Exponential decay to a line with slope

Question

I have a question about exponential decay. Suppose I have an exponential decay at point $(x,y)$. Instead of decaying to $x-$axis, I want to decay it to a line with slope $(s')$. I also want the decay line to match slope $(s)$ and curvature $(c)$ at point $(x,y)$. Can anyone please show me how to solve this problem? Thanks

Answer 1

Here's my take on this.

Exponential decay is $y = ae^{-bx}$.

A line is $y = ux+v$.

So, exponential decay to the line is $y =ae^{bx}+ux+v$.

From what you say, $u = s'$.

Since no intercept is specified, I will assume that $v = 0$.

Therefore the equation is $y =ae^{-bx}+s'x $.

The slope is $y' =-abe^{-bx}+s' $.

If the slope is $s$ at $x$, then $s =-abe^{-bx}+s' $. This is an equation involving $a$ and $b$.

The curvature is $\kappa =\dfrac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}} =\dfrac{y''}{(1+y'^2)^{3/2}} $.

For this curve, $y'' =ab^2e^{-bx} $, so $\kappa =\dfrac{ab^2e^{-bx}}{1+(-abe^{-bx}+s')^2)^{3/2}} =c $.

This is a second equation for $a$ and $b$ involving $x, y, s', $ and $c$.

Solve these for $a$ and $b$.