Answer:
[tex]x(t) = 4 cos(-t)\\y(t) = 4 sin(-t) + 3\\[/tex]
and t ∈ [0,2π]
Step-by-step explanation:
The standard equation for circle is:
[tex](x-h) ^2 + (y-k)^2 = r^2[/tex]
Comparing our equation with standard
[tex]x^2 + (y-3)^2 =16[/tex]
h= 0,
k= 3
r= 4 (as in standard r², so r =√ r²)
Required:
Parametric equations of the circle
Formula:
[tex]x = r cos(t) + h\\y= r cos(t) +k[/tex]
Putting values of r and h and k we get
[tex]x= 4 cos(t) + 0 \\x= 4 cost(t) \\y= 4 sin(t) + 3[/tex]
As we need to start the object at (4,3)
x(0) = 4 and y(0) = 3
and the period is 2π
As the rotation is starting clock wise we will invert the value of t i.e -t
[tex]x(t) = 4 cos(-t)\\y(t) = 4 sin(-t) + 3\\[/tex]
and t ∈ [0,2π]
The parametric equations for the circle are:
[tex]x(t) = 4\cos{t}[/tex]
[tex]y(t) = -4\sin{t} + 3[/tex]
The equation of a circle of radius r and center [tex](x_0, y_0)[/tex] is given by:
[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]
The parametric equations, considering a counter-clockwise movement, are given by:
[tex]x(t) = r\cos{t} + x_0[/tex]
[tex]y(t) = r\sin{t} + y_0[/tex]
Considering a clockwise movement, t is replaced by -t, thus:
[tex]x(t) = r\cos{-t} + x_0[/tex]
[tex]y(t) = r\sin{-t} + y_0[/tex]
However, considering the even cosine function and the odd sine function, we have that:
[tex]\cos{-t} = \cos{t}, \sin{-t} = -\sin{t}[/tex]
Hence:
[tex]x(t) = r\cos{t} + x_0[/tex]
[tex]y(t) = -r\sin{t} + y_0[/tex]
In this problem, the circle's equation is given by:
[tex]x^2 + (y - 3)^2 = 16[/tex]
Hence [tex]x_0 = 0, y_0 = 3, r = 4[/tex]
Thus, the parametric equations are:
[tex]x(t) = 4\cos{t}[/tex]
[tex]y(t) = -4\sin{t} + 3[/tex]
A similar problem is given at https://brainly.com/question/23719612
