Jacobians (2024)

Example 1: Compute the Jacobian ofthe polar coordinates transformation $$x \ = \ r \cos \theta,\, \qquad y = r \sin \theta\,.$$Solution: Since \begin{eqnarray*} & \frac{\partial x}{\partial r} = \cos(\theta), \quad &\frac{\partial y}{\partial r} = \sin(\theta), \\& \frac{\partial x}{\partial \theta} = -r \sin(\theta), \quad &\frac{\partial y}{\partial \theta} = r \cos(\theta),\end{eqnarray*}

our Jacobian is$$ \left|\begin{matrix} \displaystyle { \frac{\partial x}{\partial r} }& \displaystyle{\frac{\partial x}{\partial \theta} }\cr\\\displaystyle { \frac{\partial y}{\partial r} }& \displaystyle { \frac{\partial y}{\partial \theta}} \end{matrix}\right|\ = \ \left|\begin{matrix} \cos \theta & -r\sin \theta \cr\\ \sin \theta & r \cos \theta \end{matrix}\right|\ = \ r\,.$$This explains why there's an $r$ factor in polar integrals! The areaelement $dA=dx\,dy$ is notequal to $dr\,d\theta$. Instead, $dA$ is equal to $r\, dr\,d\theta$.

Let's see why the Jacobian is the distortion factor in general for a mapping$${\bf \Phi} : (u,\, v) \ \to \ (x(u,\,v),\, y(u,\,v))\ = \ x(u,\,v)\, {\bf i} +y(u,\,v)\, {\bf j}\,, $$making good use of all the vector calculus we've developed so far. Let $Q = [a,\,a+h]\times [c,\,c+k]$ be a rectangle in the $uv$-plane and ${\bf \Phi}(Q)$ its image in the $xy$-plane as shown in

Then $${\bf u} \ = \ {\bf \Phi}(a+h,\,c) - {\bf \Phi}(a,\,c)\,, \qquad{\bf v} \ = \ {\bf \Phi}(a,\,c+k) - {\bf \Phi}(a,\,c)\,.$$The area of the parallelogram spanned by ${\bf u} = u_1 {\bf i} + u_2 {\bf j}$ and ${\bf v} = v_1 {\bf i} + v_2 {\bf j}$ isthe determinant $\left | \begin{matrix} u_1 & v_1 \cr u_2 & v_2 \end{matrix}\right |$.

By the definition of partial derivatives, $$\frac{{\bf \Phi}(a+h,\,c) - {\bf \Phi}(a,\,c)}{h} \ \approx \\frac{\partial {\bf \Phi}}{\partial u}\Big|_{(a,c)}\ = \\frac{\partial x}{\partial u}\Big|_{(a,c)}\, {\bf i} + \frac{\partialy}{\partial u}\Big|_{(a,c)}\, {\bf j} \,,$$ $$\frac{{\bf \Phi}(a,\,c+k) - {\bf \Phi}(a,\,c)}{k} \ \approx \\frac{\partial {\bf \Phi}}{\partial v}\Big|_{(a,c)}\ = \\frac{\partial x}{\partial v}\Big|_{(a,c)}\, {\bf i} + \frac{\partialy}{\partial v}\Big|_{(a,c)}\, {\bf j}\,.$$We then compute $$\hbox{area}(\Phi(Q)) \approx \left | \begin{matrix} u_1 & v_1 \cr u_2 & v_2 \end{matrix}\right | \ \approx \ \left | \begin{matrix}h \frac{\partial x}{\partial u} & k \frac{\partial x}{\partial v} \crh \frac{\partial y}{\partial u} & k \frac{\partial y}{\partial v} \end{matrix} \right | \ = \ hk \left | \begin{matrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \cr\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{matrix}\right |. $$

Since $\hbox{area}(Q) \, =\, hk$, this means that the area of our regionin the $xy$ plane is given by the absolute value of the Jacobian times thearea in the $uv$ plane. Our shorthand for this is $$dA \ = \ dx\, dy \ = \ \left | \frac{\partial(x,y)}{\partial(u,v)} \right |du \, dv.$$

Areas are alwayspositive, so the area of a small parallelogram in $xy$-space is always the absolute value of the Jacobian times the area of the correspondingrectangle in $uv$-space.

So why didn't we see an absolute value in the change-of-variables formulain one dimension? This had to do with the way we write the limits of integration.

Example 2: Compute $\int_0^{10} e^{-x/5} dx$.

Solution: We did this before using $x=g(u)=5u$. Instead, let's take$x=-5u$, so $g'(u)=-5$ is negative. Now $e^{-x/5}=e^u$ and $dx= -5 du$. The map $g$ sends the interval from 0 to -2 in $u$-space to the intervalfrom 0 to 10 in $x$-space,and our change-of-variable formula says$$\int_0^{10} e^{-x/5} dx \ = \ \int_0^{-2} -5 e^{u} du.$$Of course, we usually integrate from $-2$ to $0$, not from $0$ to $-2$.

Flipping the limits of integration changes the sign of the answer, so$$\int_0^{10} e^{-x/5} dx \ = \ \int_{-2}^{0} +5 e^{u} du = 5(1-e^{-2}).$$

If we had written our 1-dimensional integrals in terms of regions insteadin terms of starting points and end points, we would have had a factor of $+5$, rather than $-5$, all along. The mapping $x=-5u$ sends the region $D^*=[-2,0]$ to the region $D=[0,10]$, and $$\int_D e^{-x/5} dx \ = \ \int_{D^*} e^u \left | \frac{dx}{du} \right| du.$$

<< Prev Next >>

FAQs

What does the Jacobian tell us? ›

The Jacobian of a vector-valued function in several variables generalizes the gradient of a scalar-valued function in several variables, which in turn generalizes the derivative of a scalar-valued function of a single variable.

Discover More ›

How to solve Jacobian? ›

Given an exact approximation x⁽^k⁾ = (x₁⁽^k⁾, x₂⁽^k⁾, x₃⁽^k⁾, …, x_n⁽^k⁾) for x, the procedure of Jacobian's method helps to use the first equation and the present values of x₂⁽^k⁾, x₃⁽^k⁾, …, x_n⁽^k⁾ to calculate a new value x₁⁽^k⁺¹⁾.

View Details ›

What is an example of a Jacobian question? ›

Example of Jacobian

Problem: Suppose x (u, v) = u² - v² and y (u, v) = 2 uv. Can you determine the Jacobian J (u, v)? Therefore, J (u, v) equals 4u² + 4v².

Read On ›

Does the Jacobian always have to be positive? ›

The Jacobian ∂(x,y)∂(u,v) may be positive or negative.

Explore More ›

What is the intuition behind the Jacobian? ›

In short, the intuition is that the Jacobian matrix will give you an approximate change in output vector if you give it a small change in input vector, at a particular point in the domain of the function. The approximation is based on the linear approximation using the classical derivative of single-variable calculus.

Explore More ›

What does it mean if the Jacobian is zero? ›

If the determinant of the Jacobian is zero, that means that there is a way to pick n linearly independent vectors in the input space and they will be transformed to linearly dependent vectors in the output space.

Show Me More ›

What is the practical application of Jacobian? ›

One of the many applications for the Jacobian matrix is to transfer mapping from one coordinate system to another, such as the transformation from a Cartesian to natural coordinate system, spherical to Cartesian coordinate system, polar to Cartesian coordinate system, and vice versa for each.

Learn More Now ›

What is the Jacobian conceptually? ›

Jacobian is the determinant of the jacobian matrix. The matrix will contain all partial derivatives of a vector function. The main use of Jacobian is found in the transformation of coordinates. It deals with the concept of differentiation with coordinate transformation.

Discover More ›

Why is it called Jacobian? ›

In mathematics, a Jacobian, named for Carl Gustav Jacob Jacobi, may refer to: Jacobian matrix and determinant (and in particular, the robot Jacobian) Jacobian elliptic functions.

Is the Jacobian a tensor? ›

In terms of shapes, the Jacobian multiplication dL/dy*dy/dx = gradient*J reduces itself to a tensor of the same shape as x .

Explore More ›

What is the ideal value of Jacobian? ›

The Jacobian value ranges from 0.0 to 1.0, where 1.0 represents a perfectly shaped element.

Learn More Now ›

What is Jacobian rule? ›

If the Jacobian matrix is a square matrix, then the number of rows and columns is same, thus it can be written as m = n, then f is a function from ℝn to itself. From the Jacobian matrix, we can form a determinant, known as the Jacobian determinant. The Jacobian determinant is sometimes called "Jacobian".

Keep Reading ›

What is the physical significance of the Jacobian? ›

Physical meaning of the Jacobian

Thus, here, the jacobian represents the transformation of one volume unit from one coordinate space to another. For instance, in 2D carthesian coordinate, a volume (surface) of 1 correspond to a volume of r in polar coordinates.

Show Me More ›

Why do we use the Jacobi method? ›

In numerical linear algebra, the Jacobi method (a.k.a. the Jacobi iteration method) is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. Each diagonal element is solved for, and an approximate value is plugged in.

Get More Info Here ›

What is the significance of Jacobian in coordinate transformation? ›

This quantity dx dt is the Jacobian of this coordinate transformation. It tells us how to change the infinitesimal. So in simple words, when we make a change of a coordinate system, the Jacobian tells us how much the infinitesimal should change.

Tell Me More ›

What does the determinant of the Jacobian represent? ›

The Jacobian determinant is a scalar value that represents the scaling factor or the amount of stretching, shrinking, or shearing that occurs when a transformation is applied to a set of variables. It provides information about how changes in the input variables collectively affect the output variables.

Tell Me More ›

Jacobians (2024)

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Modeling with Differential Equations

Euler's Method and Direction Fields

Separable Equations

Models of Growth

Linear Equations

Parametrized Curves

Calculus with Parametrized Curves

Polar Coordinates

Areas and Lengths of Polar Curves

Conic sections

Conic Sections in Polar Coordinates

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Functions of 2 and 3 variables

Partial Derivatives

Differentiability and the Chain Rule

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Double integrals in polar coordinates

Multiple integrals in physics

Integrals in Probability and Statistics

Change of Variables

FAQs

What does the Jacobian tell us? ›

What is the ideal value of Jacobian? ›

References