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Find the equation of the line tangent to the curve at the point defined by t. $t=\frac{-1}{6}$, $x=\sin(2\pi t)$, $y=cos(2\pi t)$. How would I go about solving this problem. I have taken the derivative of both $x$ and $y$. Then, I have tried to use the quotient rule. I do not know what to do next.

This is what I have done so far:

$x'=2\pi\cos(2\pi t)$

$y'= -2\pi\sin(2\pi t)$

$\frac{dy}{dx}= (-2\pi\sin(2\pi t))/(2\pi\cos(2\pi t)$

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  • $\begingroup$ Can you show what you did? $\endgroup$ – alex Mar 5 '14 at 15:58
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    $\begingroup$ You've done all the calculus you need. At this point, you need to use $t=-\frac{1}{6}$ and evaluate $\frac{dy}{dx}$, $x$ and $y$ to get the info you need to form the tangent line. $\endgroup$ – John Habert Mar 5 '14 at 16:05
  • $\begingroup$ Just plug in the value of t ($=\frac{1}{6}$) in $\frac{dy}{dx}$, using the slope-interecept form $y=\frac{dy}{dx}x+c$, and you have your answer. $\endgroup$ – Vibhav Pant Mar 5 '14 at 16:08
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When you have a curve given in parametric form $$x=f(t)\\y=g(t)$$ as you do, you can write straightforwardly the parametric equations of the tangent in the point determined by a certain value of $t$, let's say $t_0$: $$x=f'(t_0)t+f(t_0)\\y=g'(t_0)t+g(t_0).$$ In your case $f(x)=\sin(2\pi t)$, $g(x)=\cos(2\pi t)$, $t_0=-\frac 1 6$. If you want the Cartesian form of the equation you have to eliminate $t$.

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find $x$ and $y$ at $t=\frac{-1}{6}$. these is the point through which the tangent passes. now the remaining thing you need is the slope of the tangent which is equal to $$\frac{dy}{dx}_{t=\frac{-1}{6}}$$ $$\frac{dy}{dx}_{t=\frac{-1}{6}}=\frac{\frac{dy}{dt} _{t=\frac{-1}{6}} }{\frac{dx}{dt} _{t=\frac{-1}{6}} }$$ then the equation of tangent is of the form $$y=\frac{dy}{dx}x+c$$ where $c$ is a constant that can be found with the help of the point that we got that passes through it.

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