The polar equation r=7sin(2θ) graphs as a rose.
What is the length of the petals of this rose?
To find the length of the petals of the rose graphed by the polar equation r = 7sin(2θ), we can consider the following steps:
1. Recall that the equation for the length of a curve in polar coordinates is given by the integral of the square root of the sum of the squares of \(r\) and \(\frac{dr}{dθ}\)^2 with respect to \(θ\) over the interval where the curve is traced. In this case, we need to compute the integral:
\(L = \int \sqrt{r^2 + (\frac{dr}{dθ})^2} dθ\)
2. Derive the equation with respect to \(θ\). In this case, \(r = 7sin(2θ)\), so we need to find \(\frac{dr}{dθ}\):
\(\frac{dr}{dθ} = 14cos(2θ)\)
3. Substitute \(r\) and \(\frac{dr}{dθ}\) into the formula for \(L\):
\(L = \int \sqrt{(7sin(2θ))^2 + (14cos(2θ))^2} dθ\)
4. Simplify the expression:
\(L = \int \sqrt{49sin^2(2θ) + 196cos^2(2θ)} dθ\)
5. Use trigonometric identities to simplify further:
\(L = \int \sqrt{245sin^2(θ)cos^2(θ)} dθ\)
Since \(\sin^2(θ) = \frac{1}{2}(1 - \cos(2θ))\) and \(\cos^2(θ) = \frac{1}{2}(1 + \cos(2θ))\), we can substitute them:
\(L = \int \sqrt{245 \cdot \frac{1}{2}(1 - \cos(2θ)) \cdot \frac{1}{2}(1 + \cos(2θ))} dθ\)
\(L = \int \sqrt{\frac{245}{4}(1 - \cos^2(2θ))} dθ\)
\(L = \int \sqrt{\frac{245}{4}\sin^2(2θ)} dθ\)
Simplifying further, we have:
\(L = \frac{7\sqrt{5}}{2} \int |\sin(2θ)| dθ\)
6. Solve the integral using appropriate limits of integration:
Since \(|\sin(2θ)|\) has periods of size \(\frac{\pi}{2}\), the curve takes on each value twice in each period. Thus, to find the length of one full petal, we can integrate over the interval \([0, \pi]\):
\(L = \frac{7\sqrt{5}}{2} \int_0^{\pi} |\sin(2θ)| dθ\)
For the interval \([0, \pi]\), \(|\sin(2θ)| = \sin(2θ)\), so we can simplify further:
\(L = \frac{7\sqrt{5}}{2} \int_0^{\pi} \sin(2θ) dθ\)
Using the integral of \(\sin(2θ)\), we find:
\(L = \frac{7\sqrt{5}}{2} [-\frac{1}{2}\cos(2θ)]_0^{\pi}\)
\(L = \frac{7\sqrt{5}}{2} [-\frac{1}{2}\cos(2\pi) + \frac{1}{2}\cos(0)]\)
Since \(\cos(0) = 1\) and \(\cos(2\pi) = 1\), we have:
\(L = \frac{7\sqrt{5}}{2} [-\frac{1}{2} + \frac{1}{2}]\)
\(L = \frac{7\sqrt{5}}{2} \cdot 0\)
\(L = 0\)
Therefore, the length of the petals of the rose graphed by the polar equation r = 7sin(2θ) is zero. This means that the petals are actually points where they intersect at the origin.
The max of sin(anything) is 1
the max of 7sin(anything) is 7