Working with parametric equations Consider the following parametric equations. a. Eliminate the parameter to obtain an equation in x and y. b. Describe the curve and indicate the positive orientation. x = -7 cos 2t, y = -7 sin 2t, 0 <= t <= pi

Step 1
Given:
The following parametric equations, ${\displaystyle x=-7\cos \left(2t\right),y=-7\sin \left(2t\right),0\le t\le \pi }$.
Step 2
a) To eliminate the parameter to obtain an equation in x and y.
We have,
The following parametric equations,
${\displaystyle x=-7\cos \left(2t\right)}$....(1)
${\displaystyle y=-7\sin \left(2t\right)}$....(2)
Adding equations (1) and (2), we get
${\displaystyle x^2+y2={(-7\cos \left(2t\right))}^2+{(-7\sin \left(2t\right))}^2}$
${\displaystyle =49{\cos }^2\left(2t\right)+49{\sin }^2\left(2t\right)}$
${\displaystyle =49({\cos }^2\left(2t\right)+{\sin }^2\left(2t\right))}$
${\displaystyle =49\left(1\right)\dots .{}^{\prime }\text{By using trigonometricident }\text{if}y\}}$
=49
An equation in x and y is ${\displaystyle x^2+y^2=49}$.
Step 3
b) To describe the curve and indicate the positive orientation.
We have,
${\displaystyle x^2+y^2={\left(7\right)}^2}$.
The curve represents a circle at the origin, (0,0) and radius, r=7.
To indicate the positive orientation, we will use the following parametric equations,
${\displaystyle x=-7\cos \left(2t\right),y=-7\sin \left(2t\right),0\le t\le \pi }$.
The interval given is ${\displaystyle 0\le t\le \pi }$.
Therefore, for t=0,
${\displaystyle x=-7\cos \left(2\left(0\right)\right)=-7\left(\cos \left(0\right)\right)=-7\left(1\right)=-7}$
${\displaystyle y=-7\sin \left(2\left(0\right)\right)=-7\sin \left(0\right))=-7\left(0\right)=0}$.
Implies the initial point is (−7,0).
${\displaystyle F\text{or}t=\pi }$,
${\displaystyle x=-7\cos \left(2\pi \right)=-7\left(1\right)=-7}$
${\displaystyle y=-7\sin \left(2\pi \right)=-7\left(0\right)=0}$
Implies the end point is (−7,0).
The initial point and the end point are equal, that is, (−7,0).
Hence, the orientation is positive in anticlockwise direction.