Corrected exercises: arc length and area of a surface of revolution (MATH 141)

a) The chord joins $(x_{i-1}, f(x_{i-1}))$ to $(x_i, f(x_i))$. Its horizontal leg is $\Delta x$ and its vertical leg is $\Delta y_i = f(x_i) - f(x_{i-1})$, so by Pythagoras its length is $\sqrt{\Delta x^2 + \Delta y_i^2} = \sqrt{1 + (\Delta y_i/\Delta x)^2}\,\Delta x$, after factoring $\Delta x^2$ out of the root ($\Delta x > 0$). If $f$ is differentiable, the Mean Value Theorem gives a point $x_i^*$ in $(x_{i-1}, x_i)$ with $\Delta y_i = f'(x_i^*)\,\Delta x$. The polygon therefore has length $\sum_{i=1}^n \sqrt{1 + f'(x_i^*)^2}\,\Delta x$, which is a Riemann sum of the function $\sqrt{1 + (f')^2}$. When $f'$ is CONTINUOUS on $[a, b]$ (the curve is smooth), this function is continuous, the sums converge, and $L = \int_a^b \sqrt{1 + (f'(x))^2}\,dx$. In differential form, $ds^2 = dx^2 + dy^2$: the root is Pythagoras, and it is the whole chapter.

b) $|P_0P_1| = \sqrt{2^2 + 1^2} = \sqrt 5$ and $|P_1P_2| = \sqrt{2^2 + 3^2} = \sqrt{13}$, so the polygon has length $\sqrt 5 + \sqrt{13}$. It is SMALLER than $L$: a chord is the shortest path between its two endpoints, and the arc between $P_0$ and $P_1$ is one such path, so each chord is at most the piece of arc it spans. Adding points to the polygon can only make it longer (triangle inequality again), and $L$ is the limit of these increasing lengths. With $\sqrt 5 \approx 2.24$ and $\sqrt{13} \approx 3.61$, the polygon gives $5.85$, a lower bound for $L$.

c) $f'(x) = \frac{x}{2}$, so $1 + (f')^2 = 1 + \frac{x^2}{4} = \frac{4 + x^2}{4}$ and $L = \int_0^4 \frac12\sqrt{4 + x^2}\,dx$. The root does not open here, so a technique is needed: $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $\sqrt{4 + x^2} = 2\sec\theta$ (positive since $0 \le \theta < \frac{\pi}{2}$), and the bounds become $\theta = 0$ and $\theta = \arctan 2$. Then $L = \frac12\int_0^{\arctan 2} 2\sec\theta \cdot 2\sec^2\theta\,d\theta = 2\int_0^{\arctan 2}\sec^3\theta\,d\theta$. With $\int \sec^3\theta\,d\theta = \frac12(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|) + C$, and at $\theta = \arctan 2$, $\tan\theta = 2$, $\sec\theta = \sqrt{1 + 4} = \sqrt 5$: $L = \left[\sec\theta\tan\theta + \ln(\sec\theta + \tan\theta)\right]_0^{\arctan 2} = 2\sqrt 5 + \ln(2 + \sqrt 5)$. Order of magnitude: $2\sqrt 5 \approx 4.47$ and $\ln(2 + \sqrt 5)$ is a little above $\ln 4 \approx 1.39$, so $L \approx 5.9$, just above the polygon's $5.85$, as part b) demands. Two chords are already close because the parabola bends little.

d) Segment: $f' = 2$, so $L = \int_0^3 \sqrt{1 + 4}\,dx = 3\sqrt 5$. The segment runs from $(0, 1)$ to $(3, 7)$, whose distance is $\sqrt{9 + 36} = \sqrt{45} = 3\sqrt 5$. Arc of circle: $f'(x) = \frac{-x}{\sqrt{4 - x^2}}$, so $1 + (f')^2 = \frac{4 - x^2 + x^2}{4 - x^2} = \frac{4}{4 - x^2}$ and $L = \int_0^1 \frac{2}{\sqrt{4 - x^2}}\,dx = 2\left[\arcsin\frac{x}{2}\right]_0^1 = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3}$. Geometrically the arc goes from $(0, 2)$, at angle $\frac{\pi}{2}$, to $(1, \sqrt 3)$, at angle $\frac{\pi}{3}$: an angle of $\frac{\pi}{6}$ on a circle of radius $2$, of length $r\theta = \frac{\pi}{3}$. Both tests pass. Testing a formula on a segment or a circle is the fastest way to catch a misremembered one.

e) For $y = x$, $f' = 1$ and his integral gives $\int_0^1 2\,dx = 2$, while the segment from $(0, 0)$ to $(1, 1)$ has length $\sqrt 2$. In general $\int_a^b (1 + f')\,dx = (b - a) + (f(b) - f(a))$: run plus rise, the length of the STAIRCASE path that goes across then up, and it can even be negative when $f$ decreases. It needs no calculus at all, which is a sign that it cannot be a length. The error is $\sqrt{1 + (f')^2} = 1 + f'$: the square root of a sum is not the sum of the square roots. The legs of a right triangle add up to more than its hypotenuse, and that is exactly the gap between $2$ and $\sqrt 2$.

a) $f'(x) = \frac{x^2}{2} - \frac{1}{2x^2}$. Name the two terms: $A = \frac{x^2}{2}$, $B = \frac{1}{2x^2}$, and $AB = \frac14$: the root will open. Written out, $(f')^2 = \frac{x^4}{4} - \frac12 + \frac{1}{4x^4}$, so $1 + (f')^2 = \frac{x^4}{4} + \frac12 + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$. The bracket is positive, so the root is the bracket itself. $L = \int_1^2 \left(\frac{x^2}{2} + \frac{x^{-2}}{2}\right)dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 = \left(\frac86 - \frac14\right) - \left(\frac16 - \frac12\right) = \frac{13}{12} + \frac13 = \frac{17}{12}$. Check by differentiation: $\frac{d}{dx}\left(\frac{x^3}{6} - \frac{1}{2x}\right) = \frac{x^2}{2} + \frac{1}{2x^2}$. The antiderivative is NOT $f$: the sign in front of $\frac{1}{2x}$ has flipped, and that sign is the whole point.

b) $f'(x) = \frac{x}{2} - \frac{1}{2x}$, with $A = \frac{x}{2}$, $B = \frac{1}{2x}$, $AB = \frac14$. So $1 + (f')^2 = \frac{x^2}{4} + \frac12 + \frac{1}{4x^2} = \left(\frac{x}{2} + \frac{1}{2x}\right)^2$, positive on $[1, e]$. $L = \int_1^e \left(\frac{x}{2} + \frac{1}{2x}\right)dx = \left[\frac{x^2}{4} + \frac{\ln x}{2}\right]_1^e = \frac{e^2}{4} + \frac12 - \frac14 = \frac{e^2 + 1}{4}$. With $e^2 \approx 7.39$, $L \approx 2.1$, longer than the run $e - 1 \approx 1.72$, as it must be. The exact answer keeps $e$: on a MATH 141 paper, $2.1$ alone would lose the mark.

c) $f'(x) = \frac23 \cdot \frac32 (x^2 + 1)^{1/2} \cdot 2x = 2x\sqrt{x^2 + 1}$. Then $1 + (f')^2 = 1 + 4x^2(x^2 + 1) = 4x^4 + 4x^2 + 1 = (2x^2 + 1)^2$. This time the square comes from a trinomial in $x^2$ whose discriminant $16 - 16 = 0$ vanishes: same idea, the cross term $4x^2$ is exactly twice the product of $2x^2$ and $1$. $L = \int_0^1 (2x^2 + 1)\,dx = \frac23 + 1 = \frac53$. The lesson is to EXPAND $1 + (f')^2$ and look at it before reaching for a substitution.

d) $f'(x) = x^2 - \frac{c}{x^2}$, so $1 + (f')^2 = x^4 + (1 - 2c) + \frac{c^2}{x^4}$. It equals $\left(x^2 + \frac{c}{x^2}\right)^2 = x^4 + 2c + \frac{c^2}{x^4}$ exactly when $1 - 2c = 2c$, that is $c = \frac14$: the rule $AB = \frac14$ again, with $A = x^2$ and $B = \frac{c}{x^2}$. Then $L = \int_1^2 \left(x^2 + \frac{1}{4x^2}\right)dx = \left[\frac{x^3}{3} - \frac{1}{4x}\right]_1^2 = \left(\frac83 - \frac18\right) - \left(\frac13 - \frac14\right) = \frac73 + \frac18 = \frac{59}{24}$. This is how exam curves are built, and recognizing the construction is what saves the time.

e) The slip is to keep the minus sign: $\sqrt{1 + (f')^2} = \frac{x^2}{2} - \frac{1}{2x^2}$. That bracket is $f'$ itself, so the integral becomes $\int_1^2 f'(x)\,dx = f(2) - f(1) = \frac{19}{12} - \frac{8}{12} = \frac{11}{12}$: the student computed the RISE of the curve, not its length. The figure settles it without any algebra: the arc goes from $(1, \frac23)$ to $(2, \frac{19}{12})$, so it is at least as long as its chord, $\sqrt{1 + (\frac{11}{12})^2} = \frac{\sqrt{265}}{12}$, which is longer than the run $1$. Any arc over an interval of width $1$ is at least $1$ long, and $\frac{11}{12} < 1$. The check arc $\ge$ chord $\ge$ run costs five seconds and catches this error every time; the true value $\frac{17}{12} \approx 1.42$ sits just above the chord $\approx 1.36$.

a) $f'(x) = \frac{-\sin x}{\cos x} = -\tan x$, so $1 + (f')^2 = 1 + \tan^2 x = \sec^2 x$. On $[0, \frac{\pi}{3}]$, $\cos x > 0$, so $\sec x > 0$ and $\sqrt{\sec^2 x} = \sec x$. Then $L = \int_0^{\pi/3} \sec x\,dx = \left[\ln|\sec x + \tan x|\right]_0^{\pi/3} = \ln(2 + \sqrt 3) - \ln 1 = \ln(2 + \sqrt 3)$, using $\sec\frac{\pi}{3} = 2$ and $\tan\frac{\pi}{3} = \sqrt 3$. Check of the antiderivative: $\frac{d}{dx}\ln(\sec x + \tan x) = \frac{\sec x\tan x + \sec^2 x}{\sec x + \tan x} = \sec x$. Order of magnitude: $2 + \sqrt 3 \approx 3.73$ lies between $e$ and $4$, so $L$ is between $1$ and $1.39$, above the run $\frac{\pi}{3} \approx 1.05$.

b) $f'(x) = \frac{\cos x}{\sin x} = \cot x$ and $1 + \cot^2 x = \csc^2 x$. On $[\frac{\pi}{6}, \frac{\pi}{2}]$, $\sin x > 0$, so the root is $\csc x$. With $\int \csc x\,dx = \ln|\csc x - \cot x| + C$: at $\frac{\pi}{2}$, $\csc = 1$ and $\cot = 0$, giving $\ln 1 = 0$; at $\frac{\pi}{6}$, $\csc = 2$ and $\cot = \sqrt 3$, giving $\ln(2 - \sqrt 3)$. So $L = 0 - \ln(2 - \sqrt 3) = \ln\frac{1}{2 - \sqrt 3} = \ln(2 + \sqrt 3)$, since $(2 - \sqrt 3)(2 + \sqrt 3) = 4 - 3 = 1$. The coincidence: $\sin x = \cos(\frac{\pi}{2} - x)$, so the graph of $\ln(\sin x)$ is the graph of $\ln(\cos x)$ reflected in the vertical line $x = \frac{\pi}{4}$, and that reflection sends $[\frac{\pi}{6}, \frac{\pi}{2}]$ onto $[0, \frac{\pi}{3}]$. A reflection does not change lengths. Seeing this first would have given the answer in one line.

c) $\frac{d}{dx}\frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = \sinh x$. Next $\cosh^2 x - \sinh^2 x = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac44 = 1$, so $1 + \sinh^2 x = \cosh^2 x$. Since $\cosh x$ is half a sum of two positive numbers, it is positive and the root is $\cosh x$, with no absolute value to worry about. $L = \int_0^{\ln 3}\cosh x\,dx = \sinh(\ln 3) - \sinh 0 = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - \frac13}{2} = \frac43$. The only computation is $e^{\ln 3} = 3$ and $e^{-\ln 3} = \frac13$: that is why exam bounds for $\cosh$ are logarithms.

d) $f'(x) = \frac{-2x}{1 - x^2}$, so $1 + (f')^2 = \frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2} = \frac{1 + 2x^2 + x^4}{(1 - x^2)^2} = \frac{(1 + x^2)^2}{(1 - x^2)^2}$. On $[0, \frac12]$, $1 - x^2 > 0$, so the root is $\frac{1 + x^2}{1 - x^2}$. Numerator and denominator have the same degree: divide first, $\frac{1 + x^2}{1 - x^2} = -1 + \frac{2}{1 - x^2}$, then split $\frac{2}{1 - x^2} = \frac{1}{1 - x} + \frac{1}{1 + x}$. $L = \left[-x - \ln(1 - x) + \ln(1 + x)\right]_0^{1/2} = -\frac12 - \ln\frac12 + \ln\frac32 = -\frac12 + \ln 3$. With $\ln 3 \approx 1.10$, $L \approx 0.60$, just above the run $0.5$ since the curve only drops by $\ln\frac43$.

e) $\frac{d}{dx}\ln|\cos x| = \frac{-\sin x}{\cos x} = -\tan x$ again, so $1 + (f')^2 = \sec^2 x$. But on $[\frac{2\pi}{3}, \pi]$, $\cos x < 0$, so $\sec x < 0$ and $\sqrt{\sec^2 x} = |\sec x| = -\sec x$. $L = -\int_{2\pi/3}^{\pi}\sec x\,dx = -\left[\ln|\sec x + \tan x|\right]_{2\pi/3}^{\pi}$. At $\pi$: $\sec\pi = -1$, $\tan\pi = 0$, so $\ln 1 = 0$. At $\frac{2\pi}{3}$: $\sec = -2$, $\tan = -\sqrt 3$, so $\ln|-2 - \sqrt 3| = \ln(2 + \sqrt 3)$. Hence $L = -(0 - \ln(2 + \sqrt 3)) = \ln(2 + \sqrt 3)$. The student removed the root as $\sec x$ and integrated a negative function: a NEGATIVE length is never an answer, it is an alarm. Symmetry confirms: $|\cos(\pi - x)| = |\cos x|$, so this arc is the mirror image of the arc of part a) under $x \mapsto \pi - x$.

a) $\frac{dy}{dx} = \frac23 x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$. It is not defined at $x = 0$ and tends to $+\infty$ as $x \to 0^+$: the tangent at the origin is vertical, as the figure shows. The formula $L = \int_a^b \sqrt{1 + (f')^2}\,dx$ was derived for $f'$ CONTINUOUS on the closed interval; here the integrand $\sqrt{1 + \frac{4}{9x^{2/3}}}$ is unbounded near $0$, so the integral is improper and the formula cannot be applied as it stands. Writing $\int_0^8$ and carrying on without a word is the error that costs the method mark, even when the number comes out right.

b) For $x, y \ge 0$, $y = x^{2/3}$ is equivalent to $x = y^{3/2}$, and $y$ runs from $0$ to $8^{2/3} = 4$. Now $\frac{dx}{dy} = \frac32 y^{1/2}$, continuous on $[0, 4]$: the vertical tangent in $x$ is a horizontal one in $y$. $1 + (\frac{dx}{dy})^2 = 1 + \frac{9y}{4}$, and $L = \int_0^4 \sqrt{1 + \frac{9y}{4}}\,dy$. With $u = 1 + \frac{9y}{4}$, $du = \frac94\,dy$, and the bounds become $u = 1$ and $u = 10$: $L = \frac49\int_1^{10} u^{1/2}\,du = \frac49 \cdot \frac23\left[u^{3/2}\right]_1^{10} = \frac{8}{27}(10\sqrt{10} - 1)$. With $\sqrt{10} \approx 3.16$, $L \approx 9.07$, just above the chord from $(0, 0)$ to $(8, 4)$, $\sqrt{80} \approx 8.94$. The bounds are $0$ and $4$, values of $y$: writing $\int_0^8$ with $dy$ is the classic mismatch.

c) In $x$: $\sqrt{1 + \frac{4}{9x^{2/3}}} = \frac{\sqrt{9x^{2/3} + 4}}{3x^{1/3}}$ for $x > 0$. With $u = 9x^{2/3} + 4$, $du = 6x^{-1/3}\,dx$, so $\frac{dx}{3x^{1/3}} = \frac{du}{18}$. As $x \to 0^+$, $u \to 4$; at $x = 8$, $u = 9 \cdot 4 + 4 = 40$. So $L = \lim_{t \to 0^+}\int_t^8 \dots = \frac{1}{18}\int_4^{40}\sqrt u\,du = \frac{1}{27}\left(40\sqrt{40} - 8\right) = \frac{1}{27}(80\sqrt{10} - 8) = \frac{8}{27}(10\sqrt{10} - 1)$. Same value: the improper integral converges. But this route asked you to spot the problem, write the limit, and find a less obvious substitution; in $y$ nothing happened at all.

d) $\frac{dx}{dy} = \frac{y^3}{2} - \frac{1}{2y^3}$, with $A = \frac{y^3}{2}$, $B = \frac{1}{2y^3}$ and $AB = \frac14$: the perfect square of Exercise 2, in the other variable. $1 + (\frac{dx}{dy})^2 = \left(\frac{y^3}{2} + \frac{1}{2y^3}\right)^2$, positive on $[1, 2]$. $L = \int_1^2 \left(\frac{y^3}{2} + \frac{y^{-3}}{2}\right)dy = \left[\frac{y^4}{8} - \frac{1}{4y^2}\right]_1^2 = \left(2 - \frac{1}{16}\right) - \left(\frac18 - \frac14\right) = \frac{31}{16} + \frac18 = \frac{33}{16}$. Nothing new in the method: only the letters changed, and the bounds are the given values of $y$.

e) $\frac{dx}{dy} = y$, so $L = \int_0^1 \sqrt{1 + y^2}\,dy$. No perfect square this time: $y = \tan\theta$, $dy = \sec^2\theta\,d\theta$, bounds $0$ and $\frac{\pi}{4}$, and $L = \int_0^{\pi/4}\sec^3\theta\,d\theta = \frac12\left[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\right]_0^{\pi/4} = \frac12\left(\sqrt 2 + \ln(1 + \sqrt 2)\right)$. Since $\ln(1 + \sqrt 2)$ lies between $\ln 2 \approx 0.69$ and $1$, $L$ is between $1.05$ and $1.21$, above the chord $\sqrt{\frac14 + 1} = \frac{\sqrt 5}{2} \approx 1.12$. In $x$, the arc is $y = \sqrt{2x}$, $0 \le x \le \frac12$, with $\frac{dy}{dx} = \frac{1}{\sqrt{2x}}$, infinite at $x = 0$: an improper integral of $\sqrt{1 + \frac{1}{2x}}$, harder on every count. The rule: choose the variable in which the derivative stays FINITE on the closed interval.