Question

find an equation of the tangent to the curve given by

$x = 2\sin(2t)$, $y = 2\sin(t)$

at the point $(x,y)=(\sqrt{3},1)$. your answer should be in the form of $y = f(x)$ without $ts$.

$y(x)=$

Explanation:

Step1: Find the value of \( t \)

Given \( x = 2\sin(2t) \), \( y = 2\sin(t) \) and the point \( (\sqrt{3}, 1) \). We know that \( y = 2\sin(t)=1 \), so \( \sin(t)=\frac{1}{2} \). Then \( t=\frac{\pi}{6}+2k\pi \) or \( t = \frac{5\pi}{6}+2k\pi \), \( k\in\mathbb{Z} \). We check which \( t \) satisfies \( x = 2\sin(2t)=\sqrt{3} \).

For \( t=\frac{\pi}{6} \): \( \sin(2t)=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2} \), so \( x = 2\times\frac{\sqrt{3}}{2}=\sqrt{3} \), which matches. For \( t=\frac{5\pi}{6} \): \( \sin(2t)=\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2} \), \( x = 2\times(-\frac{\sqrt{3}}{2})=-\sqrt{3} \), which does not match. So \( t = \frac{\pi}{6} \).

Step2: Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

Using the chain rule, \( \frac{dx}{dt}=\frac{d}{dt}(2\sin(2t)) = 2\times2\cos(2t)=4\cos(2t) \)

\( \frac{dy}{dt}=\frac{d}{dt}(2\sin(t)) = 2\cos(t) \)

Step3: Find the slope \( \frac{dy}{dx} \)

By the formula for parametric differentiation, \( \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) (when \( \frac{dx}{dt}
eq0 \))

Substitute \( t = \frac{\pi}{6} \) into \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

\( \frac{dx}{dt}\big|_{t = \frac{\pi}{6}}=4\cos(2\times\frac{\pi}{6})=4\cos(\frac{\pi}{3})=4\times\frac{1}{2}=2 \)

\( \frac{dy}{dt}\big|_{t = \frac{\pi}{6}}=2\cos(\frac{\pi}{6})=2\times\frac{\sqrt{3}}{2}=\sqrt{3} \)

So \( \frac{dy}{dx}\big|_{t = \frac{\pi}{6}}=\frac{\sqrt{3}}{2} \)

Step4: Find the equation of the tangent line

Using the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(\sqrt{3},1) \) and \( m=\frac{\sqrt{3}}{2} \)

\( y - 1=\frac{\sqrt{3}}{2}(x - \sqrt{3}) \)

Expand the right - hand side: \( y-1=\frac{\sqrt{3}}{2}x-\frac{3}{2} \)

Then \( y=\frac{\sqrt{3}}{2}x-\frac{3}{2}+1=\frac{\sqrt{3}}{2}x-\frac{1}{2} \)

Answer:

\( y(x)=\frac{\sqrt{3}}{2}x-\frac{1}{2} \)