(adsbygoogle = window.adsbygoogle || []).push({});

Given:

[tex]

\begin{array}{l}

x = 2\cos t \\

y = 2\sin t \\

\end{array}

[/tex]

find [tex]

\frac{{dy}}{{dx}}

[/tex]

I started by finding dy/dt and dx/dt

[tex]

\begin{array}{l}

\frac{{dy}}{{dt}} = 2\cos t \\

\frac{{dx}}{{dt}} = - 2\sin t \\

\end{array}

[/tex]

Now, dy/dx = (dy/dt) / (dx/dt)

so:

[tex]

\frac{{dy}}{{dx}} = \frac{{2\cos t}}{{ - 2\sin t}} = - \cot t

[/tex]

[tex]

\begin{array}{l}

x = 2\cos t \\

\therefore t = \cos ^{ - 1} \frac{x}{2} \\

\end{array}

[/tex]

Substituting that into my dy/dx which is currently in terms of t:

[tex]

\frac{{dy}}{{dx}} = - \cot t = - \cot \cos ^{ - 1} \frac{x}{2}

[/tex]

Now, my book gives a completely different answer, which isnt even in terms of any trigonometric function. Where have i gone wrong?

Thanks,

Dan.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Differentiating Parametric Functions

**Physics Forums | Science Articles, Homework Help, Discussion**