Question

How do you find the exact length of the polar curve $r=3\sin \left( \theta \right)$ on the interval $0\le \theta \le \dfrac{\pi }{3}$ ?

Answer 1

To get the length of the given equation $r=3\sin \left( \theta \right)$ that is a polar curve on the interval $0\le \theta \le \dfrac{\pi }{3}$ , we will use the formula $\int\limits_{a}^{b}{\left[ \sqrt{{{r}^{2}}+{{\left( \dfrac{dr}{d\theta } \right)}^{2}}} \right]}d\theta$ , where the interval of the given equation is $a\le \theta \le b$ . In the formula we will replace the values with the equation and then will simplify the equation. After that we need to find the integration of the simplified value of the formula on the given interval $0\le \theta \le \dfrac{\pi }{3}$ .

Complete step by step solution:
Here, we have the equation of the polar cure that length we need to find out on the interval $0\le \theta \le \dfrac{\pi }{3}$ as:
$\Rightarrow r=3\sin \left( \theta \right)$ … $\left( 1 \right)$
Since, we know the formula for getting the length of the curve that is $\int\limits_{a}^{b}{\left[ \sqrt{{{r}^{2}}+{{\left( \dfrac{dr}{d\theta } \right)}^{2}}} \right]}d\theta$ , we need two values: first is the value of $r$ that we already have from the question and second: the differentiated value of $r$ that is $\dfrac{dr}{d\theta }$ . So, we will differentiate the above equation with respect to $\theta$ as:
$\Rightarrow \dfrac{dr}{d\theta }=3\dfrac{d\sin \left( \theta \right)}{d\theta }$
Since, we cannot differentiate a constant. So, we will take the constant as a multiple of the variable as we did in the above equation. Now, we will put the value of the differentiation of $\sin \left( \theta \right)$ that is $\cos \left( \theta \right)$ as:
$\Rightarrow \dfrac{dr}{d\theta }=3\cos \left( \theta \right)$
Now, we will use the formula. So, the formula for finding the length of the curve is:
$\Rightarrow \int\limits_{a}^{b}{\left[ \sqrt{{{r}^{2}}+{{\left( \dfrac{dr}{d\theta } \right)}^{2}}} \right]}d\theta$
Now, we will put the value of $r$ and $\dfrac{dr}{d\theta }$ in the above formula with the interval that is $0\le \theta \le \dfrac{\pi }{3}$ as:
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\left[ \sqrt{{{\left[ 3\sin \left( \theta \right) \right]}^{2}}+{{\left( 3\cos \left( \theta \right) \right)}^{2}}} \right]}d\theta$
Here, we will do the square of $3\sin \left( \theta \right)$ and $3\cos \left( \theta \right)$ as:
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{\left[ \sqrt{9{{\sin }^{2}}\left( \theta \right)+9{{\cos }^{2}}\left( \theta \right)} \right]}d\theta$
Now, we can take $9$ out side of the square root because it is a common factor in both the terms of the square root. So, we will get $3$ outside of the square root because square root of $9$ is $3$ as:
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{3\left[ \sqrt{{{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)} \right]}d\theta$
By the formula of trigonometric identities, we know that ${{\sin }^{2}}\left( \theta \right)+{{\cos }^{2}}\left( \theta \right)=1$ . So, we will use this formula in the above equation as:
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{3}}{3\times 1}d\theta$
$\Rightarrow 3\int\limits_{0}^{\dfrac{\pi }{3}}{1}d\theta$
Now, we take the integration with limits as:
$\Rightarrow 3\left[ \theta \right]_{0}^{\dfrac{\pi }{3}}$
Here, we will put the apply the interval in the bracketed term as:
$\Rightarrow 3\left[ \dfrac{\pi }{3}-0 \right]$
After solving the bracket term, we will have:
$\Rightarrow 3\times \dfrac{\pi }{3}$
Now, we will cancel out the equal like terms in the above step as:
$\Rightarrow \pi$
Hence, the length of the polar curve $r=3\sin \left( \theta \right)$ on the interval $0\le \theta \le \dfrac{\pi }{3}$ is $\pi$.

Note: Arc length is the covered distance along with the curve line that makes the arc from one point to other point continuously without any break that is also a segment of the polar and this arc length is always longer than the distance measured by a straight line between those same points. We can find the arc length for the shape that does not contain any vertex and sides.