Hello there. To solve this question, we'll have to remember some properties about polar curves and lengths of parameterized curves.
Given the following polar curve defined as:
[tex]r=7\sin(2\theta)[/tex]
First, remember that for the polar curve
[tex]r=a\sin(n\theta)[/tex]
The number of petals defined by n depends on its parity:
If n is even, we have 2n petals.
If n is odd, we have n petals.
For example, compare
[tex]r=\sin(3\theta)[/tex]
The graph is as follows:
And the graph of
[tex]r=\sin(4\theta)[/tex]
Its graph is
Okay. So in the case of the polar curve in the question, first note that it has a amplitude of 7 (that is, the distance between the origin and the farthest point from it in a petal is 7)
And it might have 4 petals, hence its graph is:
Notice that the line passing through one petal is the line representing the angle pi/4.
To solve for the arc length, we use the following formula:
[tex]\int_a^b\sqrt{(r(t))^2+\left(\dfrac{\mathrm{d}(r(t))}{\mathrm{d}t}\right)^2}\mathrm{d}t[/tex]
Where a and b are the angles in which the petal is between.
In this case, we'll calculate half of it, integrating from the angle it starts up to pi/4. Hence, doubling the result, we get the final answer.
In this case, we find the intersection of it with the origin making:
[tex]\begin{gathered} r=7\sin(2\theta)=0 \\ \\ \Rightarrow\theta=0 \end{gathered}[/tex]
Integrating it up to pi/4, we get:
[tex]\begin{gathered} \int_0^{\frac{\pi}{4}}2\cdot\sqrt{(7\sin(2\theta))^2+\left[\left(7\sin(2\theta)\right)'\right]^2}\,\mathrm{d}t \\ \\ \int_0^{\frac{\pi}{4}}2\cdot\sqrt{49\sin^2(2\theta)+196\cos^2(2\theta)}\,\mathrm{d}t \end{gathered}[/tex]
Notice we can rewrite the radical as:
[tex]\begin{gathered} 49\sin^2(2\theta)+196\cos^2(\theta)=49\sin^2(2\theta)+49\cos^2(2\theta)+147\cos^2(2\theta) \\ \\ \Rightarrow49+147\cos^2(2\theta) \end{gathered}[/tex]
Such that we get:
[tex]\int_0^{\frac{\pi}{4}}2\cdot\sqrt{49+147\cos^2(2\theta)^}\,\mathrm{d}t[/tex]
Whereas
[tex]\begin{gathered} \cos(2\theta)=2\cos^2(\theta)-1\text{ hence} \\ \\ \cos^2(2\theta)=(2\cos^2(\theta)-1)^2=4\cos^4(\theta)-4\cos^2(\theta)+1 \end{gathered}[/tex]
Therefore we get
[tex]\int_0^{\frac{\pi}{4}}2\cdot\sqrt{196+588\cos^4(\theta)-588\cos^2(\theta)}\,\mathrm{d}t[/tex]
Since we want the measure of all the petals, we multiply this result by 4, hence we get the following approximation:
[tex]\text{ Length of the petals: }67.82\text{ units of length}[/tex]
