0
$\begingroup$

I need to prove that for a full turn of a logarithmic spiral (independent from the constant tangent of it) and a straight line on the z axis, the logarithmic spiral always covers more space-time in the circle than the straight line.

spiral

line

I need to prove that;

$$ \frac{ \int_{0}^{2\pi} curve(\Theta ) \bigcap ((x-\frac{r_{max}+r_{min}}{2})^{2}+y^{2}\leq (\frac{r_{max}-r_{min}}{2})^{2}) d\theta}{length of the spiral} > \frac{\int_{0}^{2\pi} line(\Theta ) \bigcap ((x-\frac{r_{max}-r_{min}}{2})^{2}+y^{2}\leq (\frac{r_{max}-r_{min}}{2})^{2}) d\theta}{length of the line}$$

I don't know where to start here. When the spiral rotates around the z axis in clockwise you will see that it will intersect with the circle as on the right. For example, the first intersection is demonstrated when θk+1 and the second is θk+2 vs. as shown in the right figure. If we assume θk = k x ∆A I am asking for the sum of those pieces of the spiral that is in the circle when ∆A→0. That means -I guess- some king of integral but the limits of are changing with the angle. I hope I made myself clear.

How can I get this on wolfram?

$\endgroup$
  • 1
    $\begingroup$ Are asking about Wolfram|Alpha or about Wolfram Mathematica? $\endgroup$ – m_goldberg Feb 11 at 10:58
  • $\begingroup$ @m_goldberg Asking for mathematica $\endgroup$ – Alper91 Feb 11 at 13:17

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.