Intuitive understanding of tangents and graphs

Question

Consider this locus

Now a property of this is that the angle between the tangent at any point B and the line joining origin to B is constant . My question is why does this happening? I was able to prove this but don't understand why this happens.

Answer 1

Let's start with a seemingly-unrelated question. Introduce Cartesian coordinates $(u, v)$ in the Euclidean plane, and consider the set of all horizontal lines.

Question: Which curves meet all horizontal lines at the same angle?

Every non-horizontal line meets each horizontal line at the same angle. The angle is determined by the slope: The line $u = kv + u_{0}$ meets each line $v = v_{0}$ at angle $\phi$ satisfying $\cot\phi = k$.

Now, the magic: Consider the mapping $(u, v) \mapsto (x, y)$ defined by \begin{align*} x &= e^{u} \cos v, \\ y &= e^{u} \sin v. \end{align*} In complex terms, if we write $w = u + iv$ and $z = x + iy$, this is the exponential mapping $z = \exp(w)$: Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ allows us to write \begin{align*} z = x + iy &= e^{u} \cos v + ie^{u} \sin v \\ &= e^{u}(\cos v + i\sin v) \\ &= e^{u} \cdot e^{iv} = e^{u + iv} = \exp(w). \end{align*} This mapping $\exp$ satisfies three properties that answer your question:

1. $\exp$ maps the plane to the plane with the origin removed in an angle-preserving way. (This is a feature of all complex-differentiable mappings with no critical points.)
2. $\exp$ maps horizontal lines $v = v_{0}$ to rays $\theta = v_{0}$ from the origin.
3. $\exp$ maps diagonal lines $u = kv + u_{0}$ to logarithmic spirals.

Since every line meets all horizontal lines at a constant angle, every logarithmic spiral meets all rays from the origin at a constant angle.

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To see $\exp$ preserves angles, we can calculate the derivative matrix $$ d\exp(u, v) = \left[\begin{array}{@{}rr@{}} e^{u}\cos v & -e^{u}\sin v \\ e^{u}\sin v & e^{u}\cos v \\ \end{array}\right] = e^{u} \left[\begin{array}{@{}rr@{}} \cos v & -\sin v \\ \sin v & \cos v \\ \end{array}\right]. $$ At each point this is the composition of rotation and scaling, so at each point the derivative preserves angles.

To see that lines map to logarithmic spirals, we can parametrize the line $u = kv + u_{0}$ by taking $v = t$ and $u = kt + u_{0}$, i.e., $(kt + u_{0}, t)$. Applying $\exp$ to this gives a parametric description of the image curves: $$ \gamma(t) = \exp(kt + u_{0}, t) = e^{kt + u_{0}}(\cos t, \sin t), $$ or \begin{align*} x &= e^{kt + u_{0}} \cos t, \\ y &= e^{kt + u_{0}} \sin t. \end{align*} This curve traces the polar graph $r = e^{k\theta + u_{0}}$, a logarithmic spiral. To check formally that these are your curves, note that

• $y/x = \sin\theta/\cos\theta = \tan\theta$, and
• $\ln r = k\theta + u_{0}$, or $\theta = \frac{1}{k} (\ln r - u_{0}) = \frac{1}{2k} \bigl(\ln(x^{2} + y^{2}) - 2u_{0}\bigr)$. (The formula in your Desmos window has $u_{0} = 0$ and $k = 1/2$, but the plot shows multiple spirals, which correspond to different values of $u_{0}$. By contrast, changing $k$ changes the "slope" of the spirals, as in the animation loop above.)