Find the rectangular equation of the curve whose parametric equations are x=-2sin t and y=cos t, where 0 less than or equal to tis less than or equal to 2pi.
How do I do this???
Thanks for all of your help!!
x^2/4 + y^2 = 1
That would be an ellipse with semi-major axis = 2 (horizonal) and semi-minor axis = 1 (vertical).
To find the rectangular equation of a curve given its parametric equations, we can eliminate the parameter by expressing one variable in terms of the other.
In this case, we have the parametric equations:
x = -2sin(t)
y = cos(t)
To eliminate the parameter t, we can use the trigonometric identity: sin^2(t) + cos^2(t) = 1.
Rearranging the first equation, we get:
sin(t) = -(1/2)x
Substituting this value of sin(t) into the second equation, we get:
y = cos(t) = cos(arcsin(-(1/2)x))
Since the original range of t is 0 ≤ t ≤ 2π, we can consider the corresponding range of arcsin to be -π/2 ≤ arcsin(-(1/2)x) ≤ π/2.
Using the Pythagorean identity, cos^2(t) = 1 - sin^2(t), we can simplify the equation:
y = cos(arcsin(-(1/2)x))
=> y = √(1 - sin^2(arcsin(-(1/2)x)))
=> y = √(1 - (-(1/2)x)^2)
=> y = √(1 - (1/4)x^2)
Simplifying further, we get the rectangular equation in terms of x and y:
y = √(1 - (1/4)x^2)
Therefore, the rectangular equation of the curve is y = √(1 - (1/4)x^2).