Point(s) of Intersection for a Line and a Logarithmic Spiral

Question

Line: $$y = mx + b$$

Logarithmic spiral: $$x(t)=ae^{bt}\cos t\\y(t)=ae^{bt}\sin t$$

What is the equation for the intersection of these two curves?

Does the complex plane form help? And is it required to solve this problem?

$$z = ae^{(b + i)t}$$

Answer 1

$$y = mx + b$$ $$ae^{bt}\sin(t)=mae^{bt}\cos(t)+b$$ The equation for the intersections of the two curves is : $$e^{bt} (\sin(t)-m\cos(t)) =\frac{b}{a}$$ On cannot solve it explicitly for $t$. Numerical calculus is required.

For each root $t$ the coordinates of the corresponding point of intersection are $x=ae^{bt}\cos(t)\:,\: y=ae^{bt}\sin(t)$ .

Answer 2

Sketch the graphs of straight line and log spiral.

For standard spiral that makes a constant angle $\beta$ to radial lines at start we have

$$ \,r= a e^ {\cot \beta \cdot\,\theta} \tag 1 $$

The initial (minimum) radius is at

$$ \theta=0, r=a $$

The polar form of straight line where pedal length $p$ at minimum distance from origin from sketch is

$$p= = c \cos \phi = \frac{c}{\sqrt{1+m^2}} $$

$$ x \cos (\phi+\pi/2) + y \sin (\phi+\pi/2) =p; x= r \cos\theta;y= r\sin \theta$$

Put together these result in

$$ r= p \sec (\theta+ (\phi+\pi/2)) = \frac{c\sec (\theta+ (\phi+\pi/2)) }{\sqrt{1+m^2}} \tag 2$$

From 1) and 2) we have an equation to solve solve for $\theta$ in equation of form ( $P,Q$ constants)

$$ e^{a\theta} \cos (\theta +P) = Q $$

which can be solved only numerically as it is transcendental in nature.