Supplement 2.6: Differentials and derivatives (2/4)

The temperature as an example of a function of several variables

On the previous page, we already established that the temperature is a function of the Cartesian coordinates $x$ , $y$ and $z$ and on top of that it is a function of the time $t$ :

T = f (x, y, z, t)

The differential of the temperature is hence:

d T = \frac{\partial T}{\partial x} d x + \frac{\partial T}{\partial y} d y + \frac{\partial T}{\partial z} d z + \frac{\partial T}{\partial t} d t

In order to write equations in a more simple way, we combine the Cartesian coordinates within the position vector

\vec{r} = (x, y, z)

The temperature becomes then:

T = f (\vec{r}, t)

We also combine the Cartesian differentials to obtain a differential position vector

d \vec{r} = (d x, d y, d z)

The same procedure applies to the derivatives by using the spatial derivative vector

\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})

The $\nabla$ symbol is the Nabla operator.

The differential of the temperature with these vectors is then written as follows:

d T = \frac{\partial T}{\partial t} d t + d \vec{r} \cdot \nabla T

where the multiplication symbol $\cdot$ denotes the scalar product (or: dot product) of vectors. The spatial derivative of the temperature $\nabla T$ is the temperature gradient

\nabla T = (\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z})

How can we understand the two equivalent representations

d T = \frac{\partial T}{\partial x} d x + \frac{\partial T}{\partial y} d y + \frac{\partial T}{\partial z} d z + \frac{\partial T}{\partial t} d t = \frac{\partial T}{\partial t} d t + d \vec{r} \cdot \nabla T

of the temperature differential? How can they be put into practice?

Starting from a given temperature field $T = f (x, y, z, t) = f (\vec{r}, t)$ in space and time, which is known from measured data or from a numerical model, we may calculate a temperature value $T (\vec{r}', t')$ for a selected point in space $\vec{r}' = (x', y', z')$ and time $t'$ .
From the temperature, we may also calculate the partial derivatives with respect to the four variables
$\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}, \frac{\partial T}{\partial t}$ resp. $\nabla T, \frac{\partial T}{\partial t}$
From these functions, we may calculate specific values for the partial derivatives at a selected point in space and time $(\vec{r}', t')$ . These values inform us on changes in temperature at this point that we could expect from small displacements and a small time lapse.
The differentials $d x$ , $d y$ , $d z$ resp. $d \vec{r}$ such as $d t$ which act as the factors, represent such changes in space and time. They permit small displacements (within infinitesimal limits) to the location $\vec{r}$ by $d \vec{r}$ , while the time $d t$ passes.
Now everything is known to determine the differential $d T$ . The temperature at another location and a further point of time is then:
$T (x' + d x, y' + d y, z' + d z, t' + d t) = T (x', y', z', t') + d T$
resp. $T (\vec{r}' + d \vec{r}, t' + d t) = T (\vec{r}', t') + d T$

This is discussed in task 4 with a numerical example.

Task 4: temperature in space and time ↓

The temperature may be given by the function

T = T_{o} + a x - b x y^{2} + c y z^{2} + d z t

with:
$T_{o}$ =20°C
$a$ =0,3°C/m
$b$ =0,02°C/m³
$c$ =0,1°C/m³
$d$ =0,01°C/(m·s)

a) What is the temperature like at the location $\vec{r}' = (x', y', z')$ = (10 m, 5 m, 2 m) at the time $t'$ = 100 s.

b) Calculate the partial derivatives of the temperature. What values do you obtain for the location $\vec{r}'$ and at the time $t'$ ?

c) Calculate the temperature difference $d T$ for the displacement $d \vec{r} = (d x, d y, d z)$ = (1 m, 0,5 m, 0,5 m) with the time $d t$ = 2 s later.
(We treat the differentials for approximation as finite quantities)

d) What values does the temperature $T (\vec{r}' + d \vec{r}, t' + d t)$ take?

You may find the solutions on this site.

Fewer variables: snapshots and time series ↓

Temperatures do not always have to be given as a function of four variables.

Without the time as a variable one has $T = f (\vec{r})$ . This corresponds to a picture of the moment (or: a snapshot) of the temperature distribution in space. The differential of the temperature becomes:
$d T = \frac{\partial T}{\partial x} d x + \frac{\partial T}{\partial y} d y + \frac{\partial T}{\partial z} d z$
resp. $d T = d \vec{r} \cdot (\nabla T)$ .

Regarding only two resp. one spatial variable, it becomes a temperature one a surface resp. along a line.
Without the spatial variable one has $T = f (t)$ . This corresponds to a time series of the temperature at a fixed location. The differential of the temperature becomes:
$d T = \frac{\partial T}{\partial t} d t$
Time series are broadly discussed in another SEOS tutorial.

Le spectre de la Terre

Supplement 2.6: Differentials and derivatives (2/4)

The temperature as an example of a function of several variables