Find the parametrization for the curve.
-The portion of the circle $x ^ { 2 } + y ^ { 2 } = 6$ that lies in the second quadrant
A) $x = 6 \sin t , y = 6 \cos t$, for $\pi

Question

Find the parametrization for the curve.
-The portion of the circle $x ^ { 2 } + y ^ { 2 } = 6$ that lies in the second quadrant
A) $x = 6 \sin t , y = 6 \cos t$, for $\pi < t < \frac { 3 \pi } { 2 }$
B) $x = 6 \cos t , y = 6 \sin t$, for $\frac { \pi } { 2 } < t < \pi$
C) $x = 6 \cos t , y = 6 \sin t$, for $\frac { 3 \pi } { 2 } < t < 2 \pi$
D) $x = 6 \sin t , y = 6 \cos t$, for $\frac { \pi } { 2 } < t < \pi$

Quizplus · Accepted Answer

The standard parametrization for a circle centered at the origin with radius $r$ is $x = r \cos t, y = r \sin t$. Since the radius here is $\sqrt{6}$, and we want the circle $x^2 + y^2 = 6$, we adjust the parametrization to $x = \sqrt{6} \cos t, y = \sqrt{6} \sin t$. However, the options provided use a factor of 6, which is incorrect for the radius but follows the pattern of parametrization. The second quadrant corresponds to angles from $\frac{\pi}{2}$ to $\pi$, so the correct parametrization that matches the format of the options and the quadrant is $x = \sqrt{6} \cos t, y = \sqrt{6} \sin t$ for $\frac{\pi}{2} < t < \pi$, which is conceptually closest to option B, despite the incorrect radius scaling.

Find the Parametrization for the Curve x2+y2=6x ^ { 2 } + y ^ { 2 } = 6x2+y2=6

Find the Parametrization for the Curve $x ^ { 2 } + y ^ { 2 } = 6$