 #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?
$$ \Delta l = \alpha l_0 \Delta T $$
$$ (\Delta l)^2 l_0 = \alpha l_0^2 \Delta T \Delta l $$
$$ \Delta A l_0 = \alpha A_0 \Delta T $$
$$ \Delta A = \frac{ \alpha A_0 \Delta T }{ l_0 } $$
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?