- #1

- 1,123

- 4

Given:

[tex]

\begin{array}{l}

x = 2\cos t \\

y = 2\sin t \\

\end{array}

[/tex]

find [tex]

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

[/tex]

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 isn't even in terms of any trigonometric function. Where have i gone wrong?

Thanks,

Dan.