Question

prove that $\dfrac{dy}{dx}=-\sqrt[3]{\dfrac{y}{x}}$ If $x=a{{\cos }^{3}}\theta$ and $y=a{{\sin }^{3}}\theta$

Answer 1

We need to find a relation between ‘x’, ‘y’ and the first derivative of ‘y’ with respect to ‘x’. But we can’t simply differentiate ‘x’ and ‘y’ because they are given in the terms of $\theta$ . So we will first differentiate in terms of $\theta$ . Then find $\dfrac{dy}{dx}$ and $\dfrac{y}{x}$ , then finally relate each of them to get the above result.

Complete step by step answer:
Moving ahead with the question in step-wise manner;
$x=a{{\cos }^{3}}\theta$
$y=a{{\sin }^{3}}\theta$
To prove $\dfrac{dy}{dx}=-\sqrt[3]{\dfrac{y}{x}}$
First differentiate ‘x’ and ‘y’ in terms of $\theta$ .
So first differentiate x, i.e.
$\begin{aligned} & x=a{{\cos }^{3}}\theta \\\ & \dfrac{dx}{d\theta }=-3a{{\left( \cos \theta \right)}^{2}}\sin \theta \\\ \end{aligned}$ equation (i)
Similarly differentiating y, we will get;
$\begin{aligned} & y=a{{\sin }^{3}}\theta \\\ & \dfrac{dy}{d\theta }=3a{{\left( \sin \theta \right)}^{2}}\cos \theta \\\ \end{aligned}$ equation (ii)
Now to find $\dfrac{dy}{dx}$ , divide equation (ii) with equation (i), so we will get;
$\begin{aligned} & \dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}=\dfrac{3a{{\left( \sin \theta \right)}^{2}}\cos \theta }{-3a{{\left( \cos \theta \right)}^{2}}\sin \theta } \\\ & \dfrac{dy}{dx}=-\dfrac{\sin \theta }{\cos \theta } \\\ & \dfrac{dy}{dx}=-\tan \theta \\\ \end{aligned}$
So we got $\dfrac{dy}{dx}$ which is equal to $-\tan \theta$ .
Now we need relation between $\dfrac{dy}{dx}$ and $\dfrac{y}{x}$ so we got $\dfrac{dy}{dx}$ which is $-\tan \theta$ . Now find out $\dfrac{y}{x}$ so that we can relate these both two, and can get the result given in question and asked to prove it.
So as we know that;
$y=a{{\sin }^{3}}\theta$ and $x=a{{\cos }^{3}}\theta$ , so $\dfrac{y}{x}$ will be
$\begin{aligned} & \dfrac{y}{x}=\dfrac{a{{\cos }^{3}}\theta }{a{{\sin }^{3}}\theta } \\\ & \dfrac{y}{x}={{\tan }^{3}}\theta \\\ \end{aligned}$
So we got $\dfrac{y}{x}$ which is ${{\tan }^{3}}\theta$ .
Now comparing $\dfrac{dy}{dx}$ and $\dfrac{y}{x}$ , we have $\dfrac{dy}{dx}=-\tan \theta$ and $\dfrac{y}{x}={{\tan }^{3}}\theta$
If we cube root $\dfrac{y}{x}$ we will get value $\tan \theta$ which is equal to negative of $\dfrac{dy}{dx}$ and if we multiply it with negative then we will get value equal to $\dfrac{dy}{dx}$ . i.e.
$-\dfrac{dy}{dx}=\sqrt[3]{\dfrac{y}{x}}$
Now multiply with negative to have a $\dfrac{dy}{dx}$ positive, so we will get;
$\dfrac{dy}{dx}=-\sqrt[3]{\dfrac{y}{x}}$
Which is asked to prove in the question.
Hence we had proved it.

Note: For $\dfrac{dy}{dx}$ we had divided as normal division in $\dfrac{dx}{d\theta }$ and $\dfrac{dy}{d\theta }$ , in which $\dfrac{dy}{d\theta }$ will be above and $\dfrac{dx}{d\theta }$ will be in denominator because we want $dx$ in denominator.