Growth proportional to square root of worth.

[ineedhelpnow]

View attachment 3931

Hi! I'm stuck
I was thinking I need to use the equation $f=kw^2$ but I'm really not sure.

 

[Ackbach]

So that first sentence there: "A firm has a fortune that grows at a yearly rate proportional to the square root of its worth." Can you translate that sentence into a differential equation?

 

[SuperSonic4]

Hi! I'm stuck
I was thinking I need to use the equation $f=kw^2$ but I'm really not sure.

A firm has a fortune that grows at a yearly rate proportional to the square root of its worth

Suppose the firm's fortune is given by \(\displaystyle n(w)\), the firm's worth is \(\displaystyle w\) and the time taken is \(\displaystyle t\). Can you set up an differential equation to show the firm's fortune in terms of worth at a given time

Spoiler

\(\displaystyle \dfrac{dn}{dt} \propto \sqrt{w} \ \rightarrow \dfrac{dn}{dt} = k \sqrt{w}\) where \(\displaystyle k\) is some constant

edit: making it clear that \(\displaystyle n\) is a function of \(\displaystyle w\)

 

[ineedhelpnow]

Sorry for the late response. The assignment was due and I got really distracted after but I'd still like to work through the problem to see how it's done.

@Ackbach No, I don't. Well until SuperSonic showed it. :p I still don't understand where to go from there. I'm super lost.

@SuperSonic Thank you :D

 

[MarkFL]

I would let time $t=0$ be at 1/2 year ago, and let $W$ be measured in millions of dollars. Then I would state the IVP:

\(\displaystyle \frac{dW}{dt}=k\sqrt{W}\) where \(\displaystyle W(0)=1,\,W\left(\frac{1}{2}\right)=4\)

Separating variables and using the boundaries as limits, we have:

\(\displaystyle \int_1^{W} u^{-\frac{1}{2}}\,du=k\int_0^t\,dv\)

\(\displaystyle 2\left[u^{\frac{1}{2}}\right]_1^{W}=k[v]_0^t\)

\(\displaystyle 2\left(\sqrt{W}-1\right)=kt\)

Since $k$ is a constant, we can divide through by 2 and still have a constant:

\(\displaystyle \sqrt{W}-1=kt\)

\(\displaystyle W(t)=(kt+1)^2\)

Now, using the second point to determine $k$

\(\displaystyle \left(\frac{k}{2}+1\right)^2=4\)

\(\displaystyle \frac{k}{2}+1=2\)

\(\displaystyle k=2\)

Hence:

\(\displaystyle W(t)=(2t+1)^2\)