Problem 577
Medium Tangents, Normals, and Asymptotes

Find the length of the polar subtangent and the polar

Find the length of the polar subtangent and the polar subnormal of the logarithmic spiral $\rho=a^{\varphi}$, $a>0,\ a\neq1$.

Explanation of Method

The logarithmic spiral's derivative reproduces itself, and the radius vector becomes a geometric mean.

Step 1: The two formulas

\[ S_T = \left|\frac{\rho^2}{\rho'(\varphi)}\right|, \qquad S_N = \left|\rho'(\varphi)\right|. \tag{1} \]

Step 2: Differentiate

\[ \rho' = a^{\varphi}\ln a = \rho\ln a. \tag{2} \]

Step 3: Subtangent

\[ S_T = \left|\frac{\rho^2}{\rho\ln a}\right| = \frac{\rho}{|\ln a|} = \frac{a^{\varphi}}{|\ln a|}. \tag{3} \]

Step 4: Subnormal

\[ S_N = |\rho\ln a| = \rho\,|\ln a| = a^{\varphi}\,|\ln a|. \tag{4} \]

Step 5: A companion identity

Multiplying (3) and (4): \[S_T\cdot S_N = \frac{\rho}{|\ln a|}\cdot\rho|\ln a| = \rho^2, \tag{5}\] so the radius vector is the geometric mean of the polar subtangent and the polar subnormal, at every point of the spiral.

Step 6: Verification

\[a=1.5,\ \varphi=2.0\text{: } \rho\approx2.25000,\ S_T\approx5.54898,\ S_N\approx0.91259,\ S_T S_N\approx5.06250=\rho^2\ \checkmark\]


Result

Final answer:

\[\boxed{\,S_T = \frac{\rho}{|\ln a|}, \qquad S_N = \rho\,|\ln a|, \qquad S_T\cdot S_N = \rho^2\,}\]