MHB What is the growth rate of a differentiable function?

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The discussion focuses on defining the growth rate of a differentiable function, specifically how it relates to the function's logarithm. The continuous growth rate is expressed as the derivative of the natural logarithm of the function, represented mathematically as w_u(t) = d(ln u(t))/dt. Participants clarify that the growth rate can be calculated as the change in the function value divided by the function value itself over time, leading to the formula growth rate = (Δu/u)/Δt. This formula highlights that the growth rate is the proportional increase per unit of time, distinct from the slope of the function's graph. The conversation concludes with a clear understanding of the relationship between growth rate and the logarithmic representation of the function.
mathmari
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Hey! :o

We have that the differentiable function $u(t)$ is stricly positiv for all $t\in [0,\infty)$ and that $w$ is the continuous growth rate of $u(t)$.
I want to show that $$w_u(t)=\frac{d(\ln u(t))}{dt}$$How is the formula for the growth rate? Is it maybe given by $w(t)=\frac{u(t)-u(0)}{u(0)}$ ? (Wondering)
 
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mathmari said:
Hey! :o

We have that the differentiable function $u(t)$ is stricly positiv for all $t\in [0,\infty)$ and that $w$ is the constant growth rate of $u(t)$.
I want to show that $$w_u(t)=\frac{d(\ln u(t))}{dt}$$How is the formula for the growth rate? Is it maybe given by $w(t)=\frac{u(t)-u(0)}{u(0)}$ ? (Wondering)

Hey mathmari! (Smile)

That depends on the definition of growth rate, which depends on the area of expertise that it applies to.
For instance the population growth rate is the percentage growth of a population per unit of time, while in business the compound annual growth rate is the geometric progression ratio that provides a constant rate of return over the time period.

Which "growth rate" are we talking about? (Wondering)

EDIT: Actually, they are probably the same thing. I think it's the growth fraction per unit of time.
$$\text{Growth rate} = \frac {\frac{\Delta u}{u}}{\Delta t} = \frac {\frac{\Delta u}{\Delta t}}{u} = \frac{u'(t)}{u(t)}$$
(Thinking)
 
I like Serena said:
Actually, they are probably the same thing. I think it's the growth fraction per unit of time.
$$\text{Growth rate} = \frac {\frac{\Delta u}{u}}{\Delta t} = \frac {\frac{\Delta u}{\Delta t}}{u} = \frac{u'(t)}{u(t)}$$
(Thinking)

How do we get the formula $\text{Growth rate} = \frac {\frac{\Delta u}{u}}{\Delta t} $ and especially the numerator? I got stuck right now... (Wondering)
P.S. I changed at my first post the "constant growth rate" to "continuous growth rate".
 
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Suppose we start with a population $u$ of 100 and 2 days later the population is 110.

Then the population increase $\Delta u$ is:
$$\Delta u=10$$
in those 2 days.

The growth fraction or growth proportion is the increase in population divided by the population size:
$$\text{Growth fraction} = \frac{\Delta u}{u} = \frac{10}{100} = 0.1 = 10\%$$

And the growth rate is the growth fraction per unit of time:
$$\text{Growth rate} = \frac{\text{Growth fraction}}{\Delta t} = \frac{10\%}{2 \text{ days}} = 5\frac{\%}{\text{day}} = 0.05 \text{ day}^{-1}$$

In other words, the formula for growth rate is:
$$\text{Growth rate} = \frac{\frac{\Delta u}{u}}{\Delta t}$$

This growth rate can change continuously from one time to another.
To get an accurate and stable number, we take the limit $\Delta t \to 0$ to get:
$$\text{Growth rate} = \lim_{\Delta t \to 0} \frac{\frac{\Delta u}{u}}{\Delta t} = \frac 1u \lim_{\Delta t \to 0} \frac{\Delta u}{\Delta t} = \frac{u'}{u}$$
(Thinking)
 
Ah ok! (Nerd)

Is the continuous growth rate of a function the slope of its graph? (Wondering)
 
mathmari said:
Is the continuous growth rate of a function the slope of its graph? (Wondering)

Nope.
The slope of the graph is the increase per unit of time.
The growth rate is the proportional increase per unit of time - it's the slope divided by the function value. (Nerd)
 
Alternatively, it's the slope of the graph of $\ln u$, since we have:
$$\text{Growth rate} = \d{(\ln u)}{t} = \frac{u'}{u}$$
(Thinking)
 
I like Serena said:
Alternatively, it's the slope of the graph of $\ln u$, since we have:
$$\text{Growth rate} = \d{(\ln u)}{t} = \frac{u'}{u}$$
(Thinking)

Ah ok... I see! (Smile)

Thank you very much! (Mmm)
 

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