Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations.
x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π
A. x2 - y2 = 6; -6 ≤ x ≤ 6
B. x2 - y2 = 36; -6 ≤ x ≤ 6
C. x2 + y2 = 6; -6 ≤ x ≤ 6
D. x2 + y2 = 36; -6 ≤ x ≤ 6
D ?
Yes it's D.
Thank you
Can someone show the work, please?
use trig identity- cos^2 t + sin^2 t = 1
x = 6 cos t, y = 6 sin t
square both: x^2 = 6cos^2 t y^2 = 6sin^2 t
combine: x^2 + y^2 = 6cos^2 t + 6sin^2
use trig identity to eliminate sin^2 t and cos^2 t
x^2 + y^2 = 1x6x6
x^2 + y^2 = 36
To eliminate the parameter t and find a rectangular equation for the plane curve defined by the parametric equations x = 6cos(t) and y = 6sin(t), we need to express x and y in terms of each other.
Starting with the given parametric equations:
x = 6cos(t)
y = 6sin(t)
We can square both equations:
x^2 = (6cos(t))^2 = 36cos^2(t)
y^2 = (6sin(t))^2 = 36sin^2(t)
Now, let's add these two equations:
x^2 + y^2 = 36cos^2(t) + 36sin^2(t)
Using the trigonometric identity cos^2(t) + sin^2(t) = 1, we can simplify the equation to:
x^2 + y^2 = 36(1)
This equation represents a circle with radius 6 centered at the origin (0, 0) in the Cartesian coordinate system.
Comparing this result to the given options, we can see that the correct answer is:
D. x^2 + y^2 = 36; -6 ≤ x ≤ 6