Problem with differentials and relative rates of change I'm studying single variable Calculus. 
Here's a problem involving differential forms that the teacher gave us and that I don't know how to express in mathematical terms:

There's a box of height h with a square base of side length L. Assuming that L is increasing at a rate of 10% per day and h is decreasing at a rate of 10% per day, what happens if we use a linear approximation to find at what (approximate) rate the volume of the box changing?
Hint: What is the relative rate of change of the volume of the box?
Hint2: in this case it is easy to calculate the exact rate of change —8.9%—, so using linearization might seem like overkill. However, if things are set up right, there's not even need for a calculator to find out the approximate rate of change! Do you see why?

Now, he gave us the answer (10%), but I'm trying to make sense of it. I think that the relative rate of change of a function V with respect to itself could be expressed as dV/V, but I'm just guessing. I'm still wondering how to get to that result of 8.9%.
Anyone got some insight?
 A: We all seem to be answering this a bit late, but then I don't have to worry about doing too much work for somebody's homework problem, do I?

First, if a box has height $ h $ and a square base of width $ L $, then the area of the base is $ A = L ^ 2 $, so the volume of the box is $ V = h A = h L ^ 2 $.  So you want to differentiate $ V = h L ^ 2 $.
I get $$ \mathrm d V = L ^ 2 \, \mathrm d h + 2 h L \, \mathrm d L \text . $$ (If you're not sure how to get that, then I can go into it in more detail.)  And although I normally think that differentials should be used more, I'm inclined to divide this through by $ \mathrm d t $ and talk instead about derivatives with respect to time $ t $; then abbreviating $ \mathrm d x / \mathrm d t $ (for any quantity $ x $) as $ \dot x $, this becomes $$ \dot V = L ^ 2 \dot h + 2 h L \dot L \text . $$  But it doesn't really matter how you write it, because the form is the same in either case.
I agree that the relative rate of change of a quantity $ x $ is $ \mathrm d x / x $, except that (if we're talking about the relative rate of change with respect to time) I'd say $ \dot x / x $ instead.  So you are told that $ \dot L / L $ is $ 1 0 \, \% $ per day, which is $ 1 0 / 1 0 0 = 1 / 1 0 = 0 .1 $ (however you want to put it) per day; similarly, $ \dot h / h $ is $ - 1 / 1 0 $ per day.  To use these in the expression for $ \dot V $, say them as $ \dot L = \frac 1 { 1 0 } L $ and $ \dot h = - \frac 1 { 1 0 } h $ (measuring time in days).  Then $$ \dot V = L ^ 2 \Bigl ( - \frac 1 { 1 0 } h \Bigr ) + 2 h L \Bigl ( \frac 1 { 1 0 } L \Bigr ) = \frac { - L ^ 2 h + 2 h L ^ 2 } { 1 0 } = \frac 1 { 1 0 } h L ^ 2 \text . $$  Then dividing by $ V = h L ^ 2 $, we get the final answer: $$ \frac { \dot V } V = \frac 1 { 1 0 } \text . $$  Or since $ 1 / 1 0 = 1 0 \, \% $ and our unit of time is the day, the relative rate of change of the volume is $ 1 0 \, \% $ per day.

Now, about that $ 8 .9 \, \% $.
Everything that I wrote above is exact, except for my implicit interpretation of the phrase ‘L is increasing at a rate of 10% per day’ to mean that the relative rate of change of $ L $ (with respect to time) is $ 1 0 \, \% $ per day (and similarly for $ h $).  A relative rate of change is a quantity with units of reciprocal time, and $ 1 0 \, \% $ per day, or $ 0 .1 \operatorname { dy } ^ { - 1 } $ to make the units very explicit, could equally well be expressed in different units as $ 7 0 \, \% $ per week, $ \frac 5 { 1 2 } \% $ per hour, etc.  If we happen to use days as our unit of time, then this quantity comes out to $ 1 / 1 0 $.  As I calculated above, if the relative rate of change of $ L $ is $ 1 0 \, \% $ per day and the relative rate of change of $ h $ is $ - 1 0 \, \% $ per day, then the relative rate of change of $ V $ is $ 1 0 \, \% $ per day, exactly.  There is no linear approximation here.
But if the relative rate of change of a quantity $ x $ is $ 1 0 \, \% $ per day, then if you wait a day and compare the size of $ x $ at the end of the day to the size of $ x $ at the beginning of the day, then you will find that $ x $ has actually increased by a little more than $ 1 0 .5 \, \% $!  This is because, if $ \dot x / x = 1 / 1 0 $, then we can rewrite this as $ ( \mathrm d x / \mathrm d t ) / x = 1 / 1 0 $, or $ 1 0 \, \mathrm d x / x = \mathrm d t $, and since $ 1 0 \, \mathrm d x / x $ is the differential of $ 1 0 \ln x $ (check for yourself by differentiating that), $ 1 0 \ln x $ and $ t $ must be the same up to a constant term.  Dividing by $ 1 0 $ and exponentiating, $ x $ and $ \mathrm e ^ { t / 1 0 } $ must be the same up to a constant factor.  Write this as $ x = x _ 0 \mathrm e ^ { t / 1 0 } $ where $ x _ 0 $ is the original length (the value of $ x $ when $ t = 0 $).  Then a day later (when $ t = 1 $), $ x $ will be $ x _ 0 \mathrm e ^ { 1 / 1 0 } $.  Relative to $ x _ 0 $, this is $ \mathrm e ^ { 1 / 1 0 } \approx 1 .1 0 5 $ times as large.  So $ x $ increased by about $ 1 0 .5 \, \% $.
So the other interpretation of ‘L is increasing at a rate of 10% per day’ is that $ L $ grows by $ 1 0 \, \% $ (not about $ 1 0 .5 \, \% $) every day.  (In that case, it will increase by more than $ 7 0 \, \% $ in a week, but less than $ \frac 5 { 1 2 } \% $ in an hour; the day here is no longer just an arbitrary unit of measurement.)  We could figure out what relative growth rate would produce that, but that's not necessary.  We just need to know that, after a day, $ L $ will be $ 1 .1 L _ 0 $, where $ L _ 0 $ is the original length.  Similarly, $ h $ wil be $ 0 .9 h _ 0 $ after a day.  Then $ V = h L ^ 2 $ will be $ ( 0 .9 h _ 0 ) ( 1 .1 L _ 0 ) ^ 2 = 1 .0 8 9 h _ 0 L _ 0 ^ 2 = 1 .0 8 9 V _ 0 $.  In other words, the volume will increase by $ 8 .9 \, \% $.
You don't really need a calculator for this, but you probably don't want to do it in your head either.  You just have to multiply $ 0 .9 $ by $ 1 .1 ^ 2 $.  But I admit it, I used a calculator.
A: Denote the volume of the box as $V = hL^2$
so
$$ dV = (dh) L^2 + 2hL (dL) $$
Using $dL=0.1 L, dh = -0.1 h$, you see that
$dV = 0.1 V$ and thus a relative change of 10% per day.
A: I also had difficulty expressing this approximation mathematically, but this is what I came up with:
Considering the following given expressions,
$L(t) = L + (0.1*L*t), h(t) = h - (0.1*h*t), V = L^2h$,
and $dV/V=the\ relative\ rate\ of\ change\ of\ the\ volume\ of\ the\ box$,
one finds that
$dL=(0.1L)dt, dh=(-0.1h)dt, dV/V=(2dL/L) + (dh/h).$
After these considerations and substitutions, one can see an expression in the form of $dV/V=Cdt$, where C is the solution you are looking for: 0.1 (or 10%).
