Step 1: Given Information

The curve of a circle centered at the point $(a, b)$ with $t\in [0,2\pi ]$.

Step 2: Calculation

Consider a curve of a circle centered at the point $(a, b)$ with $t\in [0,2\pi ]$.

The objective is to find the parametric equations which represent the given condition.

Given that the curve is centered at the $(a, b)$, traced once in counterclockvise direction.

The parametric equations which moves counterclockwise direction centered at the point$(a, b)$ is given by

$x\left(t\right)=a+r\cos \left(kt\right),y\left(t\right)=b+r\sin \left(kt\right)$

The parametric equations which moves counterclockwise direction starting at $(a+r, b)$.

At the starting point $t=0$, if substitute t value gives the starting point.

That is,

$$\begin{gathered} \left(x\right(t),y(t\left)\right)=(a+r\cos (kt),b+r\sin (kt\left)\right) \\ (x,y)=(a+r\cos (k·0),b+r\sin (k·0\left)\right) \\ (x,y)=(a+r·1,b+r\sin 0) \\ (x,y)=(a+r,b) \end{gathered}$$

Here the coordinate is

$\left(x\right(t),y(t\left)\right)=(a+r\cos (kt),b+r\sin (kt\left)\right)$

$x\left(t\right)=a+r\cos \left(kt\right)$ and $y\left(t\right)=b+r\sin \left(kt\right)$ forms a circle with center $(a, b)$ and which starts at $(a+r, b)$ moving in counterclockwise direction.

Therefore, the required parametric equations are $x=a+r\cos \left(kt\right),y=b+r\sin \left(kt\right)$.