Q. 32 - Calculus

Question

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

one petal of the polar rose $r=\cos 3\theta$

Step-by-Step Solution

Answer

The integral can be given as $L=2{\int }_0^{\frac{\pi }{6}}\sqrt{{\left(\cos 3\theta \right)}^2+9{(-\sin 3\theta )}^2}d\theta$ and the arc length is 2.22 units

Step 1: Given information

We are given a petal of the polar rose $r=\cos 3\theta$

Step 2: Evaluate

We know that arc length of one petal of polar rose 

can be given as

$L={\int }_0^{\frac{\pi }{2n}}\sqrt{{\left(\cos n\theta \right)}^2+{\left(n^2\sin n\theta \right)}^2}d\theta$

Substituting n=3 in the above equation we get

$L=2{\int }_0^{\frac{\pi }{6}}\sqrt{{\cos }^{2}3\theta +9{\sin }^{2}3\theta }d\theta L$

Now on using CAS calculator we get

$$\begin{gathered} L=2(1.1) \\ L=2.2unit \end{gathered}$$