Derive thermal expansion of area from length

  • #1
tuki
19
1
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?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
5,095
2,106
ΔA is not (ΔL)^2. Calculate ΔA as (L+ΔL)^2 - L^2, and keep only the lowest order term.
 
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