Convert r = 5sin(2θ) to rectangular coordinates

[karush]

convert \(\displaystyle r=5\sin{2\theta}\) to rectangular coordinates

the ans to this is $\left(x^2+y^2\right)^{3/2}=10xy$

however... multiply both sides by $r$ to get $r^2=5\cdot r \cdot \sin{2\theta}$

then substitute $r^2$ with $x^2+y^2$
and $\sin{2\theta}$ with $2\sin\theta\cos\theta$
and divide each side by $r$

$$\frac{x^2+y^2}{\sqrt{x^2+y^2}}=10xy$$

how is $\left(x^2+y^2\right)^{3/2}$ derived?

 

[Evgeny.Makarov]

Re: convert r=5sin2\theta to rectangular coordinates

$\dfrac{a^2}{\sqrt{a}}=a^{3/2}$ for every $a>0$.

 

[I like Serena]

Re: convert r=5sin2\theta to rectangular coordinates

convert \(\displaystyle r=5\sin{2\theta}\) to rectangular coordinates

the ans to this is $\left(x^2+y^2\right)^{3/2}=10xy$

however... multiply both sides by $r$ to get $r^2=5\cdot r \cdot \sin{2\theta}$

then substitute $r^2$ with $x^2+y^2$
and $\sin{2\theta}$ with $2\sin\theta\cos\theta$

That is:
$$x^2+y^2 = 5\cdot r \cdot 2\sin\theta\cos\theta$$
$$x^2+y^2 = 10 \cdot \sin\theta \cdot r\cos\theta$$

and divide each side by $r$

$$\frac{x^2+y^2}{\sqrt{x^2+y^2}}=10xy$$

how is $\left(x^2+y^2\right)^{3/2}$ derived?

Let's multiply by $r$ instead of divide by it.
$$(x^2+y^2) r = 10 \cdot r\sin\theta \cdot r\cos\theta$$
Now make the substitutions:
$$(x^2+y^2) \sqrt{x^2+y^2} = 10 \cdot y \cdot x$$
$$(x^2+y^2)^{3/2} = 10 xy$$

 

[karush]

Re: convert r=5sin2\theta to rectangular coordinates

I :)should of seen that...

at least my basic steps were ok..

 

[CellCoree]

To derive $\left(x^2+y^2\right)^{3/2}$ from $r=5\sin{2\theta}$, we can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ to substitute for $\sin{2\theta}$.

First, we can square both sides of $r=5\sin{2\theta}$ to get $r^2=25\sin^2{2\theta}$. Then, we can use the identity to rewrite $\sin^2{2\theta}$ as $(1-\cos^2{2\theta})$, giving us $r^2=25(1-\cos^2{2\theta})$.

Next, we can use the double angle formula for cosine, $\cos{2\theta}=2\cos^2\theta-1$, to substitute for $\cos^2{2\theta}$ in our equation. This gives us $r^2=25(1-(2\cos^2\theta-1))$, or $r^2=25(2-2\cos^2\theta)$.

We can then use the Pythagorean identity again to rewrite $\cos^2\theta$ as $(1-\sin^2\theta)$, giving us $r^2=25(2-2(1-\sin^2\theta))$, or $r^2=50\sin^2\theta$.

Finally, we can take the square root of both sides to get $r=\sqrt{50\sin^2\theta}$, or $r=\sqrt{50}\sin\theta$. Substituting back in for $r$ in our original equation, we get $\left(x^2+y^2\right)^{3/2}=10xy$, which is equivalent to $r=5\sin{2\theta}$ in rectangular coordinates.