Question

find the length of one petal of the rose r = 3cos4theta at theta = pi/4

Answer 1

\[ r = 3\cos(4\theta) , \theta= \pi/4\]

Answer 2

That means r is 3 cos pi
cos pi is -1
So r = -3?!

Answer 3

Fundamentally, you have to go a long way to get there.

Given \(r = 2\cos(4\theta)\)

Standard definition: \(x = r\cos(\theta)\) results in \(\dfrac{dx}{d\theta} = r(-\sin(\theta)) + \dfrac{dr}{d\theta}\cos(\theta)\)

Similarly: \(y = r\sin(\theta)\) results in \(\dfrac{dy}{d\theta} = r(\cos(\theta)) + \dfrac{dr}{d\theta}\sin(\theta)\)

Wasn't that fun!? We're just getting started.

Thus, \(\left(\dfrac{dx}{d\theta}\right)^{2} + \left(\dfrac{dy}{d\theta}\right)^{2} = r^{2}+ \left(\dfrac{dr}{d\theta}\right)^{2}\)

That was just the development, because I was bored. You just have to know the last expression. That's where the integration starts. The real trick is finding the limits.

This thing has 8 petals. That makes two periods inside \(\pi/2\). With a little thought, we have \(\left[\pi/8,3\pi/8\right]\).

This concludes the fun part: \(\int\limits_{\pi/8}^{3\pi/8}\sqrt{r^{2}+ \left(\dfrac{dr}{d\theta}\right)^{2}}\;d\theta\)

Since we have \(\dfrac{dr}{d\theta} = -12\sin(4\theta)\), the final integral is:

\(\int\limits_{\pi/8}^{3\pi/8}\sqrt{(3\cos(4\theta))^{2}+ \left(-12\sin(4\theta)\right)^{2}}\;d\theta\)

or

\(\int\limits_{\pi/8}^{3\pi/8}\sqrt{135\sin^{2}(4\theta)+9}\;d\theta\)

Do you have a plan for evaluating that?

Answer 4

Estimate, first. With a radius of 3, and not a straight path, one might expect the final result to be a little over 6.