Prove that the tangent to the logarithmic spiral rho=a
Prove that the tangent to the logarithmic spiral $\rho=a\,e^{m\varphi}$ makes a constant angle with the radius vector at every point.
Explanation of Method
The defining property of the logarithmic (equiangular) spiral.
Step 1: The polar tangent–radius formula
For a polar curve $\rho=\rho(\varphi)$, the angle $\psi$ between the tangent and the radius vector satisfies
\[ \tan\psi = \frac{\rho}{\rho'}, \qquad \rho'=\frac{d\rho}{d\varphi}. \tag{1} \]
Step 2: Apply it to the spiral
\[ \rho = ae^{m\varphi} \;\Longrightarrow\; \rho' = am\,e^{m\varphi} = m\rho, \qquad \tan\psi = \frac{\rho}{m\rho} = \frac{1}{m}. \tag{2} \]
Result
Final answer:
\[\boxed{\,\tan\psi = \frac{1}{m} = \text{const}, \qquad \psi = \arctan\frac1m\,}\]
independent of $\varphi$. This constancy is exactly why $\rho=ae^{m\varphi}$ is called the equiangular spiral. $\blacksquare$