Q. 35 - Calculus

Question

In Exercises 31-36 find a definite inteyral that represents the length of the specified polar curve, and then use a graphing calculator or computer algebra system to approximate the value of the integral.

• The inner loop of the limaçon $r=1+2\sin \theta$.

Step-by-Step Solution

Answer

Ans:  ${\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\sqrt{5+4\sin \theta }d\theta =2.682$

Step 1. Given information:

$r=1+2\sin \theta$

Step 2. The formula for finding the arc length:

The arc length of the polar equation $r=1+2\sin \theta$ with the limits $\frac{7\pi }{6}$ to $\frac{11\pi }{6}$.

The formula to find the arc length of the polar curve is given by,

$L={\int }_a^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$

Step 3. Finding the derivatives:

Here $f\left(\theta \right)=r=1+2\sin \theta$

Now differentiate $f\left(\theta \right)=1+2\sin \theta$ with respect to $\theta$.

Then $f\left(\theta \right)=1+2\sin \theta$

$$\begin{gathered} f^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }(1+2\sin \theta ) \\ {f}^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }1+\frac{d}{d\theta }2\sin \theta \end{gathered}$$

Step 4. Substitute values:

Then,

$f^{\text{'}}\left(\theta \right)=2\cos \theta$

Now substitute the values of $f\left(\theta \right),f^{\text{'}}\left(\theta \right)$ in $L={\int }_a^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$.

$L={\int }_a^{\beta }\sqrt{(2\cos \theta {)}^2+(1+2\sin \theta {)}^2}d\theta$

The limits are from $\frac{7\pi }{6}$ to $\frac{11\pi }{6}$, then

$L={\int }_{\frac{7\pi }{0}}^{\frac{11\pi }{6}}\sqrt{(2\cos \theta {)}^2+(1+2\sin \theta {)}^2}d\theta$

Step 5. Simplification:

Then,

$$\begin{gathered} L={\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\sqrt{4{\cos }^2\theta +1+4{\sin }^2\theta +4\sin \theta }d\theta \\ L={\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\sqrt{5+4\sin \theta }d\theta \left[\text{ since }{\sin }^2\theta +{\cos }^2\theta =1\right] \end{gathered}$$

Then the integral ${\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\sqrt{5+4\sin \theta }d\theta =2.682$

Step 6. The required answer:

Length of the curve is equals to $2.682$.

Therefore, the required answer is ${\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\sqrt{5+4\sin \theta }d\theta =2.682$.