Differential question word problem? Joules law is a physical law expressing the relationship between heat generated by a current flowing through a semi conductor.
It states $Q=I^2Rt$ where q is heat in J joules, generated by a constant current I (in amperes A) flowing through a conductor of electrical resistance R in ohms for a time period t in seconds s.
Using a differential estimates the amount by which heat will increase if the current is increased from 16 A to 16.6 A, the resitance increased from 2 to 2.1 ohms, and time interval increased from 10 to 10.1 minutes.
For my problem I did
$\Delta q=\frac{dq}{dI} \Delta I+\frac{dq}{dt} \Delta t+\frac{dq}{dr} \Delta r$
$\frac{dq}{dI}=2IRt$
plugged in my values $2(16)(2)(1)=640$
$\frac{dq}{dt}=I^2R(1)$
$(1)^2(2)=512$
$\frac{dq}{dr}=I^2t$
$(16)^2(1)=2560$
Then I did
$640(.6)+512(.1)+2560(.1)$
$\Delta q=691.2$
But I am not sure if my solution is corrected because the time interval is in minutes but t is in seconds. 
 A: Let's not drop our units:
At $I=16 \text{ amperes}$, $R=2\text{ ohms}$ and $t=10\text{ minutes}$,
$\dfrac{\partial Q}{\partial I}=2IRt=2\cdot16 \text{ amperes}\cdot2\text{ ohms}\cdot10\text{ minutes}$
$\dfrac{\partial Q}{\partial R}=I^2t=16^2 \text{ amperes}^2\cdot10\text{ minutes}$
and $\dfrac{\partial Q}{\partial t}=I^2R=16^2 \text{ amperes}^2\cdot2\text{ ohms}$
You can simplify these units here or at the end when you reach your final answer:
On evaluating $\Delta Q=\dfrac{\partial Q}{\partial I} \Delta I+\dfrac{\partial Q}{\partial t} \Delta t+\dfrac{\partial Q}{\partial R} \Delta R$, you should get (as you already have got), $691.2 \text{ amperes$^2$ ohms minutes}$.
Physically, this unit is equivalent to that of the unit of energy, but it is not obvious that this is really the case. So, we would like to rewrite our answer in a more recognizable unit.
Since an $\text{ohm = }$ $\dfrac{\text{ volt}}{\text{ampere}}$ and an $\text{ ampere volt }=\text{ joule second$^{-1}$}$, your unit
$\text{ampere$^2$ ohm minute}= \text{ampere$^2$} \dfrac{\text{volt}}{\text{ampere}}\text{ minute}=\text{ampere volt}\text{ minute}=\text{ joule second$^{-1}$ minute.}$
So your answer is equivalent to $691.2 \text{ amperes$^2$ ohms minutes}=691.2\text{ joules second$^{-1}$ minute}$. A minute divided by a second is $60$, so the unit can be rewritten as $60\cdot \text{joules}$.
So your answer is $691.2\cdot60\cdot\text{ joules}=41472 \text{ joules}$ (you can also start out by converting all of your units to SI units, in which case you would still be led to the correct answer).
