91影视

The polar rose $r=\sin (2n+1)胃$

The objective of this problem is to show that the combined area enclosed by all the petals of the polar rose $r=\sin (2n+1)胃$is the same for every positive integer $n$.

Plot the polar rose $r=\sin (2n+1)胃$for $n=1$. For $n=1,r=\sin (2n+1)胃$shows $r=\sin 3胃$

Plot of $r=\sin 3胃$

To find the tangent at pole of polar rose $r=\sin 3胃$

Put $r=0$

$\sin 3胃=0$

This implies $3胃=n蟺$

That is $胃=\frac{n蟺}{3}$where $n=0,1,2,鈥�$

Take $n=0$and 1 for one loop. Then tangents at pole are $胃=0$and $胃=\frac{蟺}{3}$.

Area of the region bounded by the one loop of the curve can be expressed as

$A={鈭�}_0^{蟺/3}{鈭�}_0^{\sin 3胃}rdrd胃$

Integrate with respect to $r$first.

$A={鈭�}_0^{蟺/3}{\left[\frac{r^2}{2}\right]}_0^{\sin 3胃}d胃$

Put the limits

Integrate with respect to $胃$.

$A=\frac{1}{4}{\left[胃-\frac{1}{6}\sin 6胃\right]}_{胃}^{e/3}$

Put the limits

$A=\frac{1}{4}\left[\left(\frac{蟺}{3}-\frac{1}{6}\sin 2蟺\right)-0\right]$

$A=\frac{蟺}{12}$

Therefore, area of three petals

$=\frac{蟺}{12}脳3$

$=\frac{蟺}{4}$

Plot the polar rose $r=\sin (2n+1)胃$for $n=2$. For $n=2,r=\sin (2n+1)胃$shows $r=\sin 5胃$

Plot of $r=\sin 5胃$

To find the tangent at pole of polar rose $r=\sin 5胃$

Put $r=0$

$\sin 5胃=0$

This implies $5胃=n蟺$

That is $胃=\frac{n蟺}{5}$where $n=0,1,2,鈥�$

Take $n=0$and 1 for one loop. Then tangents at pole are $胃=0$and $胃=\frac{蟺}{5}$.

Area of the region bounded by the one loop of the curve can be expressed as

$A={鈭�}_0^{n/5}{鈭�}_0^{\sin 胃胃}rdrd胃$

Integrate with respect to $P$first.

$A={鈭�}_0^{*/5}{\left[\frac{r^2}{2}\right]}_0^{\text{sinse}}d胃$

Put the limits

Integrate with respect to $胃$.

$A=\frac{1}{4}{\left[\left(胃-\frac{1}{10}\sin 10胃\right)-0\right]}_0^{x/5}$

Put the limits

$A=\frac{1}{4}\left[\left(\frac{蟺}{5}-\frac{1}{10}\sin \left(2蟺\right)\right)-0\right]$

$A=\frac{蟺}{20}$

Therefore, area of five petals

Thus, the combined area enclosed by all the petals of the polar rose $r=\sin (2n+1)胃$is the same for every positive integer $n$.