# Graphical explanation of the Jacobian

The following is an illustration of the Jacobian in my text, along with part of the explanation (also see similar explanation here):

Since the side-lengths are small, by the Mean Value Theorem, we have $$x(u + ∆u, v) − x(u, v) ≈ {∂x\over {∂u}}(α_1,v)∆u,\\\ \text{where} \\u ≤ α_1 ≤ u + ∆u$$

I am trying to understand this, specifically four things (I am not sure if they are related):

1. What does the Mean Value Theorem have to do with this, and what does it have to do with the length being small?

2. What is the purpose of this new variable, $$a_1$$, that is defined as being between $$u$$ and $$u + ∆u$$?

3. What is the purpose of the partial derivative?

4. What is purpose of the $$∆u$$ that is multiplied by the partial derivative?

• The explanation seems a bit strange to me. If we're going to invoke the mean value theorem, we can have exact rather than approximate equality, even if the side lengths are not small. We do need side lengths to be small in order to say that $x(u+\Delta u, v) - x(u,v) \approx \frac{\partial x}{\partial u}(u,v) \Delta u$. Commented Oct 9, 2018 at 4:39

Remember we can think of matrices as representing linear maps from one space to another. The Jacobian is a matrix which represents the map from $$U \times V \to X \times Y$$ and approximates this distortion linearly. This means that $$(x,y)=(x(u,v), y(u,v))$$ might not be linear. It might twist and warp points in strange ways, but we can approximate that with the Jacobian. Explicitly, we can write the Jacobian as

$$J = \begin{bmatrix} \dfrac{\partial x(u,v)}{\partial u} & \dfrac{\partial x(u,v)}{\partial v}\\ \dfrac{\partial y(u,v)}{\partial u} & \dfrac{\partial y(u,v)}{\partial v} \end{bmatrix}$$

When we say approximation we mean we can find approximately where the new point maps to by $$\begin{bmatrix} x(u + \Delta u, v +\Delta v)\\ y(u+\Delta u, v+\Delta v) \end{bmatrix} \approx \begin{bmatrix} x(u,v)\\ y(u,v) \end{bmatrix} + \begin{bmatrix} \dfrac{\partial x(u,v)}{\partial u} & \dfrac{\partial x(u,v)}{\partial v}\\ \dfrac{\partial y(u,v)}{\partial u} & \dfrac{\partial y(u,v)}{\partial v} \end{bmatrix} \begin{bmatrix} \Delta u\\ \Delta v \end{bmatrix}$$

We can see how this works in one dimension (holding $$v$$ constant) by looking at the definition of (partial) derivative $$\dfrac{\partial x(u,v)}{\partial u} = \lim_{h \to 0} \dfrac{x(u+h, v) - x(u,v)}{h}$$

When $$\Delta u$$ is sufficiently small then we can approximate as $$\dfrac{\partial x(u,v)}{\partial u} \Delta u \approx x(u+\Delta u, v) - x(u,v)$$

Earlier we mentioned twisting and warping. One thing we might be interested in is how the length of a line segment changes. Say I know the length of the bottom of the rectangle, which is given by $$(u+\Delta u) - u$$. I want to know how long $$(x(u,v), x(u + \Delta u, v))$$ is.

By the mean value theorem, there is some point $$\alpha_1 \in (u, u + \Delta u)$$ such that

$$\dfrac{\partial x(u,v)}{\partial u} (\alpha_1, v) = \dfrac{x(u + \Delta u, v) - x(u,v)}{\Delta u}$$

Thus the length given by $$x(u + \Delta u, v) - x(u,v)$$ can be found with

$$\dfrac{\partial x(u,v)}{\partial u} (\alpha_1, v) \Delta u=x(u + \Delta u, v) - x(u,v)$$

I agree I'm a bit confused by the approximation if we're invoking mean value theorem, which will give exactness.

• Excellent explanation! I must say that until now have found it difficult to find a clear explanation (for a beginner) of the Jacobian, even on SE, but this graphic combined with your answer does the trick. –
– dlp
Commented Oct 9, 2018 at 12:46
• @agblt It may also be worth noting that the Jacobian is the best linear approximator. There are (infinitely) many linear approximators out there (albeit many bad ones), but the Jacobian does this best in some sense. You can see more about that here. Commented Oct 9, 2018 at 13:37