A catenary arch is the ideal structural arch form:

• Shape = the curve formed by a hanging chain or rope (catena = chain in Latin), inverted.
• Under uniform self-weight loading, a catenary arch carries loads purely in compression along its axis — no bending, no tension.
• Structurally optimal — zero bending moment throughout.
• Famous example: Antonio Gaudí used catenary arches in Sagrada Família; Eero Saarinen's Gateway Arch (St. Louis) is a weighted catenary.

Round arches and pointed arches do experience bending under non-uniform loads.

From the Lorentz force: $F = qvB$

$[B] = \dfrac{[F]}{[q][v]} = \dfrac{MLT^{-2}}{C \cdot LT^{-1}} = MT^{-1}C^{-1} = M^1L^0T^{-1}C^{-1}$

To find the ratio $\dfrac{1}{3} : \dfrac{3}{8}$, divide the first by the second:

$\dfrac{1/3}{3/8} = \dfrac{1}{3} \times \dfrac{8}{3} = \dfrac{8}{9}$

So the ratio is $\dfrac{8}{9}$, which means $8$ to $9$.

You can verify: multiply both original fractions by 24 (the LCM of 3 and 8): $\dfrac{1}{3} \times 24 = 8$ and $\dfrac{3}{8} \times 24 = 9$. Ratio is $8:9$.

$A = \begin{pmatrix}i&-i\\-i&i\end{pmatrix}$. Note $\det(A) = i^2-i^2=0$, so $A$ is singular.

$A^2$: compute $(A)^2$. Row1·Col1: $i(i)+(-i)(-i)=-1+(-1)=-2i$... Let's compute: $A^2_{11}=i\cdot i+(-i)(-i)=i^2+i^2=-1-1=-2$. So $A^2 = -2A$... check: $A^2 = -2\begin{pmatrix}i&-i\\-i&i\end{pmatrix}$.

Therefore $A^8 = (-2)^4 A^4 = 16(A^2)^2=16\cdot4A^2=64\cdot(-2A)=-128A$.

Wait: $A^2=-2A \Rightarrow A^4=4A^2=-8A \Rightarrow A^8=-8A^4... $ Correctly: $A^8=(-2)^7A = -128A$.

$-128A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8\\64\end{pmatrix}$. Since $\det(A)=0$, check consistency. The equations are linearly dependent but RHS $(8,64)$ is not in column space of $A$. Hence no solution.

Initially, beam A's plane is parallel to the polaroid ($\theta_A = 0^\circ$) and B's is perpendicular ($\theta_B = 90^\circ$).

After $30^\circ$ rotation, the angles become $30^\circ$ and $60^\circ$ respectively.

Equal brightness means: $I_A \cos^2(30^\circ) = I_B \cos^2(60^\circ)$

$I_A \left( \frac{\sqrt{3}}{2} \right)^2 = I_B \left( \frac{1}{2} \right)^2 \implies I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4} \implies \frac{I_A}{I_B} = \frac{1}{3}$.