Relating Volume to Length in an exponential function

[AmalR]

I am studying exponential functions and I was told to answer the question below:




I started off by making an equation (of the form [MATH]y=ab^x[/MATH]) to find the final volume for a snake at any given level: [MATH]V_f=V_iG^{n-1}[/MATH]
[MATH]V_i[/MATH] is the initial volume, [MATH]V_f[/MATH] is the final volume, [MATH]G[/MATH] is the "growth factor" (if it helps, my "growth factor" was 1.1), [MATH]n[/MATH] is the level that the snake is at ([MATH]n-1[/MATH] in the equation because level 1 is the initial volume). **Length and radius for level 1 are made by the student, here are mine (not sure if it can be done without values as teacher insisted we make some): Length: 5 units, Radius: 1 unit.

I am not sure how to manipulate this equation so that it will be in terms of the length of the cylinder. I feel it will be something to do with the constant [MATH]k[/MATH]. I have already tried to substitute [MATH]\pi r^2*L[/MATH] for [MATH]V_1[/MATH] but I found that I don't know both [MATH]r[/MATH] and [MATH]L[/MATH], making it impossible to solve. I see that the [MATH]k[/MATH] value could help by relating [MATH]\pi r^2[/MATH] to [MATH]L[/MATH], but I do not see how this can be done without knowing either value. Only the constant [MATH]k[/MATH] can be known since the values for radius and length on level 1 are made up.

Thanks in advance!

 

[LCKurtz]

Think of the volume as a function of [MATH]n[/MATH]. I will call your growth factor [MATH]\lambda[/MATH]. I will also try modeling it as a continuous growth as an approximation. So your growth equation is [MATH]V'(n) = \lambda V(n)[/MATH] with initial condition [MATH]V(n_0)=V_0 [/MATH]. The solution to that DE is [MATH]V(n) = V_0e^{\lambda(n - n_0)}[/MATH]. Now [MATH]V(n) = \pi r^2L[/MATH] and [MATH]\frac{L}{\pi r^2} = k[/MATH], so [MATH]\pi r^2 = \frac L k[/MATH], giving [MATH]V = \frac{L^2}{k}[/MATH]. Putting this all together gives[MATH]V(n) = V_0e^{\lambda(n - n_0)}=\frac L k[/MATH]. You can solve this for [MATH]L[/MATH] in terms of [MATH]n[/MATH].

 

[LCKurtz]

I don't seem to be able to edit it, so please note that last [MATH]\frac L k[/MATH] should be [MATH]\frac{L^2} k[/MATH].