Asked by Anonymous
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?
Answers
Answered by
Anonymous
sketch a graph
two around (1,1)
two around (1,1)
Answered by
Reiny
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
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
Answered by
Anonymous
LOL, max is 1+2
min is 1-2
:)
min is 1-2
:)
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