# Growth ```Growth
Consider any variable 𝑋 which depends on time, 𝑡. Then we can write 𝑋(𝑡). Then the rate of change of
𝑋(𝑡) is simply is the time derivative:
𝑑𝑋
= 𝑋 ′ (𝑡).
𝑑𝑡
Based on this notion of “change”, we can also define growth:
𝐺𝑟𝑜𝑤𝑡ℎ 𝑜𝑓 𝑋 = 𝑔𝑋 =
𝑋 ′ (𝑡)
.
𝑋(𝑡)
Why is this the definition of the growth rate?
Example: Suppose that GDP per capita in Turkey is \$10K this year. If the GDP per capita will rise to
\$11K next year, then
𝑑𝑋 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 = 11 − 10
𝑑𝑡 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 = 1
𝑋 = 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 = 10.
Therefore, if the definition is correct, then we should see:
𝑑𝑋
1
𝑋 ′ (𝑡)
𝑑𝑡
=
= 1 = 10%.
𝑋(𝑡)
𝑋
10
Funny fact: If you take the derivative of ln 𝑋(𝑡), then the answer is
𝑑𝑙𝑛𝑋(𝑡) 𝑋 ′ (𝑡)
=
.
𝑑𝑡
𝑋(𝑡)
That is because, if you take the derivative of an expression in logarithms, then you should first take the
derivative of the original expression and divide the derivative by the original expression.
Remark: If you want to calculate the growth rate of any variable, then you can calculate the derivative
of its logarithm.
Application: Assume that 𝑌(𝑡) = 𝐴(𝑡) &times; 𝐵(𝑡). Then the growth rate of 𝑌 can be calculated as
ln 𝑌 = ln 𝐴 + 𝑙𝑛𝐵.
Take the time derivative of both sides to find the growth rate of 𝑌:
𝑔𝑌 = 𝑔𝐴 + 𝑔𝐵 .
Solow’s Model
Consider an economy where the production technology is Cobb-Douglas:
𝑌 = (𝐾)𝑎 (𝐴𝐿)1−𝑎
where 𝑌 is output, 𝐾 is capital, 𝐿 is labor, and 𝐴 is the average (per worker) information or knowledge.
𝐴 can also be interpreted as “technology”. In fact, 𝐴𝐿 is called “effective labor”.
We assume that the growth rates of 𝐴 and 𝐿 are constant, denoted by 𝑔𝐴 &gt; 0 and 𝑔𝐿 &gt; 0. This means
technological growth and population growth are exogenous. Now let us see the growth rate of 𝐾. The
rate of change in 𝐾 is
𝑑𝐾
= 𝑠&times;𝑌−𝛿&times;𝐾
𝑑𝑡
where 𝑠 ∈ [0,1] is the saving rate, and 𝛿 ∈ [0,1] is the depreciation rate. To find the growth rate of 𝐾,
simply divide the rate of change by 𝐾 to obtain
𝑑𝐾
𝑑𝑡 = 𝑔 = 𝑠 𝑌 − 𝛿.
𝐾
𝐾
𝐾
In real life
𝑌
𝐾
≈ 1/3.
Definition: In economics, long-run is when endogenous variables are constant.
Example: In Solow’s model, the growth rate of 𝐾 is endogenous so it would be constant in the longrun.
Theorem: 𝑔𝐾 = 𝑔𝑌 in the long-run.
Proof: First note that 𝑔𝐾 is a constant in the long-run. This means
𝑠
𝑌
−𝛿
𝐾
is a constant. However, by assumption, 𝑠 and 𝛿 are constants. Deduce that
𝑌
𝐾
can be neither increasing nor decreasing over time. This means
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 =
𝑌
𝐾
ln(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) = ln
𝑌
𝐾
𝑌
dln(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝑑 ln 𝐾
=
𝑑𝑡
𝑑𝑡
0=
𝑌
𝑑 ln 𝐾
𝑑𝑡
=
𝑑(𝑙𝑛𝑌 − ln 𝐾) 𝑑𝑙𝑛𝑌 𝑑𝑙𝑛𝐾
=
−
= 𝑔𝑌 − 𝑔𝐾 .
𝑑𝑡
𝑑𝑡
𝑑𝑡
Conclude 𝑔𝑌 = 𝑔𝐾 . End of proof.
As a consequence, the following result ensues:
Theorem: 𝑔𝑌 = 𝑔𝐴 + 𝑔𝐿 in the long-run.
Proof: Recall that
𝑌 = (𝐾)𝑎 (𝐴𝐿)1−𝑎 .
Therefore, take the logarithm of this expression to see:
ln 𝑌 = 𝑎 ln 𝐾 + (1 − 𝑎)(ln 𝐴 + ln 𝐿).
Now take the time derivative to find the growth rates:
𝑔𝑌 = 𝑎𝑔𝐾 + (1 − 𝑎)(𝑔𝐴 + 𝑔𝐿 ).
Since 𝑔𝐾 = 𝑔𝑌 in the long-run, this expression simplifies to
𝑔𝑌 = 𝑎𝑔𝑌 + (1 − 𝑎)(𝑔𝐴 + 𝑔𝐿 )
𝑔𝑌 − 𝑎𝑔𝑌 = (1 − 𝑎)(𝑔𝐴 + 𝑔𝐿 )
(1 − 𝑎)𝑔𝑌 = (1 − 𝑎)(𝑔𝐴 + 𝑔𝐿 )
𝑔𝑌 = 𝑔𝐴 + 𝑔𝐿
which is the desired result in the long-run. This completes the second proof.
Our final result is this.
Theorem: 𝑔𝑌 = 𝑔𝐴 in the long run.
𝐿
Proof: Note that the growth of GDP per capita is
dln 𝑌/𝐿 dln 𝑌 dln 𝐿
=
−
= 𝑔𝑌 − 𝑔𝐿 = 𝑔𝐴 + 𝑔𝐿 − 𝑔𝐿 = 𝑔𝐴 .
𝑑𝑡
𝑑𝑡
𝑑𝑡
Conclude
𝑔𝑌 = 𝑔𝐴 .
𝐿
This completes our final result.
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