MHB -10.1.1 write polar to rectangular

[karush]

$\textrm{write polar to rectangular coordinates}$
$$r=5\sin{2\theta}$$
$\textit{Multiply both sides by $r$}$
$$r^2=5r[\sin{2\theta}]
=5\cdot2[r\cos(\theta)r\cos(\theta)]$$
$\textit{then substitute $r^2$ with $x^2+y^2$ and
$[r\cos(\theta)r\cos(\theta)$ with $xy$}\\$
$\textit{then}\\$
$$x^2+y^2=10xy$$
hopefully

 

[MarkFL]

$\textrm{write polar to rectangular coordinates}$
$$r=5\sin{2\theta}$$
$\textit{Multiply both sides by $r$}$
$$r^2=5r[\sin{2\theta}]
=5\cdot2[r\cos(\theta)r\cos(\theta)]$$

You multiplied the LHS by $r$, but in applying an $r$ to both trig functions on the RHS, you effectively multiply that side by $r^2$.

I would begin with:

$$r=5\sin(2\theta)=10\sin(\theta)\cos(\theta)$$

So, now if we multiply both sides by $r^2$, we get:

$$r^3=10r\sin(\theta)r\sin(\theta)$$

$$\left(x^2+y^2\right)^{\frac{3}{2}}=10xy$$

Now, we must observe that in this form, the LHS is always positive, and so we will miss the petals in the 2nd and 4th quadrants (where the RHS is negative), so to get those, we need to square both sides:

$$\left(x^2+y^2\right)^{3}=(10xy)^2$$

 

[karush]

I thought
$\displaystyle r=5sin(2x)$
and
$\displaystyle (x^2+y^2)^3=(10xy)^2$
would be the same graph?

one is a sine wave the other is a clover

 

[MarkFL]

I thought
$\displaystyle r=5sin(2x)$
and
$\displaystyle (x^2+y^2)^3=(10xy)^2$
would be the same graph?

The polar plot:

$$r=5\sin(2\theta)$$

And the Cartesian plot:

$$\left(x^2+y^2\right)^{3}=(10xy)^2$$

Are equivalent.

However, the equation:

$$r=5\sin(2x)$$

is assumed to be plotted on a Cartesian coordinate system.

one is a sine wave the other is a clover

Yes one is a sinusoid, while the other is referred to as a polar flower, I believe. :D