]> Jacobian

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 1: Differentiation; page 7

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Jacobian

Two variables

The change of two variables was studied on page 4. By definition

$\begin{array}{l} variables {\begin{cases} x = x (u, v) \\ y = y (u, v) \end{cases} \\ function: \\ {\begin{cases} f (x, y) = f (u, v) = \\ = f [x (u, v), y (u, v)] \end{cases} \end{array}}$	(3.1.7.1)

and the rule for differentiation, by the use of matrix notation (3.1.4.5), is reproduced here:

$(\begin{array}{l} \frac{\partial f}{\partial u} \\ \frac{\partial f}{\partial v} \end{array}) = (\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{matrix}) (\begin{array}{l} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array})$	(3.1.7.2)

One has to remember that for the differentiation, the function is dependent on one set of variables (x, y) or (u, v), without mixing them.

Our aim is now, to find a way, to invert the transformation (the change of variables). By definition, the inversion is based on the inverted relations (3.1.7.1), namely:

$\begin{array}{l} u = u (x, y) \\ v = v (x, y) \end{array}}$	(3.1.7.3)

In order to invert the transformation, we don't need to know the exact expressions of (3.1.7.3). What we need is just the derivatives $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}$ and $\frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ . This can be achieved by substituting f from (3.1.7.2) , by u or v :

$(\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{matrix}) (\begin{matrix} \frac{\partial u}{\partial x} \\ \frac{\partial u}{\partial y} \end{matrix}) = (\begin{matrix} \frac{\partial u}{\partial u} \\ \frac{\partial u}{\partial v} \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix})$	(3.1.7.4)

yielding the two linear equations

$\begin{array}{l} \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial u} \frac{\partial u}{\partial y} = 1 \| \times \frac{\partial y}{\partial v} \\ \frac{\partial x}{\partial v} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v} \frac{\partial u}{\partial y} = 0 \| \times (- \frac{\partial y}{\partial u}) \end{array}}$	(3.1.7.5)

If the equations are multiplied by the factors given on the right-hand side of (3.1.7.5) and added up, one obtains the expression

$\frac{\partial u}{\partial x} (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) = \frac{\partial y}{\partial v}$	(3.1.7.6)

that allows one to calculate $\frac{\partial u}{\partial x}$ from the known derivatives of (3.1.7.1) . The expression in brackets of (3.1.7.6) is called the Jacobian and will be denoted here as

$\frac{D (x, y)}{D (u, v)} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}$	(3.1.7.7)

Elsewhere one can find also the notations $\frac{\partial (x, y)}{\partial (u, v)}$ , or more symbolically $J (\frac{x, y}{u, v})$ . As far as the physical dimensions are concerned

$[\frac{D (x, y)}{D (u, v)}] = \frac{[x] [y]}{[u] [v]}$	(3.1.7.8)

where the square brackets denote dimensions. These dimensions are very helpful, for the verification of more complicated expressions that include the Jacobian.

The expressions of D themselves are not numbers, but similar to the expressions of differentials. Finally (3.1.7.6) can be rewritten as

$\frac{\partial u}{\partial x} \frac{D (x, y)}{D (u, v)} = \frac{\partial y}{\partial v}$	(3.1.7.9)

Notice that according to the definition (3.1.7.7):

$\frac{D (x, y)}{D (u, v)} = - \frac{D (y, x)}{D (u, v)} = - \frac{D (x, y)}{D (v, u)} = \frac{D (y, x)}{D (v, u)}$	(3.1.7.10)

It means that the exchange of two variables inverts the sign of the Jacobian.

By use of (3.1.7.9), and just by exchanging x with y and u with v , one obtains the following four relations:

$\begin{array}{l} \frac{\partial u}{\partial x} \frac{D (x, y)}{D (u, v)} = \frac{\partial y}{\partial v} \frac{\partial u}{\partial y} \frac{D (y, x)}{D (u, v)} = \frac{\partial x}{\partial v} \\ \frac{\partial v}{\partial x} \frac{D (x, y)}{D (v, u)} = \frac{\partial y}{\partial u} \frac{\partial v}{\partial y} \frac{D (y, x)}{D (v, u)} = \frac{\partial x}{\partial u} \end{array}}$	(3.1.7.11)

If the transformation (3.1.7.1) and (3.1.7.2) could be inverted, and done directly from (3.1.7.3), the exchange of the pair of variables (x,y) with (u,v) would yield, instead of (3.1.7.11), the relations

$\begin{array}{l} \frac{\partial x}{\partial u} \frac{D (u, v)}{D (x, y)} = \frac{\partial v}{\partial y} \frac{\partial x}{\partial v} \frac{D (v, u)}{D (x, y)} = \frac{\partial u}{\partial y} \\ \frac{\partial y}{\partial u} \frac{D (u, v)}{D (y, x)} = \frac{\partial v}{\partial x} \frac{\partial y}{\partial v} \frac{D (v, u)}{D (y, x)} = \frac{\partial u}{\partial x} \end{array}}$	(3.1.7.12)

with the following definition:

$\frac{D (u, v)}{D (x, y)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$	(3.1.7.13)

The relations (3.1.7.11) combined with (3.1.7.12) give a remarkable result. Let's take $\frac{\partial x}{\partial u}$ from the bottom right-hand expression of (3.1.7.11) and substitute it in the upper left-hand expression of (3.1.7.12). The result is

$\begin{array}{l} \frac{\partial v}{\partial y} \frac{D (y, x)}{D (v, u)} \frac{D (u, v)}{D (x, y)} = \frac{\partial v}{\partial y} \\ \Rightarrow \frac{D (x, y)}{D (u, v)} \frac{D (u, v)}{D (x, y)} = 1 \end{array}}$	(3.1.7.14)

which finally gives:

$\frac{D (x, y)}{D (u, v)} = \frac{1}{\frac{D (u, v)}{D (x, y)}}$	(3.1.7.15)

The relation (3.1.7.15) states that in the case of inversion, the Jacobian behaves like the full derivative, namely $\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}$ . In addition, a chain rule (as in the case of full derivatives) is also in vigour for the Jacobian, namely

$\frac{D (x, y)}{D (r, s)} = \frac{D (x, y)}{D (u, v)} \frac{D (u, v)}{D (r, s)}$	(3.1.7.16)

where

$\begin{matrix} x = x (u, v) & y = y (u, v) \\ u = u (r, s) & v = v (r, s) \end{matrix}}$	(3.1.7.17)

The proof of (3.1.7.16) is left for the reader as exercise 2.

Features

We are going to follow an example, in order to see the important features of the Jacobian. The example consists of the following change of variables:

$\begin{array}{l} x = u \exp v \\ y = v \exp u \end{array}}$	(3.1.7.18)

The particularity of the example is that in the framework of the elementary functions, the inversion

$\begin{array}{l} u = u (x, y) \\ v = v (x, y) \end{array}}$	(3.1.7.19)

cannot be solved explicitly. In spite of this, the Jacobian provides the necessary tools, for the transformation in both directions.

From (3.1.7.18) the Jacobian is

${\begin{cases} \frac{D (x, y)}{D (u, v)} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} = \\ = \exp (v + u) - u v \exp (v + u) = \\ = (1 - u v) \exp (v + u) \end{cases}}$	(3.1.7.20)

The inverse Jacobian can not be calculated directly, since (3.1.7.19) does not exist explicitly, but it can be obtained from (3.1.7.20)

$\frac{D (u, v)}{D (x, y)} = \frac{1}{\frac{D (x, y)}{D (u, v)}} = \frac{\exp [- (v + u)]}{1 - u v}$	(3.1.7.21)

Also the derivatives of u and v with respect to x and y cannot be expressed directly as functions of x and y, but can be obtained in terms of u and v, by the aid of (3.1.7.12). By the use of (3.1.7.21), we obtain

$\begin{array}{l} \frac{\partial u}{\partial x} = \frac{D (u, v)}{D (x, y)} \frac{\partial y}{\partial v} = \frac{\exp (- v)}{1 - u v} \\ \frac{\partial u}{\partial y} = \frac{D (u, v)}{D (y, x)} \frac{\partial x}{\partial v} = - \frac{u \exp (- u)}{1 - u v} \\ \frac{\partial v}{\partial x} = \frac{D (v, u)}{D (x, y)} \frac{\partial y}{\partial u} = - \frac{v \exp (- v)}{1 - u v} \\ \frac{\partial v}{\partial y} = \frac{D (v, u)}{D (y, x)} \frac{\partial x}{\partial u} = \frac{\exp (- u)}{1 - u v} \end{array}}$	(3.1.7.22)

\begin{array}{l} \frac{\partial u}{\partial x} = \frac{D (u, v)}{D (x, y)} \frac{\partial y}{\partial v} = \frac{\exp (- v)}{1 - u v} \\ \frac{\partial u}{\partial y} = \frac{D (u, v)}{D (y, x)} \frac{\partial x}{\partial v} = - \frac{u \exp (- u)}{1 - u v} \\ \frac{\partial v}{\partial x} = \frac{D (v, u)}{D (x, y)} \frac{\partial y}{\partial u} = - \frac{v \exp (- v)}{1 - u v} \\ \frac{\partial v}{\partial y} = \frac{D (v, u)}{D (y, x)} \frac{\partial x}{\partial u} = \frac{\exp (- u)}{1 - u v} \end{array}}

(3.1.7.22)

Following the same example, for a known function f(u,v), we can calculate the values of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ , with the help of (3.1.7.22):

$\begin{array}{l} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = \frac{\exp (- v)}{1 - u v} (\frac{\partial f}{\partial u} - v \frac{\partial f}{\partial v}) \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\exp (- u)}{1 - u v} (- u \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}) \end{array}}$	(3.1.7.23)

\begin{array}{l} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = \frac{\exp (- v)}{1 - u v} (\frac{\partial f}{\partial u} - v \frac{\partial f}{\partial v}) \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\exp (- u)}{1 - u v} (- u \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}) \end{array}}

(3.1.7.23)

Determinants

A determinant is a notion borrowed from linear algebra, and we are going to see just a superficial overview, in order to acquire the ability for handling it. A determinant is defined as a square pattern of numbers, and has a numerical value, obeying some rules for its evaluation.

The simplest determinant, of order one, consists of one number, and its value is the number itself. The determinant of the second order, is formed by four indexed numbers, written in two rows and two columns:

$\det A = \| \begin{matrix} A_{1, 1} & A_{1, 2} \\ A_{2, 1} & A_{2, 2} \end{matrix} \| = A_{1, 1} A_{2, 2} - A_{1, 2} A_{2, 1}$	(3.1.7.24)

The first index is the number of the row, and the second - of the column. On the left-hand side of equation (3.1.7.24) is the determinant, and on the right-hand side is the rule for evaluating the determinant. The vertical straight segments inserting the determinant, have nothing to do with absolute value, and are just an indication that the pattern of numbers forms a determinant, which should not be confused with a matrix (round brackets).

The Jacobian in the case of two variables (3.1.7.13) is a determinant of the second order:

$\frac{D (u, v)}{D (x, y)} = \| \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix} \| = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$	(3.1.7.25)

The exchange of two variables (of the same pair) in the Jacobian (3.1.7.10) is equivalent to the exchange of two rows or two columns of the determinant (3.1.7.25), and results in reversing the determinant's sign.

A determinant of order n (n is a natural number) consists of n² numbers, written in a quadratic pattern of n rows and n columns.

The evaluation of a determinant of order n can be done in different ways, without affecting the final result. For simplicity, we are going to limit ourselves to one single rule, which is a particular case of the so called Laplace expansion. For each term of the first row, we'll define a minor, a determinant of order n−1 , by removing the first row and the column c of this term, and by multiplying it with the factor (−1)^(c+1). The value of the determinant is the sum of the terms of the first row multiplied by their minors.

Let's apply this rule for a determinant of order 2 (3.1.7.24):

The minor of the term $A_{1, 1}$ is ${(- 1)}^{(1 + 1)} A_{2, 2} = A_{2, 2}$ .
The minor of the term $A_{1, 2}$ is ${(- 1)}^{(2 + 1)} A_{2, 1} = - A_{2, 1}$ .
Finally the value of the determinant is $A_{1, 1} A_{2, 2} - A_{1, 2} A_{2, 1}$

By using the same rule, the determinant of the third order becomes

$\begin{array}{l} \det A = \| \begin{matrix} A_{1, 1} & A_{1, 2} & A_{1, 3} \\ A_{2, 1} & A_{2, 2} & A_{2, 3} \\ A_{3, 1} & A_{3, 2} & A_{3, 3} \end{matrix} \| = \\ = {\begin{cases} A_{1, 1} A_{2, 2} A_{3, 3} + A_{1, 2} A_{2, 3} A_{3, 1} + A_{1, 3} A_{3, 2} A_{2, 1} - \\ - A_{1, 1} A_{2, 3} A_{3, 2} - A_{1, 2} A_{2, 1} A_{3, 3} - A_{1, 3} A_{2, 2} A_{3, 1} \end{cases} \end{array}}$	(3.1.7.26)

as it is derived and illustrated in Fig. Determinant of order 3.

One can prove by induction that the number of terms appearing in the sum, for the calculation of the determinant's value of order n, is n! . An example of a determinant of the fourth order is given in exercise 3. One can prove also that for a determinant of any order, an exchange of two rows or columns inverts the sign of the determinant. On the other hand, the exchange of all the rows with the corresponding columns conserves the value of the determinant. For instance it follows that

$\| \begin{matrix} A_{1, 1} & A_{2, 1} & A_{3, 1} \\ A_{1, 2} & A_{2, 2} & A_{3, 2} \\ A_{1, 3} & A_{2, 3} & A_{3, 3} \end{matrix} \| = \| \begin{matrix} A_{1, 1} & A_{1, 2} & A_{1, 3} \\ A_{2, 1} & A_{2, 2} & A_{2, 3} \\ A_{3, 1} & A_{3, 2} & A_{3, 3} \end{matrix} \|$	(3.1.7.27)

Many variables

The Jacobian corresponding to the change of a set of three variables

$(x, y, z) \Leftrightarrow (u, v, w)$	(3.1.7.28)

is defined as

$\frac{D (x, y, z)}{D (u, v, w)} = \| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} \| = \| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial w} & \frac{\partial y}{\partial w} & \frac{\partial z}{\partial w} \end{matrix} \|$	(3.1.7.29)

\frac{D (x, y, z)}{D (u, v, w)} = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} | = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial w} & \frac{\partial y}{\partial w} & \frac{\partial z}{\partial w} \end{matrix} |

(3.1.7.29)

where the right-hand side determinant is obtained with the aid of (3.1.7.27).

In accordance with (3.1.7.15), for three variables one has

$\frac{D (x, y, z)}{D (u, v, w)} = \frac{1}{\frac{D (u, v, w)}{D (x, y, z)}}$	(3.1.7.30)

The relations (3.1.7.29) and (3.1.7.30) can be extended to any number of variables.

The relations between the partial derivatives of one set of variables with respect to the other, like (3.1.7.9), are not so simple for more than two variable. For instance in the case of three variables the relations are of the type:

$\begin{array}{l} \frac{\partial u}{\partial x} \frac{D (x, y, z)}{D (u, v, w)} = \| \begin{matrix} \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} \| = \frac{\partial (y, z)}{\partial (v, w)} \\ \frac{\partial x}{\partial u} \frac{D (u, v, w)}{D (x, y, z)} = \| \begin{matrix} \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{matrix} \| = \frac{\partial (v, w)}{\partial (y, z)} \end{array}}$	(3.1.7.31)

\begin{array}{l} \frac{\partial u}{\partial x} \frac{D (x, y, z)}{D (u, v, w)} = | \begin{matrix} \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} | = \frac{\partial (y, z)}{\partial (v, w)} \\ \frac{\partial x}{\partial u} \frac{D (u, v, w)}{D (x, y, z)} = | \begin{matrix} \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{matrix} | = \frac{\partial (v, w)}{\partial (y, z)} \end{array}}

(3.1.7.31)

The right-hand sides of the equations (3.1.7.31) are similar to Jacobians, but since they contain only partial number of variables, they do not obey the relations corresponding to the full derivatives, namely

$\frac{\partial (y, z)}{\partial (v, w)} \neq \frac{1}{\frac{\partial (v, w)}{\partial (y, z)}}$	(3.1.7.32)

except for some coincidence. For that reason the symbol $\partial$ is used instead of D. The relations (3.1.7.31) can be extended to any number of variables more than three.

As an example of the change of three variables, we'll take the Cartesian and the cylindrical coordinates. The relations among them were given at (3.1.4.17), and are reproduced here for completeness:

$\begin{array}{l} x = r \cos φ \\ y = r \sin φ \\ z = z \\ restrictions: \\ r \geq 0 \\ 0 \leq φ < 2 π \end{array}}$	(3.1.7.33)

The Jacobian is

$\begin{array}{l} \frac{D (x, y, z)}{D (r, φ, z)} = \| \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial φ} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial φ} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial φ} & \frac{\partial z}{\partial z} \end{matrix} \| = \\ = \| \begin{matrix} \cos φ & - r \sin φ & 0 \\ \sin φ & r \cos φ & 0 \\ 0 & 0 & 1 \end{matrix} \| = r (\cos^{2} φ + \sin^{2} φ) = r \end{array}}$	(3.1.7.34)

\begin{array}{l} \frac{D (x, y, z)}{D (r, φ, z)} = | \begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial φ} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial φ} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial φ} & \frac{\partial z}{\partial z} \end{matrix} | = \\ = | \begin{matrix} \cos φ & - r \sin φ & 0 \\ \sin φ & r \cos φ & 0 \\ 0 & 0 & 1 \end{matrix} | = r (\cos^{2} φ + \sin^{2} φ) = r \end{array}}

(3.1.7.34)

One can use (3.1.7.34) for calculating the derivatives of $(r, φ, z)$ with respect to the Cartesian coordinates:

$\begin{array}{l} \frac{\partial r}{\partial x} \frac{D (x, y, z)}{D (r, φ, z)} = \frac{\partial (y, z)}{\partial (φ, z)} = \| \begin{matrix} r \cos φ & 0 \\ 0 & 1 \end{matrix} \| = r \cos φ \Rightarrow \frac{\partial r}{\partial x} = \cos φ \\ \frac{\partial r}{\partial y} \frac{D (y, x, z)}{D (r, φ, z)} = \frac{\partial (x, z)}{\partial (φ, z)} = \| \begin{matrix} - r \sin φ & 0 \\ 0 & 1 \end{matrix} \| = - r \sin φ \Rightarrow \frac{\partial r}{\partial y} = \sin φ \\ \frac{\partial r}{\partial z} \frac{D (z, y, x)}{D (r, φ, z)} = \frac{\partial (y, x)}{\partial (φ, z)} = \| \begin{matrix} r \cos φ & 0 \\ - r \sin φ & 0 \end{matrix} \| = 0 \Rightarrow \frac{\partial r}{\partial z} = 0 \\ \frac{\partial φ}{\partial x} \frac{D (x, y, z)}{D (φ, r, z)} = \frac{\partial (y, z)}{\partial (r, z)} = \| \begin{matrix} \sin φ & 0 \\ 0 & 1 \end{matrix} \| = \sin φ \Rightarrow \frac{\partial φ}{\partial x} = - \frac{\sin φ}{r} \\ \frac{\partial φ}{\partial y} \frac{D (y, x, z)}{D (φ, r, z)} = \frac{\partial (x, z)}{\partial (r, z)} = \| \begin{matrix} \cos φ & 0 \\ 0 & 1 \end{matrix} \| = \cos φ \Rightarrow \frac{\partial φ}{\partial y} = \frac{\cos φ}{r} \\ \frac{\partial φ}{\partial z} \frac{D (z, y, x)}{D (φ, r, z)} = \frac{\partial (y, x)}{\partial (r, z)} = \| \begin{matrix} \sin φ & 0 \\ \cos φ & 0 \end{matrix} \| = 0 \Rightarrow \frac{\partial φ}{\partial z} = 0 \\ \frac{\partial z}{\partial x} = 0 \frac{\partial z}{\partial y} = 0 \frac{\partial z}{\partial z} = 1 \end{array}}$	(3.1.7.35)

\begin{array}{l} \frac{\partial r}{\partial x} \frac{D (x, y, z)}{D (r, φ, z)} = \frac{\partial (y, z)}{\partial (φ, z)} = | \begin{matrix} r \cos φ & 0 \\ 0 & 1 \end{matrix} | = r \cos φ \Rightarrow \frac{\partial r}{\partial x} = \cos φ \\ \frac{\partial r}{\partial y} \frac{D (y, x, z)}{D (r, φ, z)} = \frac{\partial (x, z)}{\partial (φ, z)} = | \begin{matrix} - r \sin φ & 0 \\ 0 & 1 \end{matrix} | = - r \sin φ \Rightarrow \frac{\partial r}{\partial y} = \sin φ \\ \frac{\partial r}{\partial z} \frac{D (z, y, x)}{D (r, φ, z)} = \frac{\partial (y, x)}{\partial (φ, z)} = | \begin{matrix} r \cos φ & 0 \\ - r \sin φ & 0 \end{matrix} | = 0 \Rightarrow \frac{\partial r}{\partial z} = 0 \\ \frac{\partial φ}{\partial x} \frac{D (x, y, z)}{D (φ, r, z)} = \frac{\partial (y, z)}{\partial (r, z)} = | \begin{matrix} \sin φ & 0 \\ 0 & 1 \end{matrix} | = \sin φ \Rightarrow \frac{\partial φ}{\partial x} = - \frac{\sin φ}{r} \\ \frac{\partial φ}{\partial y} \frac{D (y, x, z)}{D (φ, r, z)} = \frac{\partial (x, z)}{\partial (r, z)} = | \begin{matrix} \cos φ & 0 \\ 0 & 1 \end{matrix} | = \cos φ \Rightarrow \frac{\partial φ}{\partial y} = \frac{\cos φ}{r} \\ \frac{\partial φ}{\partial z} \frac{D (z, y, x)}{D (φ, r, z)} = \frac{\partial (y, x)}{\partial (r, z)} = | \begin{matrix} \sin φ & 0 \\ \cos φ & 0 \end{matrix} | = 0 \Rightarrow \frac{\partial φ}{\partial z} = 0 \\ \frac{\partial z}{\partial x} = 0 \frac{\partial z}{\partial y} = 0 \frac{\partial z}{\partial z} = 1 \end{array}}

(3.1.7.35)

Since the cylindrical coordinates are a natural extension of the two dimensional polar coordinates, by adding a common Cartesian coordinate z, one can write for the relation of the polar with the two dimensional Cartesian coordinates:

$\begin{array}{l} {\begin{cases} x = r \cos φ y = r \sin φ \\ restrictions: \\ r \geq 0 0 \leq φ \leq 2 π \end{cases} \\ \frac{D (x, y)}{D (r, φ)} = r \\ {\begin{matrix} \frac{\partial r}{\partial x} = \cos φ & \frac{\partial r}{\partial y} = \sin φ \\ \frac{\partial φ}{\partial x} = - \frac{\sin φ}{r} & \frac{\partial φ}{\partial y} = \frac{\cos φ}{r} \end{matrix} \end{array}}$	(3.1.7.36)

Exercises

Exercise 1. The change of variables $(x, y) \Leftrightarrow (u, v)$ is defined by

{\begin{cases} x = \cos u \sinh v \\ y = \sin u \cosh v \end{cases}}

Obtain the Jacobian in terms of u and v !
Calculate the partial derivatives of u and of v with respect to x and to y !
For a given function $f (u, v)$ , express the derivatives $\frac{\partial f}{\partial x} and \frac{\partial f}{\partial y}$ in terms of u and v !

Exercise 2. Prove that if $\begin{matrix} x = x (u, v) & y = y (u, v) \\ u = u (r, s) & v = v (r, s) \end{matrix}}$ then $\frac{D (x, y)}{D (r, s)} = \frac{D (x, y)}{D (u, v)} \frac{D (u, v)}{D (r, s)}$

Exercise 3. Evaluate $| \begin{matrix} 1 & 1 & - 2 & 0 \\ 0 & 2 & 0 & 2 \\ - 1 & 0 & 3 & 1 \\ 0 & 2 & 1 & 0 \end{matrix} |$

Exercise 4. The spherical coordinates $(R, θ, φ)$ are related to the Cartesian by the relations

{\begin{cases} x = R \sin θ \cos φ \\ y = R \sin θ \sin φ \\ z = R \cos θ \end{cases}}

Calculate the Jacobian in terms of $(R, θ, φ)$ !
Calculate the derivatives of the spherical coordinates, with respect to the Cartesian, in terms of $(R, θ, φ)$ !
For a given function $f (R, θ, φ)$ , what are $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} and \frac{\partial f}{\partial z}$ ?

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Next topic: Section 2 Integration, page 1 Double Integral

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 1: Differentiation; page 7

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Jacobian

Two variables

Features

Determinants

Many variables

Exercises

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Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)