Pythagoras with a multivariable function (errors)? I have been looking at the formula for propagation of uncertainty in physics, but I thought my question was better suited for maths stack exchange.
The formula is 
I have been thinking about it in terms of Phythagoras' theorem, and since the variables $X, Y,\ldots$ in this formula are independent, they have orthogonal directions.
So in terms of Pythagoras' theorem, I would think that the formula should be
$$(\Delta R)^2 = (\Delta X)^2+(\Delta Y)^2 + \cdots$$
I really don't understand where the partial derivatives come from.
I have looked at the multivariable chain rule as I read that this formula was based on the multivariable chain rule, however I don't see the link!
EDIT ADDED AFTER READING THE ANSWERS
25/12/2016
So it seems to me that my confusion lies in the fact that the variation of $R$ in the orthogonal $x$ and $y$ directions is not just $\delta x$ and $\delta y$ and the orthogonality is needed to use Phythagoras' theorem. I'm not sure if this is a correct way of thinking about this, but perhaps this has to do with the orthogonal 'axes' for this being skewed or not exactly orthogonal in a certesian coordinate system, which is why the partial derivatives are needed to decompose R onto these orthogonal axes?
I don't think I'm making much sense, but in a nutshell, could linear algebra and basis sets provide an answer to my question? If so, does anyone know what topics specifically it would be useful to look up/Google search to understand this better?
Could someone please tell me where my 'intuition' has gone wrong?
 A: This question deals with the problem of estimating the uncertainty propagation for a general multivariable function, which commonly occurs in physics. In particular, in this case $R $ is a function that depends on the variables $X,Y,... $. To better understand how we get the final formula reported in the OP, it is useful to remind some basic concepts in this context. In physics, the uncertainty in the determination of a given quantity $X $ is often indicated as $\delta X$ and is usually defined as the measurement error of that quantity. It is known that every time we repeat a measurement with a sensitive instrument, we get slightly different results. Often, this error is expressed as the SD of the difference with the true value, assuming a normal distribution of the errors with zero mean (note, however, that the mean may not be zero if systematic bias is present). For example, repeatedly measuring the weight of an object using a given instrument we could obtain a value of $10 \pm 0.1$ kg. In this case, $X=10$ and $\delta X=0.1$.  
On the other hand, in many cases the situation is slightly more complex, as we have to determine not only the uncertainty of a variable $X $, but also that of a function of $X$. Calling $R(X)$ this function,  its uncertanty is defined as 
$$\delta R=\left| \frac {dR}{dX}   \right| \delta X$$
For example, let us suppose that a car travels with a constant velocity along a distance equal to $1$ km, and that an instrument measures a time of $80 \pm 0.5$ seconds for the car to cover this distance. In this scenario, we could be interested in calculating the uncertainty in the velocity. Applying the standard formula $V=S/t $, where $V $ is velocity, $S$ is the distance, and $t $ is the time, we get that, neglecting the error, the car travels at a velocity of $12.5$ m/s. To get the uncertainty, we have to consider that our function $R (x) $ in this example is given by the velocity  as a function of time, i.e. $V(t) = S/t$. So we can write
$$\delta V=\left|\frac {dV}{dt}   \right| \delta t$$
Because $\frac{dV}{dt} =-\frac{S}{t^2} $ and $\delta t=0.5$ seconds, we get
$$\delta V= \left| -\frac{S}{t^2}  \right| \cdot 0.5=  \frac {1000}{80^2}  \cdot 0.5\approx 0.078 \, \text {m/s}$$ 
which gives to us the value of $12.5 \pm 0.078$ for our velocity.
The reason why the uncertainty of a function $R (X) $ is obtained by multiplying the uncertainty of $X $ to the derivative of $R (X) $ has already been explained in the previous answers. To further clarify this issue by considering a simple coordinate system, we can remind that, in physics, the errors are usually very small in magnitude compared to the true value.  Thus, an uncertainty of $\pm \delta X$ in the measurement of $X $ at a given value $X_0$ - which can be seen as an infinitesimal segment on the $x $-axis around the point $(X_0,0)$ - produces an uncertainty of magnitude $\pm k \, \delta R $ in the measurement of $R (X)$, where  $k $ is the slope of the function in that point. This uncertainty in $R (X) $ can be geometrically visualized by drawing two vertical lines starting from the points $(X_0-\delta X,0) $ and   $(X_0+\delta X,0) $ till the line tangent to the function in $X_0$. The result is the identification of an infinitesimal segment on the tangent, whose projection on the $x $-axis is the uncertainty in $X $. This clearly leads to a magnitude of $k \, \delta X $, which corresponds to the product of $\delta X $ with the local derivative.
All these considerations are valid for univariable functions. When we deal with multivariable functions resulting from the sum of terms including different independent variables, there is a further issue to consider. Although it could apparently be intuitive to simply add the uncertainties of different terms in the sum, this approach is not correct. To understand how this approach could be misleading, it is sufficient to imagine a function of two variables where one has a positive uncertainty and the other has a negative uncertainty of similar magnitude: simply summing the uncertainties we would obtain an uncertainty of zero, which clearly is not possible in a situation that starts from two variables with definite uncertainties. To fix this, the typical solution is to square the uncertainties to make all them positive, adding them, and then taking the square root of the sum. So, for a multivariable function $R (X,Y....) $ we have
$$\delta R=\sqrt {    \left( \frac {\partial R}{\partial X}\delta X \right)^2 +    \left( \frac {\partial R}{\partial Y}\delta Y \right)^2....    }$$
which is the formula reported in the OP.
Taking into account these considerations, your thinking about this formula in terms of Phythagoras' theorem might be correct, but the presence of the partial derivatives is necessary to calculate, from the uncertainties in the independent variables $X,Y....$,  the corresponding uncertainty components in the multivariable $R (X,Y...) $, whose "projections" are $\delta X $, $\delta Y$, and so on. 
A: Your intuition is right but you have to replace $\Delta X$ with the error along $x$-axis of $R(X,Y,\ldots)$, $\Delta Y$ with the error along the y-axis of $R(X,Y,\ldots)$ and so on...hence it would be better to write:
$$\delta R=\sqrt{(\delta R_X)^2+(\delta R_Y)^2+\cdots}$$
where $$\delta R_X\approx{\partial R\over\partial X}\delta X$$ $$\delta R_Y\approx{\partial R\over\partial Y}\delta Y$$ are the first order variation of the function $R(X,Y,\ldots)$ along $X$ and $Y$ axes respectively.
A: $\newcommand{\var}{\operatorname{var}}$I think this is essentially about the variance of $\delta R$ when $\delta R$ is approximately a linear combination of $\delta X,\delta Y, \ldots\,\text{etc.},$ and those are independent random variables.
$$
\delta R = a\,\delta X + b\,\delta Y + \cdots.
$$
$$
\var(\delta R) = \var(a\,\delta X) + \var(b\,\delta Y) + \cdots = a^2 \var(\delta X) + b^2 \var(\delta Y) + \cdots.
$$
The partial derivatives are $a,b,\ldots\,\text{etc.}$ The idea is that $\delta R$ is not exactly linear in $\delta X,\delta Y,\ldots\,\text{etc.}$ but it is approximately linear if $\delta X,\delta Y,\ldots\,\text{etc.}$ are very small. Derivatives tell you the rate of change, and that's when you multiply by the infinitely small change in the independent variable to get the infinitely small change in the dependent variable. When these changes are not infinitely small, then this is approximate.
