How do you find the Cartesian equation of the curve with parametric equations #x=2cos(3t)#and #y=2sin(3t)#, and determine the domain and range of the corresponding relation?

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Answer 1 · 2017-09-13T17:46:18

The domain and range are easily obtained before converting to the Cartesian Equation.

Please observe that the range cosine function, #-1 <=cos(u)<=1#, causes the domain for x to be:

#-2 <= x <= 2#

The range for y is limited in the same way but with the sine function:

#-2 <= y <=2#

Use the equation for x to find an equation for #sin(3t)#

#x = 2cos(3t)#

Use the identity #cos(u) = +-sqrt(1-sin^2(u))#:

Because we are going to square everything, we shall only use the positive value:

#x = 2sqrt(1-sin^2(3t))#

#x^2/4 = 1 - sin^2(3t)#

#sin^2(3t) = 1 - x^2/4#

#sin(3t) = +-sqrt(1 - x^2/4)#

#y = {(2sqrt(1-x^2/4)),(-2sqrt(1-x^2/4)):}; -2 <= x <= 2#

This should look like a circle to you, therefore, a better Cartesian form is found by squaring y:

#y^2 = 4(1 - x^2/4)#

#x^2 + y^2 = 2^2#