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I have the following equations:

Eq. 1) A standard equation for a line $$y=mx + b,$$

Eq 2) A a parametric equation $$x=e^{at}-1,$$ $$y=e^{ct}-1,$$

where $a$ and $c$ are known constants.

I want to find the intercepts of Eq .1 with Eq. 2.In the context I am working in I know there are usually two intercepts. I tried substituting $x$ and $y$ to find the intercepts $t$ values, which I could use to get $x$ and $y$ (actually I really just want t anyhow). I get

$$e^{ct}-1=m(e^{at}-1)+b,$$

yet I can't see a way to solve this for t without some kind of numerical method. How do I solve this equation?

bart cubrich
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1 Answers1

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Let $z:=e^{at}$ and $\alpha:=\dfrac ca$. The equation is

$$z^\alpha-1=m(z-1)+b.$$

This shows that for most rational $\alpha$, there is no analytical solution.

The slope along the curve is given by

$$\frac{dy}{dx}=\frac cae^{(c-a)t}.$$

If you equate this to $m$, you can draw $t$, which gives the point of tangency of a parallel to the line. Then depending on the side of the line wrt the tangent and wrt the initial point of the curve $(-1,-1)$, you can determine the number of intersections ($0, 1$ or $2$).

In case there are intersections, the tangency point separates them and you can use this information to start numerical iterations.

For convenience, you can translate $x,y$ by $1$ and the equations simplify to

$$x=e^{at},\\y=e^{ct}=x^\alpha$$ and $$y-1=m(x-1)+b\iff y=m'x+b'.$$

  • This is very helpful. I am only confused exactly how to solve for the number of intersections. Knowing that would help with the numerical iterations. – bart cubrich Jun 10 '19 at 18:36
  • @bartcubrich: there are different cases depending on the position and slope of the line. Draw the curve, try different line directions and see how the number of roots varies when translating the line. –  Jun 10 '19 at 18:40