Derive thermal expansion of area from length

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SUMMARY

The discussion centers on deriving the thermal expansion of area from the linear expansion of length, specifically using the coefficient of linear expansion, α. The initial attempt involved equations that incorrectly suggested ΔA = (α)² A₀ (ΔT)², while the correct formulation is ΔA ≈ (2α)A₀ΔT. The error stems from not properly calculating the change in area as ΔA = (L + ΔL)² - L² and neglecting higher-order terms. This highlights the importance of maintaining accuracy in mathematical derivations related to thermal expansion.

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Homework Statement
Derive thermal expansion of area from length
Relevant Equations
Linear thermal expansion for length:
$$ \Delta l = \alpha l_0 \Delta T $$
I tried following:

$$ \Delta l = \alpha l_0 \Delta T $$
$$ (\Delta l)^2 l_0 = \alpha l_0^2 \Delta T \Delta l $$
$$ \Delta A l_0 = \alpha A_0 \Delta T $$
$$ \Delta A = \frac{ \alpha A_0 \Delta T }{ l_0 } $$
If we remember that:
$$ \Delta l = \alpha l_0 \Delta T $$
So we have
$$ \Delta A = \frac{ \alpha A_0 \Delta T \alpha l_0 \Delta T }{ l_0 } $$
$$ \Delta A = (\alpha)^2 A_0 (\Delta T)^2 $$

However the correct solution should be;

$$ \Delta A \approx (2 \alpha)A_0 \Delta T $$

Any suggestion on what's going wrong or what should i try next?
 
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ΔA is not (ΔL)^2. Calculate ΔA as (L+ΔL)^2 - L^2, and keep only the lowest order term.
 
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