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

Spoiler

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

edit: making it clear that $$n$$ is a function of $$w$$

 

[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:

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

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

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

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

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

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

$$\sqrt{W}-1=kt$$

$$W(t)=(kt+1)^2$$

Now, using the second point to determine $k$

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

$$\frac{k}{2}+1=2$$

$$k=2$$

Hence:

$$W(t)=(2t+1)^2$$