Question

If $x = a{\sin ^3}\theta$ and $y = a{\cos ^3}\theta$, then find the value of $\dfrac{{dy}}{{dx}}$.

Answer 1

Hint: First find the value of $\dfrac{{dx}}{{d\theta }}$ and $\dfrac{{dy}}{{d\theta }}$, then divide them to get $\dfrac{{dy}}{{dx}}$. Simplify the answer to express it in terms of x and y.

Complete step-by-step answer:

Let us start solving by finding the expression for $\dfrac{{dx}}{{d\theta }}$.
$\dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}(a{\sin ^3}\theta )$
We can use $\dfrac{d}{{dx}}(a{x^3}) = 3a{x^2}$ to simplify the equation.
$\dfrac{{dx}}{{d\theta }} = 3a{\sin ^2}\theta \dfrac{d}{{d\theta }}(\sin \theta )$
We know that $\dfrac{d}{{dx}}(\sin x) = \cos x$, hence, we have the following:
$\dfrac{{dx}}{{d\theta }} = 3a{\sin ^2}\theta \cos \theta .............(1)$
Now, let us find the expression for $\dfrac{{dy}}{{d\theta }}$.
$\dfrac{{dy}}{{d\theta }} = \dfrac{d}{{d\theta }}(a{\cos ^3}\theta )$
We can use $\dfrac{d}{{dx}}(a{x^3}) = 3a{x^2}$ to simplify the equation.
$\dfrac{{dx}}{{d\theta }} = 3a{\cos ^2}\theta \dfrac{d}{{d\theta }}(\cos \theta )$
We know that $\dfrac{d}{{dx}}(\cos x) = - \sin x$, hence, we have the following:
$\dfrac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta ............(2)$
We know that,
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}}...........(3)$
Using equation (1) and equation (2) in equation (3), we have:

$\dfrac{{dy}}{{dx}} = \dfrac{{3a{{\sin }^2}\theta \cos \theta }}{{ - 3a{{\cos }^2}\theta \sin \theta }}$
Cancelling common terms in the numerator and the denominator we have:
$\dfrac{{dy}}{{dx}} = - \dfrac{{\sin \theta }}{{\cos \theta }}$
We know that $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$, hence we have:
$\dfrac{{dy}}{{dx}} = - \tan \theta ..........(4)$
We can write equation (4) in terms of x and y.
Let us find the value of $\dfrac{x}{y}$.
$\dfrac{x}{y} = \dfrac{{a{{\sin }^3}\theta }}{{a{{\cos }^3}\theta }}$
Simplifying, we get:
$\dfrac{x}{y} = \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}$
$\dfrac{x}{y} = {\tan ^3}\theta$
Let us compute $\tan \theta$ in terms of x and y by taking the cube root on both sides.
$\tan \theta = \sqrt[3]{{\dfrac{x}{y}}}.........(5)$
Substituting equation (5) in equation (4), we get:
$\dfrac{{dy}}{{dx}} = - \sqrt[3]{{\dfrac{x}{y}}}$
Hence, the answer is $- \sqrt[3]{{\dfrac{x}{y}}}$.

Note: If you express the final answer in terms of $\theta$, it is a wrong answer. Express the final answer in terms of x and y only.