Calculating length of a parametric curve

Question

Find the length of the parametric curve $(x,y)=(2^\theta \cos(\theta), 2^\theta \sin(\theta) )$ and $0\le \theta \le \pi$.

Well, I used the formula $$ \int_a^b \sqrt{[x'(t)]^2 +[y'(t)]^2} dt $$ so if I replace, I have to compute $$ \int_0^\pi [\ln(2) 2^\theta \cos(\theta)-2^\theta \sin(\theta)]^2 + [\ln(2) 2^\theta \sin(\theta)+2^\theta \cos(\theta)]^2 d\theta $$ which is a really ugly function to integrate. Am I doing something wrong? If not, how can I integrate that thing?

Answer 1

"How can I integrate that thing?" Well, first, it's worse than you thought because you left off the square root. But second, if you square out those two terms, you get a lot of cancellation, and it turns out way better than you thought. Use $\cos^2 \theta + \sin^2 \theta = 1.$ Then what's left?

Answer 2

HINT

You are looking at simplifying $$ (a+b)^2 - (a-b)^2 = 2a^2 + 2b^2, $$

and after you are done with that, use the identity $$ \sin^2 \theta + \cos^2 \theta =1... $$

Answer 3

It might be simpler to look at the curve as $\gamma(t) = 2^t e^{it} = e^{\alpha t}$, where $\alpha = \log 2 + i$.

Then $\gamma'(t) = \alpha e^{\alpha t}$, and $|\gamma'(t)| = |\alpha| e^{(\operatorname{re} \alpha) t}$. Then $\int_0^\pi |\gamma'(t)| dt = |\alpha| {1 \over \operatorname{re} \alpha} (e^{ (\operatorname{re} \alpha)\pi}-1)$.

Assuming I have not made a mistake (big assumption), we have $\int_0^\pi |\gamma'(t)| dt = {\sqrt{\log^2 2 +1} \over \log 2}(e^{\pi \log 2}-1) = {\sqrt{\log^2 2 +1} \over \log 2}(2^\pi-1)$.

Answer 4

Your curve is given by $t\mapsto (2^t \cos t,2^t \sin t)$ for $t\in [0,\pi]$. That in the polar coordinates is simply $r(\theta)=2^{\theta}$ for $\theta\in [0,\pi]$. Now use the formula $$s=\int_0^{\pi} \sqrt{r^2(\theta)+(r')^2(\theta)}\ d\theta $$ and the inside is simplified more easily as $$\sqrt{\big(2^\theta\big)^2+\big(2^\theta\log2\big)^2}=2^{\theta}\sqrt{1+\log^2 2}$$ (and I think this is much easier to integrate).

Answer 5

The function you have to integrate is $$\sqrt{(\ln(2) 2^\theta \cos(\theta)-2^\theta \sin(\theta))^2 + (\ln(2) 2^\theta \sin(\theta)+2^\theta \cos(\theta))^2}$$ Expanding yields $$\sqrt{ \ln^2(2)2^{2\theta}\cos^2(\theta)-2\ln(2)2^{\theta}\cos(\theta)*2^{\theta}\sin(\theta)+2^{2\theta}\sin^2(\theta) + \ln^2(2)2^{2\theta}\sin^2(\theta)+2\ln(2)2^{\theta}\sin(\theta)*2^{\theta}\cos(\theta)+2^{2\theta}\cos^2(\theta)}$$

Now rearrange this and cancel to get $$\sqrt{ \ln^2(2)2^{2\theta}(\cos^2(\theta)+\sin^2(\theta))+2^{2\theta}(\sin^2(\theta) + \cos^2(\theta))}$$

Since $\cos^2(\theta) + \sin^2(\theta) = 1$

$$\sqrt{\ln^2(2)2^{2\theta} + 2^{2\theta}} = 2^{\theta}\sqrt{\ln^{2}(2)+1}$$

The integral of this is $$F(\theta) = \frac{2^{\theta}}{\ln(2)}\sqrt{\ln^{2}(2)+1} + C$$

$$F(\pi) - F(0) = \frac{\sqrt{\ln^2(2)+1}}{\ln(2)}(2^\pi-1)$$ is then the answer.