Differentiating Parametric Functions

[danago]

Given:
$$\begin{array}{l} x = 2\cos t \\ y = 2\sin t \\ \end{array}$$

find $$\frac{{dy}}{{dx}}$$



I started by finding dy/dt and dx/dt

$$\begin{array}{l} \frac{{dy}}{{dt}} = 2\cos t \\ \frac{{dx}}{{dt}} = - 2\sin t \\ \end{array}$$

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

so:

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

$$\begin{array}{l} x = 2\cos t \\ \therefore t = \cos ^{ - 1} \frac{x}{2} \\ \end{array}$$

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

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

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.

 

[neutrino]

Do you know that sin²t + cos²t = 1? ;)

 

[danago]

Do you know that sin²t + cos²t = 1? ;)

Yep i do. Not quite seeing where it comes into it though

EDIT: ahhh now I've got it. Square my x and y functions, and then add them, which eliminates the parameter :) Thanks for that.

 

[Dick]

You could also continue on your original course. cot(cos^(-1)(x/2)) can be written without use of trig functions. Draw a right triangle containing an angle cos^(-1)(x/2).