# Question # 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 leq t leq pi

Parametric equations, polar coordinates, and vector-valued functions
ANSWERED 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. $$x=\ -7\ \cos\ 2t,\ y=\ -7\ \sin\ 2t,\ 0\ \leq\ t\ \leq\ \pi$$ 2021-02-22

Step 1 Given: The following parametric equations, $$x=\ -7\ \cos(2t),\ y=\ -7\ \sin(2t),\ 0\ \leq\ t\ \leq\ \pi.$$

Step 2 a) To eliminate the parameter to obtain an equation in x and y. We have, The following parametric equations, (1)$$x =\ −7 \cos(2t)$$
$$y =\ −7 \sin(2t)$$(2) Adding equations (1) and (2), we get $$x^{2}\ +\ y^{2}=(-7\ \cos(2t))^{2}\ +\ (-7\ \sin(2t))^{2}$$
$$=49\ \cos^{2}(2t)\ +\ 49\ \sin^{2}(2t)$$
$$=49(\cos^{2}(2t)\ +\ \sin^{2}(2t))$$
$$=49(1)\ \because \text{By using trignometric identity}$$
$$=49$$ An equation in x and y is $$x^{2}\ +\ y^{2} = 49.$$ Step 3 b) To describe the curve and indicate the positive orientation. We have, $$x^{2}\ +\ y^{2} = (7)^{2}.$$ The curve represents a circle at the origin, $$(0,\ 0)\ \text{and radius,}\ r = 7.$$ To indicate the positive orientation, we will use the following parametric equations, $$x =\ −7\ \cos(2t),\ y =\ −7\ \sin(2t),\ 0\ \leq\ t\ \leq\ \pi.$$ The interval given is $$0\ \leq\ t\ \leq\ \pi.$$
$$x =\ −7\ \cos(2(0)) =\ −7(\cos(0)) =\ −7(1) =\ −7$$
$$y =\ −7\ \sin(2(0)) =\ −7(\sin(0)) =\ −7(0) = 0.$$ Implies the initial point is $$(−7,\ 0).$$ For $$t = \pi,$$
$$x =\ −7\ \cos(2\ \pi) =\ −7(1) =\ −7y =\ −7\ \sin(2\ \pi) =\ −7(0) = 0.$$ 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.