Vinyl Record Arc Length Right now the I am using this formula to calculate the arc length:
$$L=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$
The grooves on the vinyl replicate an Archimedean spiral given in the form of $r= \lambda θ$, where $\lambda $ is a constant. I am having trouble determining the exact number of complete turns and what the value for spacing between each turn should be. Currently, the vinyl record I am using is $754~\text{s}$ long and is played at $40~\text{rpm}$ .
Can someone help me formulate an equation? I am not quite sure where to go from here?
 A: Assuming the tonearm starts and ends at the same angle we have the following system of equations:
$$r = k \theta + r_0 \qquad \begin{cases} 7.12 = k\cdot 0 + r_0 \\
15.24 = k \cdot 735 \cdot 2\pi + r_0 \end{cases} \quad \to \; k = \frac{29}{5250 \pi}$$
$$\int_0^{735\cdot 2\pi} \sqrt{(k\theta+r_0)^2+k^2} \text{ d}\theta \approx 51630.8 \text{ cm}$$
A: Basically you just have to integrate the arc length of an Archimedian spiral. Let
$$r(\theta)=\lambda\theta$$
So,
$$\frac{\mathrm dr}{\mathrm d\theta}=\lambda$$
So we just have to determine
$$L(\theta_1,\theta_2)=\lambda\int_{\theta_1}^{\theta_2}\sqrt{1+\theta^2}~\mathrm d\theta$$
So we need to find an antiderivative for the function $x\mapsto \sqrt{1+x^2}$. We can use a nice trick to evaluate this as
$$L(\theta_1,\theta_2)=\frac{\lambda}{2}\left(\theta_2\sqrt{1+{\theta_2}^2}+\operatorname{arcsinh} \theta_2\right)-\frac{\lambda}{2}\left(\theta_1\sqrt{1+{\theta_1}^2}+\operatorname{arcsinh} \theta_1\right)$$
Now, you want to measure the spacing of the groove. I'll leave it as an easy exercise to determine that we can relate $\lambda$ to the spacing $s$ by
$$\lambda=\frac{s}{2\pi}$$
We first determine what our $\theta_1$ is. To make things easier, we'll actually imagine the record spinning the other way, with the tonearm going out instead of in. Assuming the tonearm started right at the center of the record, the number of turns it would need to make before it got to our starting length of $7.12~\text{cm}$ is just $7.12/s$. So,
$$\theta_1=2\pi\cdot \frac{7.12}{s}=\frac{7.12}{\lambda}$$
Similarly
$$\theta_2=\frac{15.24}{\lambda}$$
So now all that's left is to compute is
$$L\left(\frac{7.12}{\lambda},\frac{15.24}{\lambda}\right)$$
So now what you need to do is measure the spacing $s$ and compute $\lambda$.
EDIT:
Rereading the question I remembered that we figured out that it was computed that the record takes $735$ turns. In that case, we can easily solve for $\lambda$ in
$$\frac{15.24-7.12}{\lambda}=2\pi\cdot 735$$
To get
$$\lambda=0.00175828318082$$
Hence
$$L\left(\frac{7.12}{\lambda},\frac{15.24}{\lambda}\right)=51630.8192937~\text{cm}$$
Computed on Desmos.
