Given the parametric equations [tex]\( x = 2 \cos t \)[/tex] and [tex]\( y = 2 \sin t \)[/tex], we need to find [tex]\(\frac{d y}{d x}\)[/tex] at [tex]\( t = \frac{\pi}{3} \)[/tex].
To solve this problem, follow these steps:
1. Differentiate [tex]\( x \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{dx}{dt} = \frac{d}{dt}(2 \cos t) = -2 \sin t
\][/tex]
2. Differentiate [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{dy}{dt} = \frac{d}{dt}(2 \sin t) = 2 \cos t
\][/tex]
3. Evaluate [tex]\(\frac{dx}{dt}\)[/tex] and [tex]\(\frac{dy}{dt}\)[/tex] at [tex]\( t = \frac{\pi}{3} \)[/tex]:
[tex]\[
\frac{dx}{dt} \bigg|_{t = \frac{\pi}{3}} = -2 \sin \left(\frac{\pi}{3}\right) = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3}
\][/tex]
[tex]\[
\frac{dy}{dt} \bigg|_{t = \frac{\pi}{3}} = 2 \cos \left(\frac{\pi}{3}\right) = 2 \times \frac{1}{2} = 1
\][/tex]
4. Compute [tex]\(\frac{dy}{dx}\)[/tex] using the chain rule:
[tex]\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\][/tex]
[tex]\[
\frac{dy}{dx} \bigg|_{t = \frac{\pi}{3}} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}
\][/tex]
5. Rationalize the denominator:
[tex]\[
\frac{dy}{dx} = -\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
\][/tex]
Therefore, [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\( t = \frac{\pi}{3} \)[/tex] is [tex]\(-\frac{\sqrt{3}}{3}\)[/tex].
The correct answer is:
(B) [tex]\(-\frac{\sqrt{3}}{3}\)[/tex]
