Given a general equation for an exponential function: $$y=Aa^{-x}$$ I would like to find its specific form that would conform the conditions below:

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  • it starts from hight $y(0)=H$
  • at distance $w$, its height is $h$ and slope $S=-0.1$

It's been long time I haven't touched math and I find quite difficult to solve this equation :/ I think, as the slope is involved, I must include a differential equation.

I have found a derivative of the function: $$\frac{\mathrm{d}}{\mathrm{d}x}Aa^{-x}=Aa^{-x}\ln(a)$$ Now, the set of differential equations is following: $$\begin{cases} y=Aa^{-x} \\ \dot{y}=Aa^{-x}\ln{a} \end{cases}$$ I have worked out that $A=H$ given $y=0$.

For now onward I don't really know how to proceed. I don't know how to handle unknown $a$ in two forms $a^{-x}$ and $\ln{a}$. Any help is always appreciated :)



1 Answer 1


Express the two remaining constraints:

  • $y(w)=h=Ha^{-w}$,
  • $y'(w)=S=-Ha^{-w}\ln(a)=-h\ln(a)$.

They give you two ways of computing $a$, namely

$$a=\sqrt[w]{\frac Hh}$$and $$a=e^{-S/h}.$$

Unless these two values coincide by chance, what you are asking is not possible (three constraints on two unknowns).


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