KevinReed8292
21.09.2019 •
Physics
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. x^2 - y^2 = 6; -6 ≤ x ≤ 6
b. x^2 - y^2 = 36; -6 ≤ x ≤ 6
c. x^2 + y^2 = 36; -6 ≤ x ≤ 6
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Ответ:
Take square
x^2 = 36cos^2t
y^2 = 36sin^2t , add both
x^2+y^2= 36( cos^2t+sin^2t)
x^2+y^2= 36 (1)
x^2+y^2=36 a circle with radius R^2=36 , ie, R=+-6 , so -6<x<6 ANSWER C
The other way ,
cos t = x/6 in a right triangle , adjacent side is x , hypotenuse is 6
so opposite side is b^2= 6^2-x^2
b^2 = 36-x^2
thus, sin t = b/6 ,
sint = sqrt ( 36-x^2) /6
We know that
y = 6sint
y = 6 sqrt (36-x^2) /6
y = sqrt (36-x^2)
taking square
x^2+y^2=36
Ответ: