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

Hint: 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{align}
  & x=a{{\cos }^{3}}\theta \\
 & \dfrac{dx}{d\theta }=-3a{{\left( \cos \theta \right)}^{2}}\sin \theta \\
\end{align} $ equation (i)
Similarly differentiating y, we will get;
 $ \begin{align}
  & y=a{{\sin }^{3}}\theta \\
 & \dfrac{dy}{d\theta }=3a{{\left( \sin \theta \right)}^{2}}\cos \theta \\
\end{align} $ equation (ii)
Now to find \[\dfrac{dy}{dx}\] , divide equation (ii) with equation (i), so we will get;
 $ \begin{align}
  & \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{align} $
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{align}
  & \dfrac{y}{x}=\dfrac{a{{\cos }^{3}}\theta }{a{{\sin }^{3}}\theta } \\
 & \dfrac{y}{x}={{\tan }^{3}}\theta \\
\end{align} $
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.