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Question 14

Given that [tex]\( x = 2 \cos t \)[/tex] and [tex]\( y = 2 \sin t \)[/tex], where [tex]\( 0 \leq t \leq 2\pi \)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\( t = \frac{\pi}{3} \)[/tex].

(A) [tex]\(\frac{1}{2}\)[/tex]

(B) [tex]\(-\frac{\sqrt{3}}{3}\)[/tex]

(C) [tex]\(\frac{3}{\sqrt{3}}\)[/tex]

(D) [tex]\(-\frac{1}{2}\)[/tex]

(E) [tex]\(\frac{\sqrt{3}}{3}\)[/tex]


Sagot :

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]