I have a question for homework I am suffering with, the question is:
A steel bar with square cross sections $5 \operatorname{cm}$ by $5 \operatorname{cm}$ and length $4 \operatorname{m} $ is being heated. For each dimension, the bar expands $13\cdot10^{-6} \operatorname{m} $ for each $1\operatorname{C°} $ rise in temperature. The answer will be in the format of $\frac{\partial V}{\partial t} = \underline{ } \cdot 10^{-6} \operatorname {m^3/C°}$. A hint given reads: Let the square cross section have side length $x(T)$ and the bar have length $L(T)$ at temperature $T$ degrees Celsius. Letting $V$ be the volume of the bar, the chain rule gives $\frac{\partial V} {\partial T} =\frac{\partial V} {\partial x} \frac{\partial x}{\partial T} + \frac{\partial V}{\partial L}\frac{\partial L}{\partial T}$
I didn't really know where to start so I found the base volume $(5\cdot5\cdot400) = 10000$, and the updated volume after a change of $1$ degree $(5\cdot (13\cdot 10^{-6} )\cdots) = 10000.39001$, and subtracted the second from the first, and got $0.39$ but that seems too big given the answer format given.
Any help would be appreciated!
You were told where to start in the first paragraph.
Just apply the Chain Rule to the formula for the volume of a square-based rectangular prism.
To shorten typing, let $c=13\cdot 10^{-6}~\textrm m~{^\circ \mathrm C}^{-1}$
$\begin{align}x(T) &= (0.05+c~T)~\text{m} \\[2ex]\ell(T)&=(4+c~T)~\text{m}\\[2ex]V(T) &= x^2(T)~\ell(T)\\&=(0.05+cT)^2~(4+cT)~\text{m}^3 \\[2ex]\left.\dfrac{\mathrm d V}{\mathrm d T}\right\rvert_{T=1} &= \end{align}$