Exercises — Arc length formula — derivation

Move 1 — differentiate. $\frac{dy}{dx}=3x^2$ (power rule). Move 2 — build the integrand. ${\left(\frac{dy}{dx}\right)}^2=9x^4$, so $1+(y^{\prime }{)}^2=1+9x^4$. Move 3 — assemble the integral between the given limits: $L={\int }_1^4\sqrt{1+9x^4}dx.$ That is a correct setup. (This one has no nice closed form — recognition problems only test that you can build the integral, not evaluate it.)

${\left(\frac{1}{2x}\right)}^2=\frac{1}{4x^2}$, so $1+(y^{\prime }{)}^2=1+\frac{1}{4x^2}=\frac{4x^2+1}{4x^2}.$ Why write it over a common denominator? Because the next step is always $\sqrt{\cdot }$, and roots of fractions split as $\frac{\sqrt{4x^2+1}}{2x}$ — cleaner to spot whether the top is a perfect square.

Move 1. $\frac{dy}{dx}=\frac{2}{3}\cdot \frac{3}{2}(x-1{)}^{1/2}=(x-1{)}^{1/2}$. Why so clean? The constants $\frac{2}{3}$ were chosen precisely so the $\frac{3}{2}$ from the power rule cancels. Move 2. $(y^{\prime }{)}^2=x-1$, so $1+(y^{\prime }{)}^2=1+(x-1)=x$. A perfect... well, just $x$. Even nicer. Move 3. $L={\int }_1^5\sqrt{x}dx={\left[\frac{2}{3}{x}^{3/2}\right]}_1^5=\frac{2}{3}\left(5^{3/2}-1\right)=\frac{2}{3}\left(5\sqrt{5}-1\right).$ Numerically $\approx 6.79$. Sanity check: endpoints are $(1,0)$ and $(5,\frac{2}{3}\cdot 8)=(5,\frac{16}{3})\approx (5,5.33)$; chord $=\sqrt{16+28.4}\approx 6.66<6.79$. The curve is longer than its chord ✓. See the figure below.

Why the $x=g(y)$ formula? Because the curve is handed to us as $x$ depending on $y$, so the natural leg to pull out of the root is $dy$. Move 1. $\frac{dx}{dy}=\frac{1}{3}\cdot \frac{3}{2}{y}^{1/2}=\frac{1}{2}{y}^{1/2}$. Move 2. ${\left(\frac{dx}{dy}\right)}^2=\frac{y}{4}$, so $1+\frac{y}{4}=\frac{4+y}{4}$, and $\sqrt{\cdot }=\frac{\sqrt{4+y}}{2}$. Move 3. $L={\int }_0^3\frac{\sqrt{4+y}}{2}dy=\frac{1}{2}\cdot \frac{2}{3}(4+y{)}^{3/2}{|}_0^3=\frac{1}{3}\left(7^{3/2}-4^{3/2}\right)=\frac{1}{3}\left(7\sqrt{7}-8\right).$ Numerically $\approx 3.51$.

Move 1. $\frac{dy}{dx}=\frac{x^2}{2}-\frac{1}{2x^2}$. Why term-by-term? Sum rule; $\frac{1}{2x}=\frac{1}{2}{x}^{-1}$ differentiates to $-\frac{1}{2}{x}^{-2}$. Move 2 — the analysis step. Write the slope as $P-Q$ with $P=\frac{x^2}{2}$ and $Q=\frac{1}{2x^2}$. Square it, watching the cross term: $(P-Q{)}^2=P^2-2PQ+Q^2=\frac{x^4}{4}-2\cdot \frac{x^2}{2}\cdot \frac{1}{2x^2}+\frac{1}{4x^4}=\frac{x^4}{4}-\frac{1}{2}+\frac{1}{4x^4}.$ The cross term is $-2PQ=-\frac{1}{2}$. Now add the mandatory $1$ from the formula: $1+(y^{\prime }{)}^2=\frac{x^4}{4}+\left(-\frac{1}{2}+1\right)+\frac{1}{4x^4}=\frac{x^4}{4}+\frac{1}{2}+\frac{1}{4x^4}.$ Why is this now a perfect square? The perfect square $(P+Q{)}^2=P^2+2PQ+Q^2$ needs a middle term of $+2PQ=+\frac{1}{2}$. We started with $-\frac{1}{2}$, and adding $1$ lands us exactly on $+\frac{1}{2}$ — because $-\frac{1}{2}+1=+\frac{1}{2}$. So the "$+1$" is precisely the amount that flips the middle term from $-2PQ$ to $+2PQ$. Hence $1+(y^{\prime }{)}^2={\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)}^2=(P+Q{)}^2.$ This only works because these curves are engineered so that $2PQ=\frac{1}{2}$ exactly; then the formula's $+1$ does the sign flip for free. Move 3 — take the root (positive on $[1,2]$): $L={\int }_1^2\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx={\left[\frac{x^3}{6}-\frac{1}{2x}\right]}_1^2=\left(\frac{8}{6}-\frac{1}{4}\right)-\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{13}{12}+\frac{1}{3}=\frac{17}{12}.$ Numerically $\approx 1.417$.

Move 1. $\frac{dy}{dx}=\frac{-\sin x}{\cos x}=-\tan x$. Why? Chain rule: derivative of $\ln u$ is $u^{\prime }/u$ with $u=\cos x$. Move 2. $(y^{\prime }{)}^2={\tan }^{2}x$, and the Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ gives $1+(y^{\prime }{)}^2={\sec }^{2}x$. Move 3 — mind the absolute value. $\sqrt{{\sec }^{2}x}=|\sec x|$. On $[0,\frac{\pi }{4}]$, $\cos x>0$ so $\sec x>0$ and $|\sec x|=\sec x$. Thus $L={\int }_0^{\pi /4}\sec xdx=[\ln |\sec x+\tan x|{]}_0^{\pi /4}=\ln (\sqrt{2}+1)-\ln (1)=\ln (1+\sqrt{2}).$ Numerically $\approx 0.881$.

Why $ds$ appears here. A surface of revolution is swept by a tiny arc piece of length $ds$ travelling in a circle of radius $y$; each band has area $2\pi yds$. So we reuse the same $ds=\sqrt{1+(y^{\prime }{)}^2}dx$ built in the parent derivation. Move 1. $y=x^{1/2}\Rightarrow \frac{dy}{dx}=\frac{1}{2}{x}^{-1/2}=\frac{1}{2\sqrt{x}}$. Move 2. $1+(y^{\prime }{)}^2=1+\frac{1}{4x}=\frac{4x+1}{4x}$, so $ds=\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$. Move 3 — assemble the surface integral with radius $y=\sqrt{x}$: $S={\int }_0^{1}2\pi yds={\int }_0^{1}2\pi \sqrt{x}\cdot \frac{\sqrt{4x+1}}{2\sqrt{x}}dx=\pi {\int }_0^1\sqrt{4x+1}dx.$ The $\sqrt{x}$ from radius and the $\sqrt{x}$ in the denominator cancel — a single clean square root remains. (Evaluating: $\pi \cdot \frac{1}{4}\cdot \frac{2}{3}(4x+1{)}^{3/2}{|}_0^1=\frac{\pi }{6}(5\sqrt{5}-1)\approx 5.33$.)

Why a new-looking formula, same idea. Over a tiny $dt$, the point moves $dx=\frac{dx}{dt}dt$ across and $dy=\frac{dy}{dt}dt$ up. Pythagoras on that infinitesimal triangle: $ds=\sqrt{d{x}^2+d{y}^2}=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$. Identical to the parent, just parameterised by $t$. Move 1. $\frac{dx}{dt}=2t,\frac{dy}{dt}=3t^2$. Move 2. ${\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=4t^2+9t^4=t^2(4+9t^2)$, so $ds=t\sqrt{4+9t^2}dt$ (with $t\ge 0$ on $[0,1]$, so $\sqrt{t^2}=t$, not $|t|$ trouble). Move 3. Substitute $u=4+9t^2$, $du=18tdt$: $L={\int }_0^{1}t\sqrt{4+9t^2}dt=\frac{1}{18}{\int }_4^{13}\sqrt{u}du=\frac{1}{18}\cdot \frac{2}{3}{u}^{3/2}{|}_4^{13}=\frac{1}{27}\left(13\sqrt{13}-8\right).$ Numerically $\approx 1.44$.

Why that polar $ds$. In polar coordinates a tiny step has a radial part $dr$ and a sideways arc part $rd\theta$ (arc = radius × angle). These two are perpendicular, so Pythagoras gives $ds=\sqrt{(dr{)}^2+(rd\theta {)}^2}=\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}d\theta$ — same Pythagorean skeleton. See the figure below. Move 1. $\frac{dr}{d\theta }=-\sin \theta$. Move 2. $r^2+{\left(\frac{dr}{d\theta }\right)}^2=(1+\cos \theta {)}^2+{\sin }^2\theta =1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta =2+2\cos \theta .$ Use the half-angle identity $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$, so $2+2\cos \theta =4{\cos }^2\frac{\theta }{2}$ and $\sqrt{2+2\cos \theta }=2\left|\cos \frac{\theta }{2}\right|.$ Move 3 — check the sign. On $\theta \in [0,\pi ]$, $\frac{\theta }{2}\in [0,\frac{\pi }{2}]$, where $\cos \frac{\theta }{2}\ge 0$, so the bars drop safely: $L={\int }_0^{\pi }2\cos \frac{\theta }{2}d\theta =2\cdot \frac{\sin (\theta /2)}{1/2}{|}_0^{\pi }=4\sin \frac{\theta }{2}{|}_0^{\pi }=4(\sin \frac{\pi }{2}-0)=4.$ The full cardioid ($0$ to $2\pi$) would be twice this, $8$, by symmetry.

Move 1 — differentiate. With $y=\frac{1}{2(a+1)}{x}^{a+1}+\frac{1}{2(a-1)}{x}^{-(a-1)}$: $\frac{dy}{dx}=\frac{a+1}{2(a+1)}{x}^a+\frac{-(a-1)}{2(a-1)}{x}^{-a}=\frac{1}{2}{x}^a-\frac{1}{2}{x}^{-a}.$ Why the constants matter: the denominators $2(a\pm 1)$ are chosen so the exponent factors $(a\pm 1)$ cancel, leaving the clean $\frac{1}{2}{x}^{\pm a}$. Move 2 — square, watch the cross term. With $P=\frac{1}{2}{x}^a$, $Q=\frac{1}{2}{x}^{-a}$: $(y^{\prime }{)}^2=(P-Q{)}^2=P^2-2PQ+Q^2,2PQ=2\cdot \frac{1}{2}{x}^a\cdot \frac{1}{2}{x}^{-a}=\frac{1}{2}.$ So $(y^{\prime }{)}^2=\frac{1}{4}{x}^{2a}-\frac{1}{2}+\frac{1}{4}{x}^{-2a}$. Add 1: the middle term $-\frac{1}{2}$ becomes $-\frac{1}{2}+1=+\frac{1}{2}=+2PQ$: $1+(y^{\prime }{)}^2=\frac{1}{4}{x}^{2a}+\frac{1}{2}+\frac{1}{4}{x}^{-2a}=(P+Q{)}^2={\left(\frac{1}{2}{x}^a+\frac{1}{2}{x}^{-a}\right)}^2.$ Conclusion. $\sqrt{1+(y^{\prime }{)}^2}=\frac{1}{2}{x}^a+\frac{1}{2}{x}^{-a}$ (positive for $x>0$). This is why textbook arc-length curves always split into "power plus reciprocal power": the mandatory $+1$ turns the difference-of-squares cross term $-\frac{1}{2}$ into the sum-of-squares $+\frac{1}{2}$. That's the fingerprint you now recognise instantly.

1. Did I differentiate first, then square? (L1 trap)
2. Is the curve given as $y(x)$, $x(y)$, $(x(t),y(t))$, or $r(\theta )$? Pick the matching $ds$. (L2/L4 traps)
3. After $\sqrt{(\text{stuff}{)}^2}$, is stuff $\ge 0$ on the whole interval? If not, split the integral where the sign changes, or carry $|\text{stuff}|$. (L3/L5 traps)
4. Sanity: is my answer $\ge$ the straight chord between endpoints? Length can never be shorter than a straight line.