The graph of the parametric equations x=1+2cos3t and y=1-2sin3t, for [0, 2pi], is a circle of a radius whose magnitude is what?
sketch a graph
two around (1,1)
Wolfram shows a nice circle with centre (1,1)
http://www.wolframalpha.com/input/?i=plot+x%3D1%2B2cos3t+and+y%3D1-2sin3t
To find the radius, use one of the x-intercepts
let y = 0
0 = 1-2sin 3t
2sin 3t = 1
sin 3t = 1/2
3t = π/6
then x = 1 + 2cos π/6
= 1 + 2(√3/2) = 1 + √3 = appr 2.73 , the graph confirms that
radius is the distance between (1,1) and (1+√3,0)
= √[(1+√3 - 1)^2 + 1^2) ]
= √(3+1) = 2
you are correct
LOL, max is 1+2
min is 1-2
:)
To find the magnitude of the radius of the circle represented by the given parametric equations, we need to understand the relationship between the equations and the properties of a circle.
The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle, and r represents the radius.
In this case, we have x = 1 + 2cos(3t) and y = 1 - 2sin(3t). By comparing these equations with the general equation of a circle, we can determine the center coordinates and the radius.
Comparing the x-coordinate equation with the general equation, we have:
(x - 1)^2 = (2cos(3t))^2 = 4cos^2(3t)
Comparing the y-coordinate equation, we have:
(y - 1)^2 = (-2sin(3t))^2 = 4sin^2(3t)
Adding both equations together, we have:
(x - 1)^2 + (y - 1)^2 = 4cos^2(3t) + 4sin^2(3t)
Using the trigonometric identity cos^2(θ) + sin^2(θ) = 1, we simplify the equation to:
(x - 1)^2 + (y - 1)^2 = 4
From the equation above, we can see that the center of the circle is at (1, 1), and the radius squared is 4. Therefore, the magnitude of the radius is √4, which is 2.
Hence, the magnitude of the radius of the circle represented by the given parametric equations is 2.