Derivative calculation of parametric curves

Question

I have some problem in the calculation of the derivative $\frac{dy}{dx}$ of the following parametric functions: $$x(\theta)=a\cos(\theta)\left[ 1-\frac{1}{2}\cos(\theta)^2\right],y(\theta)=\frac{1}{2}a\sin(\theta)\cos(\theta)^2$$ I tried to obtain $\theta$ from $x(\theta)$, but the functions seems to be not easily invertible. I would appreciate solutions or suggestions. Thanks

Answer 1

Indeed, a route to give an implicit formula for $f(x,y)=0$ exists. The ratio between the two equations give $$ \frac{y(\theta)}{x(\theta)}=\frac{1}{2}\tan\theta\left(\tan^2\theta+\frac{1}{2}\right). $$ Putting $t=\tan\theta$, this is just an algebraic equation easy to invert. The only real solution is $$ \tan\theta=\frac{1}{6}P^\frac{1}{3}\left(\frac{y}{x}\right)-\frac{1}{P^\frac{1}{3}\left(\frac{y}{x}\right)} $$ being $P(z)=216z+6\sqrt{1296z+6}$. Now, it is enough to yield $\sin\theta$ and $\cos\theta$ as functions of $\tan\theta$ and, using either $x(\theta)$ or $y(\theta)$ you will be able to get an implicit form for your equation. This does not make easier computations but has the advantage to permit working with just $x$ and $y$. Please, note that $P(z)$ is always different from zero but the ratio given above is singular wherever $\tan\theta$ is and there you cannot use it. These are the points where $x(\theta)$ and $y(\theta)$ are 0 and their ratio is meanignless.

Answer 2

You don't need to express $\theta$ as a function of $x$. You just calculate $$ \frac{dy}{dx} = \frac{dy}{d\theta}\Bigg/\frac{dx}{d\theta} $$

Answer 3

The best hint is to use the chain rule. In fact, $$\frac{dx}{dy}=\frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}$$