Derive thermal expansion of area from length

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
tuki
Messages
19
Reaction score
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?
 
Physics news on Phys.org
ΔA is not (ΔL)^2. Calculate ΔA as (L+ΔL)^2 - L^2, and keep only the lowest order term.
 
  • Like
Likes   Reactions: CWatters and Chestermiller