<p><span class="math-container">$k(t)=\dfrac{K(t)}{L(t)}$</span></p>
<p>Taking log both sides, we get</p>
<p><span class="math-container">$\ln k(t)=\ln K(t)- \ln L(t)$</span></p>
<p>Differentiation with respect to <span class="math-container">$t$</span>, we get</p>
<p><span class="math-container">$\dfrac{1}{k}\dfrac{dk}{dt}=\dfrac{1}{K}\dfrac{dK}{dt}-\dfrac{1}{L}\dfrac{dL}{dt}$</span></p>
<p>which can also be written as
<span class="math-container">$\dfrac{\dot{k}}{k}=\dfrac{\dot{K}}{K}-\dfrac{\dot{L}}{L}$</span></p>

CC BY-SA 4.0
2023-01-09T18:01:36.810Z
2023-01-09T18:01:36.810Z
Amit
<p>This only answers your first question, but the quotient rule <em>does</em> apply here. The right hand side of your relation is more naturally derived using the log formulation (as in other answers), but consider the following:</p>
<p><span class="math-container">$k = K/L$</span></p>
<p><span class="math-container">$\dot{k} = \frac{L \dot{K} - K\dot{L}}{L^2}$</span></p>
<p><span class="math-container">$\frac{\dot{k}}{k} = \frac{\dot{k}L}{K} = \frac{L}{K} \left( \frac{L \dot{K} - K\dot{L}}{L^2} \right) = \frac{L^2\dot{K}}{K L^2} - \frac{LK\dot{L}}{KL^2} = \frac{\dot{K}}{K} - \frac{\dot{L}}{L}$</span></p>

CC BY-SA 4.0
2023-01-10T21:04:44.337Z
What is the more natural way of doing that calculation is a subjective matter. I think that it could be a test: If one prefers the quotient rule is a mathematician, if one prefers the logaritm is an economist :). It is a joke, but has a kernel of truth: economist are more used to use logarithms, mathematicians think at first at derivation rules.
BakerStreet
CC BY-SA 4.0
2023-01-10T20:24:56.960Z
2023-01-10T20:24:56.960Z
user43240
<p>There is another way to formulate the issue, equivalent to the formulation of the preceding answers, to explain why <span class="math-container">$\frac {\dot L}{L}=n$</span>.</p>
<p>In the Solow growth model, as in your case, the usual assumption is that the growth rate of the population is a constant, <span class="math-container">$n$</span>.</p>
<p>The growth rate (the equivalent of the percentage growth rate in discrete time)  in continuos time is, by definition, <span class="math-container">$\frac {\dot L}{L}$</span>.</p>
<p>A system with a constant growth rate is ruled by an equation<span class="math-container">$^1$</span> as:</p>
<p><span class="math-container">$$ L(t)= L_0 e^{nt}\;\;\;(1),$$</span></p>
<p>where <span class="math-container">$L_0$</span> is the population at time <span class="math-container">$0$</span>.</p>
<p>Indeed, taking the derivative of <span class="math-container">$(1)$</span> with respect to time  we have:
<span class="math-container">$$\dot L(t)= nL_0 e^{nt}.$$</span></p>
<p>Dividing both sides by <span class="math-container">$L(t)$</span>, we have (neglecting <span class="math-container">$t$</span>):</p>
<p><span class="math-container">$$\frac {\dot L}{L}= \frac{nL_0 e^{nt}}{L_0 e^{nt}}= n.$$</span></p>
<hr />
<p><span class="math-container">$^1$</span> <sup> This is the formalization of the assumption of constant growth rate of population in the Solow model in continuous time </sup></p>

CC BY-SA 4.0
2023-01-09T18:33:51.603Z
2023-01-09T20:14:29.100Z
2023-01-09T20:14:29.100Z
BakerStreet
<p>No, you shouldn't use quotient rule here. These are the steps:</p>
<ol>
<li>log linearize <span class="math-container">$$k=K/L \implies \ln k = \ln K - \ln L$$</span></li>
<li>Take time derivative vrt <span class="math-container">$t$</span> (treat any variable <span class="math-container">$x$</span> as <span class="math-container">$x(t)$</span>) use chain rule.</li>
</ol>
<p><span class="math-container">$$ \frac{1}{k}k'(t)= \frac{1}{K} K'(t) -\frac{1}{L} L'(t)$$</span></p>
<ol>
<li>Simplify and adjust notation (<span class="math-container">$x'(t) = \dot{x}$</span>):</li>
</ol>
<p><span class="math-container">$$ \frac{\dot{k}}{k} = \frac{\dot{K}}{K}  -\frac{\dot{L}}{L} $$</span></p>
<p>As you see there is no reason for using quotient rule here.</p>
<p>Regarding to second question. A continuous time equivalent of <span class="math-container">$(L_1 - L_0)/L_0$</span> is <span class="math-container">$\dot{L}/L$</span> since <span class="math-container">$\dot{L}= L_1 - L_0$</span> as <span class="math-container">$\Delta t \to 0$</span>. So if <span class="math-container">$L_1=(1+n)L_0$</span> you would expect that <span class="math-container">$\dot{L}/L=n$</span> although note <span class="math-container">$n$</span> in discrete and continuous time are not 100% the same since in continuous time it is instantaneous growth and in discrete time its over period of time, but if you think of it in terms of methodological continuous time equivalent then they are equivalent.</p>

CC BY-SA 4.0
2023-01-09T17:59:31.827Z
2023-01-09T18:05:07.730Z
2023-01-09T18:05:07.730Z
1muflon1
<p>Given that the per-capita capital <span class="math-container">$ k = \frac{K}{L}\ $</span> (total capita divided by labor force), and we want to find <span class="math-container">$ \dot{k}\ $</span>, it seems that
<span class="math-container">$ \frac{\dot{k}}{k}\ =   \frac{\dot{K}}{K}\  -  \frac{\dot{L}}{L}\ $</span></p>
<p>Why is this? Shouldnt't the derivative with respect to time <span class="math-container">$ \dot{k}\ $</span>yield something like a quotient rule?</p>
<p>Furthermore, why does <span class="math-container">$ \frac{\dot{L}}{L}\ = n $</span> ? Shouldn't the derivative with respect to time of the labor force be equal to <span class="math-container">$ n $</span> (the growth rate of L, in discrete time, <span class="math-container">$ L_1 = (1+n)L_0 $</span> ) only, so that <span class="math-container">$ \dot{L}\ = n $</span> ?</p>

<mathematical-economics><solow>
Solow model continuous time - definitions
user43188
# Solow model continuous time - definitions

(Asked by: username_0 on 2023-01-09)

Given that the per-capita capital $ k = \frac{K}{L}\ $ (total capita divided by labor force), and we want to find $ \dot{k}\ $, it seems that
$ \frac{\dot{k}}{k}\ = \frac{\dot{K}}{K}\ - \frac{\dot{L}}{L}\ $

Why is this? Shouldnt't the derivative with respect to time $ \dot{k}\ $yield something like a quotient rule?

Furthermore, why does $ \frac{\dot{L}}{L}\ = n $ ? Shouldn't the derivative with respect to time of the labor force be equal to $ n $ (the growth rate of L, in discrete time, $ L\_1 = (1+n)L\_0 $ ) only, so that $ \dot{L}\ = n $ ?

**Tags:** mathematical-economics, solow

---

## 4 answers

### Answer by username_1 on 2023-01-09. Score: 3

$k(t)=\dfrac{K(t)}{L(t)}$

Taking log both sides, we get

$\ln k(t)=\ln K(t)- \ln L(t)$

Differentiation with respect to $t$, we get

$\dfrac{1}{k}\dfrac{dk}{dt}=\dfrac{1}{K}\dfrac{dK}{dt}-\dfrac{1}{L}\dfrac{dL}{dt}$

which can also be written as
$\dfrac{\dot{k}}{k}=\dfrac{\dot{K}}{K}-\dfrac{\dot{L}}{L}$


---

### Answer by username_2 on 2023-01-10. Score: 3

This only answers your first question, but the quotient rule *does* apply here. The right hand side of your relation is more naturally derived using the log formulation (as in other answers), but consider the following:

$k = K/L$

$\dot{k} = \frac{L \dot{K} - K\dot{L}}{L^2}$

$\frac{\dot{k}}{k} = \frac{\dot{k}L}{K} = \frac{L}{K} \left( \frac{L \dot{K} - K\dot{L}}{L^2} \right) = \frac{L^2\dot{K}}{K L^2} - \frac{LK\dot{L}}{KL^2} = \frac{\dot{K}}{K} - \frac{\dot{L}}{L}$

> **Comments:**
> **username_3**: What is the more natural way of doing that calculation is a subjective matter. I think that it could be a test: If one prefers the quotient rule is a mathematician, if one prefers the logaritm is an economist :). It is a joke, but has a kernel of truth: economist are more used to use logarithms, mathematicians think at first at derivation rules.  

---

### Answer by username_3 on 2023-01-09. Score: 2

There is another way to formulate the issue, equivalent to the formulation of the preceding answers, to explain why $\frac {\dot L}{L}=n$.

In the Solow growth model, as in your case, the usual assumption is that the growth rate of the population is a constant, $n$.

The growth rate (the equivalent of the percentage growth rate in discrete time) in continuos time is, by definition, $\frac {\dot L}{L}$.

A system with a constant growth rate is ruled by an equation$^1$ as:

$$ L(t)= L\_0 e^{nt}\;\;\;(1),$$

where $L\_0$ is the population at time $0$.

Indeed, taking the derivative of $(1)$ with respect to time we have:
$$\dot L(t)= nL\_0 e^{nt}.$$

Dividing both sides by $L(t)$, we have (neglecting $t$):

$$\frac {\dot L}{L}= \frac{nL\_0 e^{nt}}{L\_0 e^{nt}}= n.$$

---

$^1$  This is the formalization of the assumption of constant growth rate of population in the Solow model in continuous time


---

### Answer by username_4 on 2023-01-09. Score: 1

No, you shouldn't use quotient rule here. These are the steps:

1. log linearize $$k=K/L \implies \ln k = \ln K - \ln L$$
2. Take time derivative vrt $t$ (treat any variable $x$ as $x(t)$) use chain rule.

$$ \frac{1}{k}k'(t)= \frac{1}{K} K'(t) -\frac{1}{L} L'(t)$$

1. Simplify and adjust notation ($x'(t) = \dot{x}$):

$$ \frac{\dot{k}}{k} = \frac{\dot{K}}{K} -\frac{\dot{L}}{L} $$

As you see there is no reason for using quotient rule here.

Regarding to second question. A continuous time equivalent of $(L\_1 - L\_0)/L\_0$ is $\dot{L}/L$ since $\dot{L}= L\_1 - L\_0$ as $\Delta t \to 0$. So if $L\_1=(1+n)L\_0$ you would expect that $\dot{L}/L=n$ although note $n$ in discrete and continuous time are not 100% the same since in continuous time it is instantaneous growth and in discrete time its over period of time, but if you think of it in terms of methodological continuous time equivalent then they are equivalent.


---

