# Posts on Mathematics

## The intersection area of two circles

Posted by Diego Assencio on 2017.07.12 under Mathematics (Geometry)

Let $C_1$ and $C_2$ be two circles of radii $r_1$ and $r_2$ respectively whose centers are at a distance $d$ from each other. Assume, without loss of generality, that $r_1 \geq r_2$. What is the intersection area of these two circles?

If $d \geq r_1 + r_2$, the circles intersect at most up to a point (when $d = r_1 + r_2$) and therefore the intersection area is zero. On the other extreme, if $d + r_2 \leq r_1$, circle $C_2$ is entirely contained within $C_1$ and the intersection area is the area of $C_2$ itself: $\pi r_2^2$. The challenging case happens when both $d \lt r_1 + r_2$ and $d + r_2 \gt r_1$ are satisfied, i.e., when the the circles intersect only partially but the intersection area is more than simply a point. Rearranging the second inequality, we obtain $r_1 - r_2 \lt d \lt r_1 + r_2$, so we will assume this to be the case from now on.

To solve this problem, we will make use of a Cartesian coordinate system with origin at the center of circle $C_1$ such that the center of $C_2$ is at $(d,0)$ as shown on figure 1.

 Fig. 1: Two intersecting circles $C_1$ (blue) and $C_2$ (red) of radii $r_1$ and $r_2$ respectively. The distance between the centers of the circles is $d = d_1 + d_2$, where $d_1$ is the $x$ coordinate of the intersection points and $d_2 = d - d_1$. Notice that $d_1 \geq 0$ since these points are always located to the right of the center of $C_1$, but $d_2$ may be negative when $r_2 \lt r_1$ since, in this case, the intersection points will eventually fall to the right of the center of $C_2$ as we move $C_2$ to the left, making $d \lt d_1$ and therefore $d_2 \lt 0$.

The circles $C_1$ and $C_2$ are described by the following equations respectively: $$\begin{eqnarray} x^2 + y^2 &=& r_1^2 \label{post_8d6ca3d82151bad815f78addf9b5c1c6_c1}\\[5pt] (x - d)^2 + y^2 &=& r_2^2 \\[5pt] \end{eqnarray}$$ At the intersection points, we have $x = d_1$. To determine $d_1$, we can replace $x$ with $d_1$ and isolate $y^2$ on both equations above to get: $$r_1^2 - d_1^2 = r_2^2 - (d_1 - d)^2$$ Solving for $d_1$ is a simple task: $$r_1^2 - d_1^2 = r_2^2 - d_1^2 + 2d_1d - d^2 \Longrightarrow d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}$$ From equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}, we can see that $d_1 \geq 0$ since $r_1 \geq r_2$. The intersection area is the sum of the blue and red areas shown on figure 1, which we refer to as $A_1$ and $A_2$ respectively. We then have that: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \label{%INDEX_eq_A1_def} \\[5pt] A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \end{eqnarray}$$ where the factors of $2$ come from the fact that each integral above accounts for only half of the area of the associated region (only points on and above the $x$ axis are taken into account); the results must then be multiplied by two so that the areas below the $x$ axis are taken into account as well.

The computation of these integrals is straightforward. Before we proceed, notice first that: $$\begin{eqnarray} A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \nonumber \\[5pt] &=& 2\int_{- r_2}^{d_1 - d} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d - d_1}^{r_2} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d_2}^{r_2} \sqrt{r_2^2 - x^2}dx \label{%INDEX_eq_A2} \end{eqnarray}$$ where above we used the fact that $d_2 = d - d_1$. This is the same as equation \eqref{%INDEX_eq_A1_def} if we apply the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$. Therefore, by computing $A_1$, we will immediately obtain $A_2$ as well. Let's then compute $A_1$ first: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \nonumber\\[5pt] &=& 2r_1 \int_{d_1}^{r_1} \sqrt{1 - \left(\frac{x}{r_1}\right)^2}dx \nonumber\\[5pt] &=& 2r_1^2 \int_{d_1/r_1}^{1} \sqrt{1 - x^2}dx \label{%INDEX_eq_A1} \end{eqnarray}$$ All we need to do now is to integrate $\sqrt{1 - x^2}$. The process is straightforward if we use integration by parts: $$\begin{eqnarray} \int \sqrt{1 - x^2}dx &=& x \sqrt{1 - x^2} - \int x \left(\frac{-x}{\sqrt{1 - x^2}}\right) dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} + \int \frac{x^2 - 1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} - \int \sqrt{1 - x^2} dx + \sin^{-1}(x) \end{eqnarray}$$ Therefore: $$\int \sqrt{1 - x^2}dx = \frac{1}{2}\left( x \sqrt{1 - x^2} + \sin^{-1}(x) \right) \label{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2}$$ Using equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} on equation \eqref{%INDEX_eq_A1} yields: $$\begin{eqnarray} A_1 &=& r_1^2 \left( \frac{\pi}{2} - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} - \sin^{-1}\left(\frac{d_1}{r_1}\right) \right) \nonumber\\[5pt] &=& r_1^2 \left( \cos^{-1}\left(\frac{d_1}{r_1}\right) - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} \right) \nonumber\\[5pt] &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1 \sqrt{r_1^2 - d_1^2} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final} \end{eqnarray}$$ where above we used the fact that $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$ for any $\alpha$ in $[-1,1]$. This fact is easy to prove: $$\cos\left(\frac{\pi}{2} - \sin^{-1}(\alpha)\right) = \cos\left(\frac{\pi}{2}\right)\cos(\sin^{-1}(\alpha)) + \sin\left(\frac{\pi}{2}\right)\sin(\sin^{-1}(\alpha)) = \alpha$$ and therefore $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$. As discussed above, we can now obtain $A_2$ directly by doing the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$ on the expression for $A_1$ on equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final}: $$A_2 = r_2^2 \cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2 \sqrt{r_2^2 - d_2^2}$$ The sum of $A_1$ and $A_2$ is the intersection area of the circles: $$\boxed{ \begin{eqnarray} A_{\textrm{intersection}} &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1\sqrt{r_1^2 - d_1^2} \nonumber \\[5pt] &+& r_2^2\cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2\sqrt{r_2^2 - d_2^2} \nonumber \end{eqnarray} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection}$$ where: $$\boxed{ d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} } \quad \textrm{ and } \quad \boxed{ d_2 = d - d_1 = \displaystyle\frac{r_2^2 - r_1^2 + d^2}{2d} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1_final}$$

### Summary

Given two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ respectively (with $r_1 \geq r_2$) whose center points are at a distance $d$ from each other, the intersection area of the circles is:

 1 zero, if $d \geq r_1 + r_2$, since in this case the circles intersect at most up to a point. 2 $\pi r_2^2$, if $d \leq r_1 - r_2$, since in this case $C_2$ is entirely contained within $C_1$. 3 given by equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} in all other cases.

## An easy derivation of 3D rotation matrices

Posted by Diego Assencio on 2016.09.23 under Mathematics (Linear Algebra)

In this post, we will derive the components of a rotation matrix in three dimensions. Our derivation favors geometrical arguments over a purely algebraic approach and therefore requires only basic knowledge of Analytic Geometry.

Given a vector ${\bf x} = (x,y,z)$, our goal is to rotate it by an angle $\theta \gt 0$ around a fixed axis represented by a unit vector $\hat{\bf n} = (n_x, n_y, n_z)$; we call ${\bf x}'$ the result of rotating ${\bf x}$ around $\hat{\bf n}$. The rotation is such that if we look into $\hat{\bf n}$, the vector ${\bf x}$ will be rotated along the counter-clockwise direction (see figure 1).

 (a)
 (b)
 Fig. 1: The vector ${\bf x}$ is rotated by an angle $\theta$ around $\hat{\bf n}$. Figure (a) shows the components ${\bf x}_{\parallel}$ and ${\bf x}_{\perp}$ of ${\bf x}$ which are parallel and perpendicular to $\hat{\bf n}$ respectively. Figure (b) shows the rotation as seen from top to bottom, i.e., from the perspective of an observer looking into the head of $\hat{\bf n}$: ${\bf x}_{\parallel}$ remains unchanged after the rotation; it is only ${\bf x}_{\perp}$ which changes. The unit vector $\hat{\bf q}$ is parallel to $\hat{\bf n} \times {\bf x} = \hat{\bf n} \times {\bf x}_{\perp}$. The rotation is in the counterclockwise direction for $\theta \gt 0$.

Even though we already anticipated the fact that the transformation which rotates ${\bf x}$ into ${\bf x}'$ can be represented as a matrix, we will prove this explicitly by showing that ${\bf x}' = R(\hat{\bf n},\theta){\bf x}$ for a $3 \times 3$ matrix $R(\hat{\bf n},\theta)$ whose components depend only on $\hat{\bf n}$ and $\theta$.

As a first step, notice that ${\bf x}$ can be decomposed into two components ${\bf x}_{\parallel}$ and ${\bf x}_{\perp}$ which are parallel and perpendicular to $\hat{\bf n}$ respectively as shown in figure 1a. This means: $${\bf x} = {\bf x}_{\parallel} + {\bf x}_{\perp}$$ Since $\hat{\bf n}$ is a unit vector, then: $$\begin{eqnarray} {\bf x}_{\parallel} &=& (\hat{\bf n}\cdot{\bf x})\hat{\bf n} \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_parallel} \\[5pt] {\bf x}_{\perp} &=& {\bf x} - {\bf x}_{\parallel} = {\bf x} - (\hat{\bf n}\cdot{\bf x})\hat{\bf n} \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_perp} \end{eqnarray}$$ When we rotate ${\bf x}$ around $\hat{\bf n}$, its parallel component ${\bf x}_{\parallel}$ remains unchanged; it is only the perpendicular component ${\bf x}_{\perp}$ that actually rotates around $\hat{\bf n}$. This gives us: $${\bf x}_{\parallel}' = {\bf x}_{\parallel} = (\hat{\bf n}\cdot{\bf x})\hat{\bf n} \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_parallel}$$

Let us now define a unit vector $\hat{\bf q}$ which is orthogonal to both $\hat{\bf n}$ and ${\bf x}$ as shown in figure 1b. We can do this by computing and normalizing the cross product of $\hat{\bf n}$ and ${\bf x}$ (below, we implicitly assume that $\hat{\bf n}$ and ${\bf x}$ are not parallel to each other, but if they are, we have trivially that ${\bf x}' = {\bf x} = {\bf x}_{\parallel}$; our final results will be compatible with this corner case as well): $$\displaystyle\hat{\bf q} = \frac{\hat{\bf n} \times {\bf x}}{\|\hat{\bf n} \times {\bf x}\|} \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_def_q}$$ Since a rotation does not change the length of a vector, we have that $\|{\bf x}'\| = \|{\bf x}\|$; in particular, $\|{\bf x}_{\perp}'\| = \|{\bf x}_{\perp}\|$ as shown in figure 1b. When we rotate ${\bf x}_{\perp}$ by an angle $\theta$ around $\hat{\bf n}$, a component proportional to $\|{\bf x}_{\perp}\|\cos\theta$ remains parallel to ${\bf x}_{\perp}$, and a component proportional to $\|{\bf x}_{\perp}\|\sin\theta$ which is parallel to $\hat{\bf q}$ is generated. Therefore: $${\bf x}_{\perp}' = \cos\theta\,{\bf x}_{\perp} + \|{\bf x}_{\perp}\|\sin\theta\,\hat{\bf q} = \cos\theta\,{\bf x}_{\perp} + \sin\theta\,(\hat{\bf n}\times{\bf x}) \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_perp}$$ where above we used the definition of $\hat{\bf q}$ from equation \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_def_q} as well as the fact that: $$\|\hat{\bf n} \times {\bf x}\| = \|\hat{\bf n}\|\|{\bf x}\|\sin\alpha = \|{\bf x}_{\perp}\|$$ with $\alpha$ being the angle between $\hat{\bf n}$ and ${\bf x}$ as shown in figure 1a.

We can now obtain an expression relating ${\bf x}'$ and ${\bf x}$ in terms of $\hat{\bf n}$ and $\theta$. Since ${\bf x}' = {\bf x}_{\parallel}' + {\bf x}_{\perp}'$, and using equations \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_parallel} and \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_perp}, we obtain: $${\bf x}' = (\hat{\bf n}\cdot{\bf x})\hat{\bf n} + \cos\theta\,{\bf x}_{\perp} + \sin\theta\,(\hat{\bf n}\times{\bf x})$$ and now using equation \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_perp}, we get: $${\bf x}' = (\hat{\bf n}\cdot{\bf x})\hat{\bf n} + \cos\theta\,({\bf x} - (\hat{\bf n}\cdot{\bf x})\hat{\bf n}) + \sin\theta\,(\hat{\bf n}\times{\bf x})$$ Rearranging terms, we obtain a very useful expression for computing ${\bf x}'$: $$\boxed{ {\bf x}' = \cos\theta\,{\bf x} + (1 - \cos\theta)(\hat{\bf n}\cdot{\bf x})\hat{\bf n} + \sin\theta\,(\hat{\bf n}\times{\bf x}) } \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_x}$$

As we claimed earlier, equation \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_x} can be expressed as ${\bf x}' = R(\hat{\bf n},\theta){\bf x}$, where $R(\hat{\bf n},\theta)$ is a $3 \times 3$ matrix. To see that this is true, notice that: $$\cos\theta\,{\bf x} = \cos\theta \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right)$$ and that: $$(\hat{\bf n}\cdot{\bf x})\hat{\bf n} = \left( \begin{matrix} (\hat{\bf n}\cdot{\bf x})n_x \\ (\hat{\bf n}\cdot{\bf x})n_y \\ (\hat{\bf n}\cdot{\bf x})n_z \end{matrix} \right) = \left( \begin{matrix} n_x^2 & n_y n_x & n_z n_x \\ n_x n_y & n_y^2 & n_z n_y \\ n_x n_z & n_y n_z & n_z^2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_mat1}$$ and that: $$\hat{\bf n}\times{\bf x} = \left( \begin{matrix} n_y z - n_z y \\ n_z x - n_x z \\ n_x y - n_y x \end{matrix} \right) = \left( \begin{matrix} 0 & -n_z & n_y \\ n_z & 0 & -n_x \\ -n_y & n_x & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) \label{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_mat2}$$ Therefore ${\bf x'} = R(\hat{\bf n},\theta){\bf x}$, with: $$\boxed{ \begin{eqnarray} R(\hat{\bf n},\theta) &=& \cos\theta \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) + (1 - \cos\theta) \left( \begin{matrix} n_x^2 & n_y n_x & n_z n_x \\ n_x n_y & n_y^2 & n_z n_y \\ n_x n_z & n_y n_z & n_z^2 \end{matrix} \right) \nonumber \\[5pt] &+& \sin\theta \left( \begin{matrix} 0 & -n_z & n_y \\ n_z & 0 & -n_x \\ -n_y & n_x & 0 \end{matrix} \right) \nonumber \end{eqnarray} }$$

Whenever $\hat{\bf n}$ and ${\bf x}$ are parallel, we have ${\bf x} = {\bf x}_{\parallel}$ and $\hat{\bf n} \times {\bf x} = {\bf 0}$, so equation \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_prime_x} together with equation \eqref{post_b155574a293a5cbfdd0fbe82a9b8bf28_eq_x_parallel} gives us ${\bf x}' = {\bf x}$, as expected. Additionally, our derivation did not actually rely on the assumption that $\theta \gt 0$, so it is valid for arbitrary values of $\theta$. Finally, notice that by changing $\hat{\bf n} \rightarrow -\hat{\bf n}$ and $\theta \rightarrow -\theta$, $R(\hat{\bf n},\theta)$ does not change, i.e., $R(\hat{\bf n},\theta) = R(-\hat{\bf n},-\theta)$, so we can always convert a rotation with $\theta \lt 0$ into an equivalent one having $\theta \gt 0$ by inverting the direction of $\hat{\bf n}$ and negating $\theta$.

## Surface normals and linear transformations

Posted by Diego Assencio on 2016.01.15 under Mathematics (Linear Algebra)

Suppose we have a surface $S$ and that ${\bf n}$ is a unit vector normal to $S$ at a point ${\bf x} \in S$. There are many common transformations we can perform on $S$: rotation, scaling, shear etc. After we transform $S$ into a new surface $S'$, the point ${\bf x}$ from $S$ will end up at a new position ${\bf x}'$ in $S'$. This post answers the following question: when we apply an invertible linear transformation to all points of $S$, what is the relation between ${\bf n}$ and the corresponding ${\bf n}'$ which is a unit vector normal to $S'$ at ${\bf x}'$?

We will illustrate our points in this post using a one-dimensional surface on a two-dimensional plane, but everything here applies to any surface on an $n$-dimensional space which undergoes an invertible linear transformation $M$. A natural but incorrect assumption would be that ${\bf n}$ is converted to ${\bf n}' = M{\bf n}$ since every point ${\bf x}$ in $S$ is converted to ${\bf x}' = M{\bf x}$ in $S'$. In fact, as we will show, ${\bf n}'$ is actually parallel to $(M^{-1})^T{\bf n}$, which is in general not parallel to $M{\bf n}$.

Consider first what happens when we apply a shear transformation to a surface $S$ in a two-dimensional space, with the transformation matrix $M$ having the following form (here we assume $a \gt 0$): $$M = \left(\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}\right) \label{post_a2386030d8d82e457a8c7dc124d00564_shear_matrix}$$ A point ${\bf x}$ in $S$ is transformed into ${\bf x}' = M{\bf x}$: $${\bf x} \;\longrightarrow\; {\bf x}' = M{\bf x} = \left(\begin{matrix} x + ay \\ y \end{matrix}\right)$$ The transformation $M$ preserves the $y$ coordinates of all points in $S$, but moves the $x$ coordinates by $ay$. Figure 1 shows an example of how a unit square is transformed by $M$.

 Fig. 1: A shear transformation $M$ given in equation \eqref{post_a2386030d8d82e457a8c7dc124d00564_shear_matrix} is applied to all points of a unit square $S$, yielding a new surface $S'$. Surface normals are shown as red arrows. Notice how the direction of a surface normal ${\bf n}_{\textrm{t}}$ on the top edge stays the same, but the direction of a surface normal ${\bf n}_{\textrm{r}}$ on the right edge changes under $M$.

A normal vector ${\bf n}_{\textrm{t}}$ on the top edge of $S$ has coordinates $(0,1)$, and $M{\bf n}_{\textrm{t}} = (a,1)$, which is not parallel to ${\bf n}_{\textrm{t}}'$ since ${\bf n}_{\textrm{t}}' = {\bf n}_{\textrm{t}}$ and $a \neq 0$. Also, ${\bf n}_{\textrm{r}}$ on the right edge has coordinates $(1,0)$, $M{\bf n}_{\textrm{r}} = (1,0) = {\bf n}_{\textrm{r}}$, which is not parallel to ${\bf n}_{\textrm{r}}'$ because ${\bf n}_{\textrm{r}}' \neq {\bf n}_{\textrm{r}}$. In other words, surface normals do not in general transform like points from $S$, so we cannot just take a normal vector ${\bf n}$ and expect that $M{\bf n}$ has the same direction as the corresponding normal ${\bf n}'$ in $S'$.

Fortunately, finding an expression for ${\bf n}'$ in terms of ${\bf n}$ and $M$ is easy. All we need to do is use the fact that tangent vectors in $S$ are transformed into tangent vectors in $S'$. In fact, consider a tangent vector ${\bf t}$ connecting ${\bf x}$ and an infinitesimally close point $\tilde{\bf x} = {\bf x} + \Delta{\bf x}$ from $S$, i.e., ${\bf t} = \tilde{\bf x} - {\bf x} = \Delta{\bf x}$ (${\bf t}$ is not a unit vector, but we only care about its direction here). Since ${\bf x}$ and $\tilde{\bf x}$ are converted into ${\bf x}' = M{\bf x}$ and $\tilde{\bf x}' = M({\bf x} + \Delta{\bf x}) = {\bf x}' + M\Delta{\bf x}$ respectively, the infinitesimal tangent vector ${\bf t}'$ connecting ${\bf x}'$ and $\tilde{\bf x}'$ is then: $${\bf t}' = \tilde{\bf x}' - {\bf x}' = M{\Delta{\bf x}} = M{\bf t}$$ In other words, an infinitesimal vector ${\bf t}$ tangent to ${\bf x} \in S$ is transformed into an infinitesimal vector ${\bf t}' = M{\bf t}$ which is tangent to ${\bf x}' \in S'$. As a matter of fact, since $S$ undergoes a linear transformation, the argument applies even if ${\bf t}$ is not an infinitesimal vector, in which case ${\bf t}' = M{\bf t}$ is also not infinitesimal. Representing vectors as columns, we have that: $${\bf t}\cdot{\bf n} = {\bf t}^T{\bf n} = 0 \label{post_a2386030d8d82e457a8c7dc124d00564_dotp1}$$ because ${\bf t}$ and ${\bf n}$ are orthogonal. Equivalently, we also have that: $${\bf t}' \cdot {\bf n}' = (M{\bf t})^T {\bf n}' = 0 \label{post_a2386030d8d82e457a8c7dc124d00564_dotp2}$$ Equation \eqref{post_a2386030d8d82e457a8c7dc124d00564_dotp2} is satisfied if we take ${\bf n}' = (M^T)^{-1}{\bf n}$. Indeed: $$(M{\bf t})^T {\bf n}' = (M{\bf t})^T (M^T)^{-1}{\bf n} = {\bf t}^T M^T (M^T)^{-1}{\bf n} = {\bf t}^T{\bf n} = 0$$ Therefore, we have that $(M^T)^{-1}{\bf n}$ is normal to the surface $S'$ at ${\bf x}'$. All we need to do now is normalize it: $$\boxed{ {\bf n}' = (M^T)^{-1}{\bf n} \; \big/ \; \big\| (M^T)^{-1}{\bf n} \big\| }$$ It is important to say, however, that if $M$ is an orthogonal matrix (e.g. a rotation), then $(M^T)^{-1} = M$ and $\|M{\bf n}\| = \|{\bf n}\| = 1$, so in this case surface normal vectors transform just like regular points from $S$.