# Length of Polar Curve

#### opus

Gold Member
1. The problem statement, all variables and given/known data
Use Desmos to graph the spiral $r=\theta$ on the interval $0\leq\theta\leq4\pi$, and then determine the exact length of the curve and a four decimal approximation.

Hint: $\int \sec^3(x)dx=\frac{1}{2}\sec(x)tan(x)+\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C$

2. Relevant equations
Arc Length of a Polar Curve is given as:
\begin{align} S & = \int_\alpha^{\beta}\sqrt{\left[f(\theta)\right]^2+\left[f'(\theta)\right]^2}d\theta\\ & = \int_\alpha^{\beta}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta \end{align}

3. The attempt at a solution
I have attached the required graph from Desmos.

Right off the bat, I am stuck here.
I start off with $S=\int_0^{4\pi}\sqrt{(\theta)^2 + (\theta')^2}$ and this makes no sense.
I also tried using the identity $r^2=x^2+y^2$ but then it of course gave me a nasty polynomial in two variables as the integrand.

I'm not seeing where this hint is coming into play.
Any ideas?

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#### Orodruin

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What is $d\theta/d\theta$?
Also: Change of integration variables...

• opus

#### opus

Gold Member
Ok I see it now I think, but I'm a little shaky on the nuances here.

From your hint, I got rid of the $S = \int_\alpha^{\beta}\sqrt{\left[f(\theta)\right]^2+\left[f'(\theta)\right]^2}d\theta$ and replaced it with $\int_\alpha^{\beta}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$
From here, $\int_\alpha^{\beta}\sqrt{r^2+\left(\frac{d\theta}{d\theta}\right)^2}d\theta = \int_\alpha^{\beta}\sqrt{r^2+1}d\theta$
Then I let $r=\tan(\theta)$ and make the appropriate trigonometric substitutions and I get what's given in the hint.

Now for the "nuances" I mentioned, I am unsure exactly what to do with the interplay between the $r$ and $\theta$ here. What I did in my work, which I will post as an image, is kind of swap the r's for the theta's, and change the limits of integration from $\alpha$ and $\beta$ to $a$ and $b$. I'm really not sure how to handle this in a mathematically proper way.

If you wouldn't mind having a look at my work in the image to see what I mean (kind of hard to explain in words), could you guide me in the right direction on how the notation/change of variables needs to be handled appropriately?

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#### Orodruin

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No, that is not correct, $r = \theta$, not $\tan\theta$.

• Delta2 and opus

#### opus

Gold Member
Ahhh I see what you mean. It wasn't pretty, but looks like my solution matches the answer key now. Thank you!

#### Chestermiller

Mentor
• opus

#### opus

Gold Member
To show my work in helps of connecting the two threads:
Starting from $\int_0^{4\pi}\sqrt{\theta^2+1}d\theta$,
Using u-substitution,
Let $\theta=\tan(u)$ and $d\theta=\sec^2(u)du$

$\int_0^{\tan^{-1}{4\pi}}\sqrt{\tan^2(u)+1}\sec^2(u)du$
$=\int_0^{\tan^{-1}{4\pi}}\sec^3(u)du$
Which is followed by some messy algebra and trigonometry that evaluates approximately to 80.8.

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