Question

Transform to Cartesian coordinates the equation:
$r\left( {\cos 3\theta + \sin 3\theta } \right) = 5k\sin \theta \cos \theta$

Answer 1

In order to transform the given equation into Cartesian coordinates, we first expand the given equation $\left( {\sin 3\theta {\text{ and }}\cos 3\theta } \right)$ and then substitute some variable in such a way that it expresses all the trigonometric and the variables in the given equation in terms of itself. Thus, we obtain a quadratic equation.

Complete step by step answer:
Given Data,
$r\left( {\cos 3\theta + \sin 3\theta } \right) = 5k\sin \theta \cos \theta$
The given Cartesian equation is, $r\left( {\cos 3\theta + \sin 3\theta } \right) = 5k\sin \theta \cos \theta$ --- (1)

Now we know the formula of sin 3θ and cos 3θ are given as follows:
$\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$ and $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$
Substituting the above formulae in equation (1), we get
$\Rightarrow r\left( {\left( {4{{\cos }^3}\theta - 3\cos \theta } \right) + \left( {3\sin \theta - 4{{\sin }^3}\theta } \right)} \right) = 5k\sin \theta \cos \theta$
$\Rightarrow r\left[ {4\left( {{{\cos }^3}\theta - {{\sin }^3}\theta } \right) + 3\left( {\sin \theta - \cos \theta } \right)} \right] = 5k\sin \theta \cos \theta$ ……… (2)
Now let us consider two variables x and y such that $x = r\sin \theta$ and $y = r\cos \theta$, substitute them in equation (2), we get
$\Rightarrow r\left[ {4\left( {{{\left( {\dfrac{y}{r}} \right)}^3} - {{\left( {\dfrac{x}{r}} \right)}^3}} \right) + 3\left( {\dfrac{x}{r} - \dfrac{y}{r}} \right)} \right] = 5k\left( {\dfrac{{xy}}{{{r^2}}}} \right)$
$\Rightarrow r\left[ {4\left( {\dfrac{{{y^3} - {x^3}}}{{{r^3}}}} \right) + 3\left( {\dfrac{{x - y}}{r}} \right)} \right] = 5k\left( {\dfrac{{xy}}{{{r^2}}}} \right)$
$\Rightarrow \dfrac{1}{{{r^2}}}\left[ {4\left( {{y^3} - {x^3}} \right) + 3{r^2}\left( {x - y} \right)} \right] = 5k\left( {\dfrac{{xy}}{{{r^2}}}} \right)$
$\Rightarrow \left[ {4\left( {{y^3} - {x^3}} \right) + 3{r^2}\left( {x - y} \right)} \right] = 5k\left( {xy} \right)$ ……. (3)
As we know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$, and here $\sin \theta = \dfrac{x}{r}$ and $\cos \theta = \dfrac{y}{r}$
Putting this, we will get
$\Rightarrow {\left( {\dfrac{x}{r}} \right)^2} + {\left( {\dfrac{y}{r}} \right)^2} = 1$
$\Rightarrow {x^2} + {y^2} = {r^2}$
Put this value of ${r^2}$ in equation (3),
$\Rightarrow \left[ {4\left( {{y^3} - {x^3}} \right) + 3\left( {{x^2} + {y^2}} \right)\left( {x - y} \right)} \right] = 5k\left( {xy} \right)$
Now, solving this, we will get
$\Rightarrow \left[ {4{y^3} - 4{x^3} + 3\left( {{x^3} - {x^2}y + {y^2}x - {y^3}} \right)} \right] = 5k\left( {xy} \right)$
$\Rightarrow \left[ {4{y^3} - 4{x^3} + 3{x^3} - 3{x^2}y + 3{y^2}x - 3{y^3}} \right] = 5k\left( {xy} \right)$
$\Rightarrow \left[ {{y^3} - {x^3} - 3{x^2}y + 3{y^2}x} \right] = 5k\left( {xy} \right)$
Therefore, the given equation $r\left( {\cos 3\theta + \sin 3\theta } \right) = 5k\sin \theta \cos \theta$ is expressed in Cartesian form as
$\Rightarrow \left[ {{y^3} - {x^3} - 3{x^2}y + 3{y^2}x} \right] = 5k\left( {xy} \right)$

Note: In order to solve this type of questions the key is to know how to consider a variable that eliminates all the trigonometric terms in the equation and converts them into Cartesian form. This can be achieved as we keep on solving problems like this.
Here the way we consider the variable x and y, we end up eliminating all the r terms and θ terms in the equation in order to obtain a pure quadratic equation.
It is very essential to know a few useful trigonometric formulae such as sin 3θ, cos 3θ and the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ which help us reach to the answer in problems like these.