Answers to Selected Exercises
Worked answers to selected (mostly odd-numbered) exercises across all 40 chapters, chosen to span the difficulty tiers (⭐ to ⭐⭐⭐⭐). Each gives the final result plus the key step — use them to check your work, not to replace it. Attempt every problem before reading the answer.
Full step-by-step solutions for all odd-numbered exercises are available to instructors through the instructor companion.
Chapter 1 — Why Calculus
1.3 A secant line passes through two distinct points $(a, f(a))$ and $(b, f(b))$ on a curve; its slope is the average rate of change $\frac{f(b)-f(a)}{b-a}$. A tangent line touches the curve at a single point and matches its direction there; its slope is the limiting value of secant slopes as the second point slides into the first. Pictures: a chord cutting the curve at two points vs. a line grazing it at one.
1.11 Secant slopes $\frac{(2+h)^2-4}{h} = 4+h$ give $4.1,\ 4.01,\ 4.001$ for $h=0.1,0.01,0.001$. They approach $\mathbf{4}$, the slope of $y=x^2$ at $x=2$.
1.13 $\frac{\sqrt{4+h}-2}{h}$ rationalizes to $\frac{1}{\sqrt{4+h}+2}$, giving $\approx 0.2485,\ 0.24984,\ 0.249984$. They approach $\frac14 = \mathbf{0.25}$ (which is $\frac{1}{2\sqrt 4}$).
1.15 $\frac{(a+h)^3-a^3}{h} = \frac{3a^2h + 3ah^2 + h^3}{h} = 3a^2 + 3ah + h^2$. As $h\to 0$ the last two terms vanish, leaving slope $= \mathbf{3a^2}$.
1.17 Right-endpoint sum for $y=x$ on $[0,1]$ is $\sum_{i=1}^n \frac{i}{n}\cdot\frac1n = \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2n} = \frac12 + \frac{1}{2n}$. For $n=10,100,1000$: $0.55,\ 0.505,\ 0.5005$, approaching the true area $\mathbf{1/2}$ (overestimating by $\frac{1}{2n}$).
1.31 Right secant ($h=0.1$): $\frac{|0.1|-0}{0.1} = +1$; left secant ($h=-0.1$): $\frac{|-0.1|-0}{-0.1} = -1$. The one-sided slopes disagree ($+1$ vs $-1$), so $|x|$ has no well-defined slope at the corner $x=0$ — the secant lines do not settle on a single limiting direction.
1.32 Right-endpoint sum $S_n = \sum_{i=1}^n \frac in\cdot\frac1n = \frac{1}{n^2}\sum_{i=1}^n i = \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2n} = \frac12 + \frac{1}{2n}$. Taking $n\to\infty$, the term $\frac{1}{2n}\to 0$, so the area $= \mathbf{1/2}$, proven by Riemann sum rather than geometry.
Chapter 2 — Functions and Models
2.3 $f(0)=3$, $f(1)=1-4+3=0$, $f(2)=4-8+3=-1$, $f(-1)=1+4+3=8$. And $f(a+1)=(a+1)^2-4(a+1)+3 = a^2+2a+1-4a-4+3 = \mathbf{a^2-2a}$.
2.5 $f(x)=x^2+3$: since $x^2\ge 0$, range is $[3,\infty)$. $g(x)=2\sin x+1$: since $\sin x\in[-1,1]$, range is $[-1,3]$.
2.9 $\sin 0 = 0$, $\sin(\pi/4)=\frac{\sqrt2}{2}$, $\sin(\pi/2)=1$, $\sin\pi = 0$, $\cos(\pi/3)=\frac12$, $\cos\pi = -1$.
2.11 $f(g(8)) = 2^{\log_2 8} = 2^3 = 8$; $g(f(5)) = \log_2(2^5) = 5$. This illustrates that inverse functions undo each other: $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$.
2.21 Swap and solve $x = \frac{2y-1}{y+3}$: $x(y+3)=2y-1 \Rightarrow xy+3x = 2y-1 \Rightarrow y(x-2) = -1-3x \Rightarrow y = \frac{3x+1}{2-x}$. So $f^{-1}(x)=\dfrac{3x+1}{2-x}$, domain all $x\ne 2$.
2.23 (a) $g=x^2+1$, $f=\sqrt u$; (b) $g=3x$, $f=e^u$; (c) $g=\sin x$, $f=u^2$; (d) $g=\cos x$, $f=\ln u$.
2.25 Half-life $4$ h, $D_0=200$ mg. (a) $D(t)=200\cdot 2^{-t/4} = 200\,e^{-(\ln 2/4)t}$. (b) $t=12$ is $3$ half-lives: $200/2^3 = \mathbf{25}$ mg. (c) $25 = 200\cdot 2^{-t/4} \Rightarrow 2^{-t/4}=\frac18 \Rightarrow t/4 = 3 \Rightarrow t=\mathbf{12}$ h.
2.27 $h(t)=2+15t-4.9t^2$. (b) Vertex at $t = \frac{15}{2(4.9)} = \frac{15}{9.8}\approx 1.53$ s; max height $h(1.53)\approx 2 + 15(1.531) - 4.9(1.531)^2 \approx \mathbf{13.5}$ m. (c) $h=0$: $t=\frac{15+\sqrt{225+39.2}}{9.8} = \frac{15+\sqrt{264.2}}{9.8}\approx \frac{15+16.25}{9.8}\approx \mathbf{3.19}$ s (positive root).
Chapter 3 — The Limit
3.7 (a) $8-8+1 = 1$; (b) $e^0+5 = 6$; (c) $\cos(\pi/4)=\frac{\sqrt2}{2}$; (d) $\ln 3$.
3.9 (a) Rationalize: $\frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} = \frac{1}{\sqrt{x+9}+3}\to \frac{1}{6}$. (b) $\frac{\sqrt x-2}{x-4} = \frac{1}{\sqrt x+2}\to \frac14$. (c) $\frac{x-1}{\sqrt x-1} = \sqrt x+1 \to 2$.
3.13 (a) $\lim_{x\to2^-}(x^2+1)=5$; (b) $\lim_{x\to2^+}(5-x)=3$; (c) one-sided limits differ ($5\ne 3$), so $\lim_{x\to2}f(x)$ does not exist; $f(2)=5-2=3$.
3.16 (a) $\frac{3x+1}{x-2}\to 3$ (ratio of leading coeffs); (b) $\frac{x^2+5x}{2x^2-7}\to \frac12$; (c) $\frac{x^2+1}{x^3-2}\to 0$ (lower-degree numerator); (d) $\frac{x^3+2}{x^2+5}\to +\infty$; (e) $\frac{4x^2-3}{2x^2+x+1}\to 2$.
3.23 For $x\ne 0$, $-1\le\sin(1/x)\le 1$, so $-|x|\le x\sin(1/x)\le |x|$. Since $\lim_{x\to0}(-|x|)=\lim_{x\to0}|x|=0$, the Squeeze Theorem forces $\lim_{x\to0}x\sin(1/x)=\mathbf{0}$.
3.27 $\frac{s(1+h)-s(1)}{h} = \frac{4.9(1+h)^2-4.9}{h} = \frac{4.9(2h+h^2)}{h} = 4.9(2+h)\to 9.8$. Instantaneous velocity at $t=1$ is $\mathbf{9.8}$ m/s.
3.29 $\lim_{t\to\infty}\frac{500t}{t+4} = \lim_{t\to\infty}\frac{500}{1+4/t} = \mathbf{500}$. The population saturates at a carrying capacity of $500$ individuals.
Chapter 4 — Continuity
4.6 (a) Removable ($\frac{x^2-4}{x-2}=x+2$, hole at $x=2$); (b) infinite ($\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}$ blows up at $x=1$); (c) jump (greatest-integer step); (d) oscillating; (e) at $x=1$, left value $1^2=1$, right $2(1)-2=0$ — jump; (f) $\frac{|x-3|}{x-3}$ is $+1$ right, $-1$ left — jump.
4.8 Factor: $\frac{x^2-x-6}{x^2-4} = \frac{(x-3)(x+2)}{(x-2)(x+2)}$. At $x=-2$ the $(x+2)$ cancels: removable. At $x=2$ the factor remains in the denominator: infinite. So discontinuous at $x=-2$ (removable) and $x=2$ (infinite).
4.11 Continuity at $x=1$ requires the two pieces to agree: $1^2-c = 2c(1) \Rightarrow 1-c = 2c \Rightarrow 1 = 3c \Rightarrow c = \mathbf{1/3}$.
4.13 Continuity at $x=1$: $1+1 = a+b$, so $a+b=2$. Continuity at $x=3$: $3a+b = 7$. Subtract: $2a=5 \Rightarrow a=\frac52$, $b=2-\frac52=-\frac12$. Solution $a=\frac52,\ b=-\frac12$ is unique (two linear equations, two unknowns).
4.15 $f(x)=x^3+x-1$ is continuous; $f(0)=-1<0$ and $f(1)=1+1-1=1>0$. Since $f$ changes sign on $[0,1]$, IVT guarantees a root in $(0,1)$.
4.24 Bisection on $f(x)=x^2-2$, $[1,2]$: (1) $m=1.5$, $f=0.25>0$, keep $[1,1.5]$. (2) $m=1.25$, $f=-0.4375<0$, keep $[1.25,1.5]$. (3) $m=1.375$, $f=-0.109<0$, keep $[1.375,1.5]$. (4) $m=1.4375$, $f=0.066>0$, keep $[1.375,1.4375]$. After four steps $\sqrt2$ is bracketed in $[1.375,1.4375]$, width $\mathbf{0.0625}$ (= $1/16$).
Chapter 5 — Rates of Change
5.1 Average rate of $f(x)=x^2$ on $[1,3]$: $\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = \mathbf{4}$.
5.10 $f'(2)=\lim_{h\to0}\frac{(2+h)^2+3(2+h)-(4+6)}{h} = \lim_{h\to0}\frac{4+4h+h^2+6+3h-10}{h} = \lim_{h\to0}(7+h) = \mathbf{7}$.
5.11 $f(x)=1/x$ at $a=-1$: $\frac{1/(-1+h)-(-1)}{h} = \frac{\frac{1}{-1+h}+1}{h} = \frac{\frac{1+(-1+h)}{-1+h}}{h} = \frac{h}{h(-1+h)} = \frac{1}{-1+h}\to \frac{1}{-1} = \mathbf{-1}$.
5.13 $f(x)=x^3$: $\frac{(x+h)^3-x^3}{h} = \frac{3x^2h+3xh^2+h^3}{h} = 3x^2+3xh+h^2 \to \mathbf{3x^2}$.
5.19 Tangent to $y=1/x$ at $(1,1)$: $f'(x)=-1/x^2$ (from §5/def), so $f'(1)=-1$. Line: $y-1 = -1(x-1)$, i.e. $\mathbf{y = -x+2}$.
5.21 $y=x^2-4x+7$: $f'(x)=2x-4=0 \Rightarrow x=2$, $y = 4-8+7 = 3$. Horizontal tangent at $\mathbf{(2,3)}$.
5.26 Difference quotient of $|x|$ at $0$: from the right $\frac{|h|}{h}=\frac{h}{h}=1$; from the left $\frac{|h|}{h}=\frac{-h}{h}=-1$. The one-sided limits $1$ and $-1$ disagree, so $f'(0)$ does not exist — $|x|$ is not differentiable at $0$.
5.32 $C(q)=1000+5q+0.01q^2$. $\frac{C(q+h)-C(q)}{h} = \frac{5h + 0.01(2qh+h^2)}{h} = 5+0.02q+0.01h \to C'(q)=5+0.02q$. The 101st widget costs about $C'(100)=5+2 = \mathbf{\$7}$.
Chapter 6 — The Derivative
6.5 (a) $f'(x)=3$ (slope of a line); (b) $\frac{(x+h)^2+5(x+h)-(x^2+5x)}{h}=2x+5+h\to 2x+5$; (c) $\frac{1/(x+h)-1/x}{h}=\frac{-1}{x(x+h)}\to -1/x^2$; (d) rationalize $\Rightarrow \frac{1}{\sqrt{x+1+h}+\sqrt{x+1}}\to \frac{1}{2\sqrt{x+1}}$; (e) $f'(x)=4x^3$.
6.6 $f(x)=1/(x+1)$: $f'(x)=-1/(x+1)^2$ (same form as 6.5c shifted), so $f'(2) = -1/(3)^2 = \mathbf{-1/9}$.
6.16 $f(x)=|x-3|$: for $x>3$, $f=x-3$ so $f'=+1$; for $x<3$, $f=3-x$ so $f'=-1$. The right slope $+1$ and left slope $-1$ disagree at $x=3$, so $f'(3)$ does not exist (a corner).
6.20 $f(x)=x^5$: $f'=5x^4$, $f''=20x^3$, $f'''=60x^2$, $f^{(4)}=120x$, $f^{(5)}=120$, $f^{(6)}=\mathbf{0}$. The 6th derivative is zero: each differentiation drops the degree by one, and after $5$ steps a degree-$5$ polynomial becomes constant, then $0$.
6.23 $f(x)=x^3-3x$: $f''(x)=6x$. Concave down on $(-\infty,0)$ (where $f''<0$), concave up on $(0,\infty)$; inflection point at $\mathbf{(0,0)}$ where concavity changes.
6.25 $C(q)=0.01q^2+5q+200$. (a) $C'(q)=0.02q+5$. (b) Cost of 51st unit $\approx C'(50)=0.02(50)+5 = \mathbf{\$6}$. (c) $C''(q)=0.02>0$: marginal cost is increasing.
6.29 $L(m)=(3-m)^2+(5-2m)^2$. (a) $L'(m)=-2(3-m)-4(5-2m)=-6+2m-20+8m = 10m-26$. (b) $L'(m)=0 \Rightarrow m=\mathbf{2.6}$. (c) $L''(m)=10>0$, confirming a minimum.
Chapter 7 — Differentiation Rules
7.1 (a) $7x^6$; (b) $\frac13 x^{-2/3}$; (c) $-2x^{-3}$; (d) $\frac52 x^{3/2}$; (e) $0$; (f) $\frac{1}{2\sqrt x}$; (g) $-3x^{-4}$; (h) $\pi x^{\pi-1}$.
7.3 (a) $(x^2+1)(x^3-2)$: $2x(x^3-2)+(x^2+1)(3x^2) = 5x^4+3x^2-4x$; (b) $x^2e^x$: $2xe^x+x^2e^x = e^x(x^2+2x)$; (c) $x\sin x$: $\sin x + x\cos x$; (d) $(2x+3)(5-x^2)$: $2(5-x^2)+(2x+3)(-2x) = -6x^2-6x+10$; (e) $\sqrt x\cos x$: $\frac{\cos x}{2\sqrt x}-\sqrt x\sin x$.
7.5 (a) $7(x^2+3)^6(2x)=14x(x^2+3)^6$; (b) $5\cos(5x)$; (c) $3e^{3x}$; (d) $\frac{2x}{x^2+1}$; (e) $\frac{x}{\sqrt{x^2+1}}$; (f) $-\sin(\ln x)\cdot\frac1x$; (g) $\cos x\, e^{\sin x}$; (h) $4(\tan x)^3\sec^2 x$.
7.9 (a) $y=x^x$: $\ln y = x\ln x$, $\frac{y'}{y}=\ln x+1$, so $y'=x^x(\ln x+1)$. (b) $y=(\sin x)^x$: $\ln y = x\ln\sin x$, $\frac{y'}{y}=\ln\sin x+x\cot x$, so $y'=(\sin x)^x(\ln\sin x+x\cot x)$. (c) $y=x^{\ln x}$: $\ln y=(\ln x)^2$, $\frac{y'}{y}=\frac{2\ln x}{x}$, so $y'=x^{\ln x}\cdot\frac{2\ln x}{x}$.
7.11 $y=x^3-2x$, $y'=3x^2-2$; at $x=1$: $y(1)=-1$, slope $=1$. Tangent: $y-(-1)=1(x-1)$, i.e. $\mathbf{y=x-2}$.
7.13 $y=\ln x$, $y'=1/x$; at $x=1$: $y(1)=0$, slope $=1$. Tangent: $\mathbf{y=x-1}$ (so $\ln(1+h)\approx h$ for small $h$).
7.34 $\frac{d}{dx}(\sin x)^3 = 3\sin^2x\cos x$ (outer is cubing, inner is $\sin x$). $\frac{d}{dx}\sin(x^3)=\cos(x^3)\cdot 3x^2$ (outer is $\sin$, inner is $x^3$). They differ because the order of composition is reversed: which function is outer determines the form.
Chapter 8 — Implicit and Related Rates
A5 $x^2+xy+y^2=7$: $2x+(y+xy')+2yy'=0 \Rightarrow y'(x+2y)=-(2x+y) \Rightarrow y'=-\frac{2x+y}{x+2y}$. At $(1,2)$: $y'=-\frac{4}{5}$.
B3 Circle $x^2+y^2=25$ with slope $y'=-x/y=\frac34$: so $x=-\frac34 y$. Substitute: $\frac{9}{16}y^2+y^2=25 \Rightarrow \frac{25}{16}y^2=25 \Rightarrow y^2=16$, $y=\pm4$. Points: $(-3,4)$ and $(3,-4)$.
D2 $y=\arccos x \Leftrightarrow x=\cos y$ with $\sin y\ge0$ on $[0,\pi]$. Differentiate: $1=-\sin y\,y' \Rightarrow y'=-\frac{1}{\sin y}=-\frac{1}{\sqrt{1-\cos^2 y}}=-\frac{1}{\sqrt{1-x^2}}$.
F5 Cone with $r=2h$: $V=\frac13\pi r^2h=\frac43\pi h^3$, so $\frac{dV}{dt}=4\pi h^2\frac{dh}{dt}$. At $h=5$, $\frac{dV}{dt}=10$: $10=100\pi\frac{dh}{dt} \Rightarrow \frac{dh}{dt}=\frac{1}{10\pi}\approx 0.0318$ ft/min.
G1 Ladder $x^2+y^2=169$; at $x=5$, $y=12$. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{5}{12}(2)=-\frac56\approx -0.833$ ft/s (top descending).
H1 $\tan\theta=h/100$, so $\sec^2\theta\frac{d\theta}{dt}=\frac{1}{100}\frac{dh}{dt}=\frac{5}{100}=0.05$. At $h=50$, $\tan\theta=0.5$, $\sec^2\theta=1.25$, so $\frac{d\theta}{dt}=\frac{0.05}{1.25}=\mathbf{0.04}$ rad/s.
I3 $D=\sqrt{a^2+b^2}$; with $\frac{da}{dt}=-600$, $\frac{db}{dt}=-800$, at $a=40,b=30$: $D=50$. From $D\frac{dD}{dt}=a\frac{da}{dt}+b\frac{db}{dt}=-24000-24000=-48000$, so $\frac{dD}{dt}=-960$ km/h — closing at $960$ km/h.
Chapter 9 — Applications of Derivatives
A1 $f(x)=x^3-12x+5$: $f'(x)=3x^2-12=3(x-2)(x+2)$, critical points $x=\pm2$.
A5 $f(x)=x+\frac1x$ on $[\frac12,3]$: $f'=1-\frac{1}{x^2}=0\Rightarrow x=1$. Values $f(\frac12)=2.5$, $f(1)=2$, $f(3)=\frac{10}{3}\approx 3.33$. Absolute min $=2$ at $x=1$; absolute max $=\frac{10}{3}$ at $x=3$.
B3 $f(x)=3x^4-8x^3+6$: $f'=12x^3-24x^2=12x^2(x-2)$, critical points $x=0,2$. $f'$ stays negative through $x=0$ (no sign change → neither) and goes $-\to+$ at $x=2$ (local minimum).
C3 $f(x)=x^4-8x^2+3$: $f'=4x(x^2-4)$, critical points $0,\pm2$; $f''=12x^2-16$. $f''(0)=-16<0$ → local max ($f(0)=3$); $f''(\pm2)=32>0$ → local minima ($f(\pm2)=-13$). Inflections at $x=\pm\frac{2}{\sqrt3}$.
D1 MVT for $f(x)=x^2-2x$ on $[0,4]$: average rate $=\frac{f(4)-f(0)}{4}=\frac{8}{4}=2$. Set $f'(c)=2c-2=2 \Rightarrow c=2\in(0,4)$.
E3 $\lim_{x\to\infty}\frac{\ln x}{x}$, form $\frac\infty\infty$. L'Hôpital: $\lim\frac{1/x}{1}=0$. So $\ln x$ grows slower than any positive power of $x$.
E4 $\lim_{x\to0}\frac{1-\cos x}{x^2}$, form $\frac00$. L'Hôpital once: $\frac{\sin x}{2x}$ (still $\frac00$); again: $\frac{\cos x}{2}\to \frac12$.
G4 $L(w)=(w-3)^2+\frac12 w^2$: $L'(w)=2(w-3)+w=3w-6=0 \Rightarrow w^*=2$. $L''(w)=3>0$ everywhere, so $L$ is convex and $w^*=2$ is the global minimum.
Chapter 10 — Optimization
A1 $f(x)=x^2-6x+5$ on $[0,5]$: $f'=2x-6=0\Rightarrow x=3$. Candidates $f(0)=5$, $f(3)=-4$, $f(5)=0$. Global max $=5$ at $x=0$; global min $=-4$ at $x=3$.
B1 Perimeter $2x+2y=40\Rightarrow y=20-x$. $A=x(20-x)=20x-x^2$, $A'=20-2x=0\Rightarrow x=10$, $y=10$. $A''=-2<0$, a max. Optimum is a $10\times10$ square, area $100$ m².
B5 $V(x)=x(18-2x)^2=4x^3-72x^2+324x$ on $[0,9]$. $V'=12(x-3)(x-9)=0\Rightarrow x=3$ (or $9$). $V(3)=3(12)^2=432$ in³ (max); endpoints give $0$. Optimum: $3$-in cuts, $432$ in³.
C3 Open box, $x^2h=32\Rightarrow h=32/x^2$. Material $S=x^2+4xh=x^2+\frac{128}{x}$. $S'=2x-\frac{128}{x^2}=0\Rightarrow x^3=64\Rightarrow x=4$ ft, $h=2$ ft. $S''>0$, a min; material $=16+32=48$ ft².
D5 $R(\theta)=\frac{v^2\sin(2\theta)}{g}$: $R'(\theta)=\frac{2v^2}{g}\cos(2\theta)=0\Rightarrow 2\theta=\frac\pi2\Rightarrow \theta=\frac\pi4=45^\circ$. $R''<0$ there, a max — the classic optimal launch angle.
E1 $P(x)=200x-2x^2-40x-300=-2x^2+160x-300$. $P'=-4x+160=0\Rightarrow x=40$. $P''=-4<0$, a max. $P(40)=2900$. Check MR$=R'(40)=200-160=40=$MC. Max profit $\$2900$ at $x=40$.
E7 $x$ passengers above $100$, $0\le x\le 80$. $R(x)=(100+x)(800-4x)=80000+400x-4x^2$. $R'=400-8x=0\Rightarrow x=50$, i.e. $150$ passengers at $\$600$. $R(50)=\$90{,}000$; endpoints $R(0)=\$80{,}000$, $R(80)=\$86{,}400$. Optimum: $150$ passengers.
Chapter 11 — Linear Approximation, Differentials, and Newton's Method
A1. $f(9)=3$, $f'(x)=\tfrac{1}{2\sqrt x}$ so $f'(9)=\tfrac16$. Thus $L(x)=3+\tfrac16(x-9)$, and $\sqrt{9.2}\approx L(9.2)=3+\tfrac16(0.2)=3.0\overline{3}$ (true $3.03315$).
A3. $f(1)=0$, $f'(x)=1/x$ so $f'(1)=1$, giving $L(x)=x-1$. Then $\ln(0.97)\approx L(0.97)=-0.03$ (true $-0.030459$).
A5. Anchor $f(x)=x^{1/3}$ at $a=27$: $f(27)=3$, $f'(27)=\tfrac13(27)^{-2/3}=\tfrac13\cdot\tfrac19=\tfrac1{27}$. So $\sqrt[3]{27.5}\approx 3+\tfrac1{27}(0.5)=3.01852$ (true $3.01849$).
A9. $f(0)=\cos 0=1$, $f'(x)=-\sin x$ so $f'(0)=0$; hence $L(x)=1+0\cdot x=1$. The linear term vanishes because $x=0$ is a critical point of $\cos$ (the graph is flat there, so the tangent line is horizontal).
C1. Absolute error: $dA=2s\,ds=2(20)(0.1)=4\ \mathrm{cm^2}$. Area $A=400\ \mathrm{cm^2}$, so relative error $=4/400=1\%$ (twice the $0.5\%$ side error — the square law).
C5. $g=4\pi^2\ell/T^2$, so $\tfrac{\Delta g}{g}\approx\tfrac{\Delta\ell}{\ell}+2\tfrac{\Delta T}{T}=0.5\%+2(0.3\%)=1.1\%$. The period enters squared, so timing $T$ deserves more care — each $1\%$ in $T$ costs $2\%$ in $g$.
D3. $f(x)=x^2-x-1$, $f'(x)=2x-1$. From $x_0=1.5$: $f=-0.25$, $f'=2$, $x_1=1.5-(-0.25)/2=1.625$. Then $f(1.625)=0.015625$, $f'=2.25$, $x_2=1.625-0.006944=1.61806$ (true $\varphi=1.61803$).
E3. $f(x)=x^3-2x+2$, $f'(x)=3x^2-2$. From $x_0=0$: $x_1=0-2/(-2)=1$; $x_2=1-1/1=0$; $x_3=1$. The iterates cycle $0\to1\to0\to1$, a period-2 cycle — Newton's method fails to converge.
Chapter 12 — Antiderivatives
A7. $\int x^{1/3}\,dx=\dfrac{x^{4/3}}{4/3}+C=\dfrac34 x^{4/3}+C$.
A9. Divide first: $\dfrac{2x^3-x+5}{x}=2x^2-1+\dfrac5x$, so $\int=\dfrac23 x^3-x+5\ln|x|+C$.
C3. $\int 3^x\,dx=\dfrac{3^x}{\ln 3}+C$ (since $(3^x)'=3^x\ln 3$).
E3. Power rule with linear interior, divide by inner slope $3$: $\int(3x+1)^4\,dx=\dfrac{(3x+1)^5}{15}+C$.
F3. $f''=6x\Rightarrow f'=3x^2+C_1$; $f'(0)=C_1=2$. Then $f=x^3+2x+C_2$; $f(0)=C_2=1$. So $f(x)=x^3+2x+1$.
G1. $a(t)=6t\Rightarrow v(t)=3t^2+C_1$, $v(0)=1$ gives $v=3t^2+1$. Then $s(t)=t^3+t+C_2$, $s(0)=4$ gives $s(t)=t^3+t+4$.
H1. $C(q)=\int(100-0.1q)\,dq=100q-0.05q^2+1000$ (fixed cost is the constant). $C(200)=20000-0.05(40000)+1000=\$19{,}000$.
Chapter 13 — The Definite Integral (pre-FTC: sums and geometry only)
A1. $f(x)=2x+1$, $\Delta x=1$, left endpoints $0,1,2,3$ give $f$-values $1,3,5,7$. $L_4=1(1+3+5+7)=16$.
A5. $f(x)=x^2$ on $[1,3]$, $\Delta x=0.5$. Right endpoints $1.5,2,2.5,3$: $R_4=0.5(2.25+4+6.25+9)=0.5(21.5)=10.75$. Left endpoints $1,1.5,2,2.5$: $L_4=0.5(1+2.25+4+6.25)=0.5(13.5)=6.75$. Since $f$ is increasing on $[1,3]$, $R_4$ overestimates and $L_4$ underestimates the true integral.
B3. $1+4+9+\cdots+100=\displaystyle\sum_{i=1}^{10}i^2=\dfrac{10\cdot11\cdot21}{6}=385$.
C3. $\int_0^3 2x\,dx$: $\Delta x=3/n$, $x_i=3i/n$, $f(x_i)=6i/n$. $R_n=\sum_{i=1}^n\tfrac{6i}{n}\cdot\tfrac3n=\tfrac{18}{n^2}\cdot\tfrac{n(n+1)}{2}=9\cdot\tfrac{n+1}{n}\to 9$. Geometrically a triangle of base $3$, height $6$: area $\tfrac12(3)(6)=9$. ✓
D3. $\int_{-2}^2 x\,dx=0$. The triangle below the axis on $[-2,0]$ (signed area $-2$) exactly cancels the triangle above on $[0,2]$ (area $+2$); equivalently $x$ is odd, so the integral over a symmetric interval is $0$.
F3. From F2, $\overline f=\tfrac1{2-0}\int_0^2 x^2\,dx=\tfrac12\cdot\tfrac83=\tfrac43$. The MVT point solves $c^2=\tfrac43$, so $c=\sqrt{4/3}=\tfrac{2}{\sqrt3}\approx1.155\in[0,2]$. ✓
G3. $v(t)=4-2t$, zero at $t=2$. (a) Net displacement: triangle above on $[0,2]$ has area $\tfrac12(2)(4)=4$; triangle below on $[2,4]$ has signed area $-\tfrac12(2)(4)=-4$. Sum $=0$. (b) Total distance $=4+4=8$ m. They differ because displacement counts the backward leg negatively, while distance counts every leg positively.
Chapter 14 — Fundamental Theorem of Calculus
A1. $\int_0^3 x^2\,dx=\big[\tfrac{x^3}{3}\big]_0^3=\tfrac{27}{3}=9$.
A5. $\int_0^1(4x^3-2x+1)\,dx=[x^4-x^2+x]_0^1=1-1+1=1$.
A9. $\int_1^{e^2}\tfrac1x\,dx=[\ln x]_1^{e^2}=\ln(e^2)-\ln 1=2$.
B1. By FTC Part 1, $F(x)=\int_2^x t^3\,dt\Rightarrow F'(x)=x^3$ (the upper-limit integrand, no need to compute $F$).
C1. $\dfrac{d}{dx}\int_0^{x^2}\sin t\,dt=\sin(x^2)\cdot\dfrac{d}{dx}(x^2)=2x\sin(x^2)$ (Leibniz chain factor).
C3. Variable in the lower limit: $\dfrac{d}{dx}\int_x^5\cos(t^2)\,dt=-\cos(x^2)$.
D1. $v(t)=2t-6$; net displacement $=\int_0^5(2t-6)\,dt=[t^2-6t]_0^5=25-30=-5$ m.
E1. $\overline f=\tfrac13\int_0^3 x^2\,dx=\tfrac13\cdot 9=3$. Attained where $c^2=3$, i.e. $c=\sqrt3\approx1.732\in[0,3]$.
F3. $\int_0^5 30e^{-0.2t}\,dt=\big[-150e^{-0.2t}\big]_0^5=-150e^{-1}+150=150(1-e^{-1})\approx 94.82$ mg.
Chapter 15 — Integration Techniques I
A1. $u=x^2+7$, $du=2x\,dx$: $\int u^4\,du=\tfrac15(x^2+7)^5+C$.
A5. $u=5x+2$: $\int\tfrac1u\cdot\tfrac15\,du=\tfrac15\ln|5x+2|+C$.
B3. $u=1-x^2$, $x\,dx=-\tfrac12 du$: $-\tfrac12\int u^{-1/2}\,du=-u^{1/2}+C=-\sqrt{1-x^2}+C$.
C5. $u=x^2+1$, $x\,dx=\tfrac12 du$, limits $1\to2$: $\tfrac12\int_1^2\tfrac1u\,du=\tfrac12\ln 2$.
D1. Parts, $u=x$, $dv=e^{2x}dx$, $v=\tfrac12 e^{2x}$: $\tfrac{x}{2}e^{2x}-\tfrac12\int e^{2x}\,dx=\tfrac{x}{2}e^{2x}-\tfrac14 e^{2x}+C=\tfrac14(2x-1)e^{2x}+C$.
D7. Tabular ($x^2\!\to\!2x\!\to\!2\!\to\!0$; integrate $e^{-x}$ to $-e^{-x},e^{-x},-e^{-x}$; signs $+,-,+$): $-x^2e^{-x}-2xe^{-x}-2e^{-x}+C=-(x^2+2x+2)e^{-x}+C$.
E1. $I=\int e^{2x}\cos x\,dx$. Two parts (keep $u=e^{2x}$) give $I=\tfrac15 e^{2x}(2\cos x+\sin x)+C$.
F3. Parts twice give $E[X^2]=\dfrac{2}{\lambda^2}$. Then $\operatorname{Var}(X)=\dfrac{2}{\lambda^2}-\Big(\dfrac1\lambda\Big)^2=\dfrac{1}{\lambda^2}$.
Chapter 16 — Integration Techniques II
16.1 Power reduction $\sin^2 x=\tfrac{1-\cos 2x}{2}$: $\int\sin^2 x\,dx=\tfrac{x}{2}-\tfrac14\sin 2x+C$.
16.3 $\tan^2 x=\sec^2 x-1$: $\int\tan^2 x\,dx=\tan x-x+C$.
16.11 $\int\dfrac{dx}{\sqrt{9-x^2}}=\arcsin\dfrac{x}{3}+C$ (standard $\arcsin$ form with $a=3$).
16.19 Partial fractions $\dfrac{1}{x^2-4}=\dfrac{1/4}{x-2}-\dfrac{1/4}{x+2}$: $\int=\tfrac14\ln\left|\dfrac{x-2}{x+2}\right|+C$.
16.21 $\dfrac{2x+3}{(x-1)(x+2)}=\dfrac{5/3}{x-1}+\dfrac{1/3}{x+2}$ (cover-up: at $x=1$, $5/3$; at $x=-2$, $(-1)/(-3)=1/3$). $\int=\tfrac53\ln|x-1|+\tfrac13\ln|x+2|+C$.
16.23 Long-divide-style split $\dfrac{x^2}{(x-1)^3}=\dfrac{1}{x-1}+\dfrac{2}{(x-1)^2}+\dfrac{1}{(x-1)^3}$ (from $x^2=(x-1)^2+2(x-1)+1$). $\int=\ln|x-1|-\dfrac{2}{x-1}-\dfrac{1}{2(x-1)^2}+C$.
16.43 All three give $\int\dfrac{dx}{x(x+1)}=\ln|x|-\ln|x+1|+C=\ln\left|\dfrac{x}{x+1}\right|+C$. (a) partial fractions $\tfrac1x-\tfrac1{x+1}$; (c) the identity is literally that decomposition; (b) $u=x+1$ gives $\int(\tfrac1{u-1}-\tfrac1u)\,du$, the same up to the constant. ✓
Chapter 17 — Improper Integrals
17.1 $\int_1^\infty x^{-3}\,dx=\lim_{b\to\infty}\big[-\tfrac{1}{2x^2}\big]_1^b=0-(-\tfrac12)=\tfrac12$ (converges; $p=3>1$).
17.3 $\int_1^b x^{-1/2}\,dx=[2\sqrt x]_1^b=2\sqrt b-2\to\infty$. Diverges: with $p=\tfrac12\le 1$, the tail $1/\sqrt x$ decays too slowly for its accumulated area to stay finite.
17.5 $u=x^2$, $du=2x\,dx$: $\int_0^\infty xe^{-x^2}\,dx=\tfrac12\int_0^\infty e^{-u}\,du=\tfrac12[-e^{-u}]_0^\infty=\tfrac12$.
17.9 Parts ($u=x$, $dv=e^{-x}dx$): $\int_0^\infty xe^{-x}\,dx=[-(x+1)e^{-x}]_0^\infty=0-(-1)=1$. This equals $\Gamma(2)=1!=1$. ✓
17.17 (a) $p=\tfrac32>1$ converges; (b) $p=\tfrac12\le1$ diverges; (c) at $0$, $p=\tfrac32\ge1$ diverges; (d) at $0$, $p=\tfrac12<1$ converges. (Tail rule: converge iff $p>1$; near-zero rule: converge iff $p<1$.)
17.29 Using $\Gamma(s+1)=s\Gamma(s)$ and $\Gamma(\tfrac12)=\sqrt\pi$: $\Gamma(\tfrac32)=\tfrac12\sqrt\pi$; $\Gamma(\tfrac52)=\tfrac32\Gamma(\tfrac32)=\tfrac34\sqrt\pi$; $\Gamma(\tfrac72)=\tfrac52\Gamma(\tfrac52)=\tfrac{15}{8}\sqrt\pi$.
17.31 $u=x/\sqrt2$, $dx=\sqrt2\,du$: $\int_{-\infty}^\infty e^{-x^2/2}\,dx=\sqrt2\int_{-\infty}^\infty e^{-u^2}\,du=\sqrt2\cdot\sqrt\pi=\sqrt{2\pi}$. The normal density divides by exactly this total area, so $\tfrac{1}{\sqrt{2\pi}}e^{-x^2/2}$ integrates to $1$.
Chapter 18 — Applications of Integration
18.1 $y=2x$ on top, $y=x^2$ below; intersect at $x=0,2$. $A=\int_0^2(2x-x^2)\,dx=[x^2-\tfrac{x^3}{3}]_0^2=4-\tfrac83=\tfrac43$.
18.3 $y=x^3-x=x(x-1)(x+1)$ crosses at $-1,0,1$. By symmetry total area $=2\int_0^1|x^3-x|\,dx=2\int_0^1(x-x^3)\,dx=2[\tfrac{x^2}{2}-\tfrac{x^4}{4}]_0^1=2(\tfrac14)=\tfrac12$.
18.9 Disks about the $x$-axis: $V=\pi\int_0^4(\sqrt x)^2\,dx=\pi\int_0^4 x\,dx=\pi[\tfrac{x^2}{2}]_0^4=8\pi$.
18.15 Shells about the $y$-axis: $V=\int_0^1 2\pi x\cdot x^2\,dx=2\pi\int_0^1 x^3\,dx=2\pi\cdot\tfrac14=\tfrac{\pi}{2}$.
18.27 $W=\int_0^{0.2}kx\,dx=\tfrac12 k L^2=\tfrac12(100)(0.2)^2=\tfrac12(100)(0.04)=2\ \mathrm{J}$.
18.29 $W=\rho g\pi\int_0^4(\tfrac34 y)^2(4-y)\,dy=\rho g\pi\cdot\tfrac{9}{16}\int_0^4(4y^2-y^3)\,dy$. The integral $=[\tfrac{4y^3}{3}-\tfrac{y^4}{4}]_0^4=\tfrac{256}{3}-64=\tfrac{64}{3}$. So $W=\rho g\pi\cdot\tfrac{9}{16}\cdot\tfrac{64}{3}=12\pi\rho g$. With $\rho=1000$, $g=9.81$: $W=12\pi(9810)\approx 3.70\times10^5\ \mathrm{J}$.
18.39 $D(q^*)=p^*$: $100-5q=50\Rightarrow q^*=10$. Consumer surplus $=\int_0^{10}[(100-5q)-50]\,dq=\int_0^{10}(50-5q)\,dq=[50q-\tfrac52 q^2]_0^{10}=500-250=\$250$.
Chapter 19 — Differential Equations
19.3 Separable: $y\,dy=x\,dx\Rightarrow\tfrac{y^2}{2}=\tfrac{x^2}{2}+C$. With $y(0)=1$: $C=\tfrac12$, so $y^2=x^2+1$, $y=\sqrt{x^2+1}$. Check: $y'=x/\sqrt{x^2+1}=x/y$. ✓
19.9 $y'+y=e^{-x}$; integrating factor $\mu=e^{\int 1\,dx}=e^x$. Then $(e^x y)'=e^x e^{-x}=1$, so $e^x y=x+C$ and $y=(x+C)e^{-x}$.
19.15 $y'=x-y$ slopes: at $(0,0)$, $0$; at $(1,0)$, $1$; at $(0,1)$, $-1$; at $(2,1)$, $1$. The slope is $0$ exactly on the line $y=x$ (where $x-y=0$) — not $y=x-1$; the isocline for slope $0$ is $y=x$.
19.19 Euler, $y'=y$, $h=0.5$: $y_1=1+0.5(1)=1.5$; $y_2=1.5+0.5(1.5)=2.25$. Versus exact $e\approx2.718$, Euler under-shoots because the convex solution curves away from each straight tangent step.
19.25 $0.25=e^{-kt}$ with $k=\ln2/5730$. So $kt=\ln 4=2\ln 2$, giving $t=2\cdot5730=11{,}460$ yr (two half-lives, since $25\%=(\tfrac12)^2$).
19.31 $T-20=(70-20)$ at $t=10$ from gap $90-20=70$: $70e^{-10k}=50\Rightarrow k=\tfrac1{10}\ln(7/5)$. Reach $30^\circ$: $70e^{-kt}=10\Rightarrow e^{-kt}=1/7\Rightarrow t=\dfrac{\ln 7}{k}=\dfrac{10\ln 7}{\ln(7/5)}\approx\dfrac{10(1.9459)}{0.3365}\approx57.8$ min.
19.37 $R_0=\beta N/\gamma=(3\times10^{-4})(1000)/0.1=0.3/0.1=3$. Since $R_0=3>1$, the epidemic grows initially (each infective produces more than one new case).
Chapter 20 — Sequences
20.1 Terms: $1,-\tfrac14,\tfrac19,-\tfrac1{16},\tfrac1{25}$. Since $|a_n|=1/n^2\to0$, the sequence converges to $0$.
20.3 (a) converges (constant $5$); (b) diverges ($(-1)^n n$ unbounded); (c) converges ($\tfrac{n}{n+1}\to1$); (d) diverges ($\cos\pi n=(-1)^n$ oscillates between $\pm1$).
20.7 Divide by $n$: $\dfrac{3n+1}{2n-5}=\dfrac{3+1/n}{2-5/n}\to\dfrac{3}{2}$ (limit laws, $1/n\to0$).
20.13 $n^{1/n}$: take logs, $\tfrac{\ln n}{n}\to0$ (standard limit, log grows slower than $n$), so $n^{1/n}\to e^0=1$.
20.15 $-\tfrac1n\le\dfrac{\cos n}{n}\le\tfrac1n$ since $|\cos n|\le1$. Both bounds $\to0$, so by the Squeeze Theorem $\dfrac{\cos n}{n}\to0$.
20.19 $a_n=\tfrac{n}{n+1}=1-\tfrac1{n+1}$ increases (the subtracted term shrinks) and is bounded above by $1$. By the MCT it converges; the limit is $\lim(1-\tfrac1{n+1})=1$. ✓
20.21 $a_{n+1}=\sqrt{2+a_n}$, $a_1=1$. (a) If $a_n<2$ then $a_{n+1}=\sqrt{2+a_n}<\sqrt4=2$ (induction). (b) $a_{n+1}-a_n=\sqrt{2+a_n}-a_n>0$ on $[1,2)$ (check $2+a>a^2$ there), so increasing. (c) Bounded above and increasing $\Rightarrow$ converges; $L=\sqrt{2+L}\Rightarrow L^2-L-2=0\Rightarrow L=2$.
20.29 (a) $r_n=F_{n+1}/F_n=(F_n+F_{n-1})/F_n=1+F_{n-1}/F_n=1+1/r_{n-1}$. (b) If $r_n\to L$: $L=1+1/L\Rightarrow L^2-L-1=0\Rightarrow L=\tfrac{1+\sqrt5}{2}=\varphi$ (taking the positive root).
Chapter 21 — Series
21.1 Converges. $\lim_{N\to\infty}\dfrac{2N}{N+3}=2$, so the sum is $2$.
21.3 Diverges, since $S_N=\dfrac{N^2}{N+1}\to\infty$. For $n\ge 2$, $a_n=\dfrac{n^2}{n+1}-\dfrac{(n-1)^2}{n}=\dfrac{n^3-(n-1)^2(n+1)}{n(n+1)}=\dfrac{n^2+n-1}{n(n+1)}$.
21.7 Geometric with first term $a=-\tfrac14$, $r=-\tfrac14$: sum $=\dfrac{-1/4}{1-(-1/4)}=\dfrac{-1/4}{5/4}=-\dfrac15$.
21.11 Split: $\sum_{n\ge1}(2/6)^n+\sum_{n\ge1}(3/6)^n=\sum(1/3)^n+\sum(1/2)^n=\dfrac{1/3}{1-1/3}+\dfrac{1/2}{1-1/2}=\tfrac12+1=\dfrac32$.
21.13 Geometric in $(x-2)$; converges for $|x-2|<1$, i.e. $1 21.19 Partial fractions: $\dfrac{2}{(n+1)(n+3)}=\dfrac{1}{n+1}-\dfrac{1}{n+3}$. Telescoping leaves $\big(\tfrac12+\tfrac13\big)=\dfrac56$. 21.21 $S_N=\sum_{n=1}^N(\sqrt{n+1}-\sqrt n)=\sqrt{N+1}-1\to\infty$, so the series diverges even though $a_n\to0$. 21.25 Converges; $p=2>1$ by the $p$-series rule (value $\pi^2/6$). 21.33 $PV=C/r=500/0.04=\$12{,}500$. The underlying geometric series has ratio $r=\dfrac{1}{1.04}\approx0.9615$. 21.37 Length removed $=\sum_{k=0}^\infty 2^k\cdot 3^{-(k+1)}=\tfrac13\sum (2/3)^k=\tfrac13\cdot\dfrac{1}{1-2/3}=1$. So the Cantor set has measure $1-1=0$. 22.1 (a) $p=3$ converges; (b) $p=\tfrac12$ diverges; (c) $p=1.001>1$ converges; (d) $p=1$ diverges (harmonic). 22.3 $\int_2^\infty\dfrac{dx}{x\ln x}$, $u=\ln x$: $\int \dfrac{du}{u}=\ln u\to\infty$. Diverges. 22.9 For $n\ge3$, $\dfrac{\ln n}{n}\ge\dfrac1n$ and $\sum\tfrac1n$ diverges, so by direct comparison the series diverges. 22.15 Ratio: $\dfrac{a_{n+1}}{a_n}=\dfrac{n+1}{n}\cdot\dfrac{1}{3}\to\dfrac13<1$. Converges. 22.17 $\dfrac{a_{n+1}}{a_n}=\dfrac{((n+1)!)^2}{(2n+2)!}\cdot\dfrac{(2n)!}{(n!)^2}=\dfrac{(n+1)^2}{(2n+2)(2n+1)}\to\dfrac14<1$. Converges. 22.21 $b_n=1/\sqrt n$ is positive, decreasing, and $\to0$, so by the alternating series test it converges (conditionally). 22.23 $S_4=1-\tfrac18+\tfrac1{27}-\tfrac1{64}=0.8964$ (approx). Error $\le b_5=1/125=0.008$; since the next term is $+$, the true value exceeds $S_4$, so the error is positive. 22.27 $\sum 1/n^2$ converges ($p=2$), so the series is absolutely convergent. 22.29 $\sum 1/n^{3/2}$ converges ($p=3/2>1$), so absolutely convergent. 22.35 $a_n=\dfrac{n^2+1}{n^3+5}\sim\dfrac1n$; limit comparison with $1/n$ gives ratio $1$, and $\sum 1/n$ diverges, so the series diverges. 22.37(c) Root test on $\big(\tfrac{\ln n}{n}\big)^n$: $\sqrt[n]{a_n}=\dfrac{\ln n}{n}\to0<1$. Converges. A1. $R=\lim|c_n/c_{n+1}|=\lim\big|3^{-n}/3^{-(n+1)}\big|=3$. A3. $R=\lim\dfrac{n!}{(n+1)!}=\lim\dfrac{1}{n+1}=0$; the series converges only at $x=0$. B1. $\sin(3x)=\sum_{n=0}^\infty\dfrac{(-1)^n(3x)^{2n+1}}{(2n+1)!}$. B3. Substitute $-x^3$ into $\tfrac{1}{1-u}$: $\dfrac{1}{1+x^3}=\sum_{n=0}^\infty(-1)^n x^{3n}$, radius $R=1$. C3. $f=\sqrt x$, $a=4$: $f(4)=2$, $f'=\tfrac{1}{2\sqrt x}\Rightarrow\tfrac14$, $f''=-\tfrac14 x^{-3/2}\Rightarrow-\tfrac{1}{32}$. $T_2=2+\tfrac14(x-4)-\tfrac{1}{64}(x-4)^2$. D3. $\ln(1.1)$: $x=0.1$, first two terms $0.1-\tfrac{0.01}{2}=0.095$. Error $<$ first omitted term $\dfrac{(0.1)^3}{3}\approx3.3\times10^{-4}$. E4. $\dfrac{1-\cos x}{x^2}=\dfrac{1}{x^2}\Big(\tfrac{x^2}{2}-\tfrac{x^4}{24}+\cdots\Big)=\tfrac12-\tfrac{x^2}{24}+\cdots\to\dfrac12$. F1. Substitute $-x^2$ into $e^u$: $e^{-x^2}=\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n}}{n!}$. G3. All derivatives of $\cos$ are $\pm\sin,\pm\cos$, so $|f^{(N+1)}|\le1=M$. Then $|R_N(x)|\le\dfrac{|x|^{N+1}}{(N+1)!}\to0$ as $N\to\infty$ for every fixed $x$ (factorial beats any power). The bound $M=1$ is used precisely because all derivatives are bounded by $1$. A1. $\theta=0\!:1$; $\theta=\pi/2\!:i$; $\theta=\pi\!:-1$; $\theta=3\pi/2\!:-i$ — the four points $1,i,-1,-i$ on the unit circle. A3. $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1,\ i^5=i,\ i^6=-1$. The period-4 cycle reproduces the alternating sign/swap pattern of the $\cos/\sin$ series, which is what lets $e^{i\theta}$ split into $\cos\theta+i\sin\theta$. A5. $e^{2\pi i}=1$ (at angle $0$); $e^{-i\pi}=-1$ (at angle $\pi$). Both have modulus $1$. A7. Since $e^{i\theta}=-1$ whenever $\theta=\pi+2\pi k$, $\ln(-1)=i\pi(2k+1)$ for any integer $k$ — infinitely many values, hence multi-valued. B3. $|z|=\sqrt2$, $\arg z=\pi/4$, so $1+i=\sqrt2\,e^{i\pi/4}$. C1. $\sin(0.3)\approx0.3-\tfrac{0.3^3}{6}+\tfrac{0.3^5}{120}-\tfrac{0.3^7}{5040}\approx0.295520$. Alternating-series error $<$ next term $\tfrac{0.3^9}{9!}\approx5\times10^{-11}$. C5. $\dfrac{x^{11}}{11!}$ at $x=0.5$: $\dfrac{0.5^{11}}{39916800}=\dfrac{4.88\times10^{-4}}{3.99\times10^7}\approx1.2\times10^{-11}<10^{-9}$, confirming terms through $x^9$ suffice. D5. Two sentences: the Gibbs overshoot of about $9\%$ of the jump height persists no matter how many harmonics are added (it narrows but does not shrink). At the jump itself the series converges to the midpoint of the two one-sided limits. E1. $p=2>1$, so $\sum 1/n^2$ converges. Integral-test tail bound: $R_N<\int_N^\infty x^{-2}dx=1/N$; for two-decimal accuracy ($<5\times10^{-3}$) take $N\ge200$. E3. $G(s)=e^{\lambda(s-1)}$, $G'(s)=\lambda e^{\lambda(s-1)}$, so $G'(1)=\lambda=E[X]$; and $G(1)=e^{0}=1$. A3. $t=x-1$, so $y=(x-1)^2$, a parabola; $x$ ranges over all reals (no restriction), $y\ge0$. A5. Eliminating gives $y=x$, but since $\sin t\in[-1,1]$, only the segment from $(-1,-1)$ to $(1,1)$ is traced; the parameter range caps both coordinates. B1. $\dfrac{dy}{dx}=\dfrac{\dot y}{\dot x}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}$ (for $t\ne0$). B3. $\dot x=-\sin t$, $\dot y=\cos t$, so $\dfrac{dy}{dx}=-\cot t$; at $t=\pi/6$ this is $-\cot(\pi/6)=-\sqrt3$. C3. $\dot x=1$, $\dot y=t^{1/2}$, integrand $\sqrt{1+t}$; $L=\int_0^3\sqrt{1+t}\,dt=\tfrac23[(1+t)^{3/2}]_0^3=\tfrac23(8-1)=\dfrac{14}{3}$. C5. $\dot x=r(1-\cos t)$, $\dot y=r\sin t$; speed $=r\sqrt{2-2\cos t}=2r\big|\sin\tfrac t2\big|$. $L=\int_0^{2\pi}2r\sin\tfrac t2\,dt=2r[-2\cos\tfrac t2]_0^{2\pi}=8r$. E1. $A=\int_0^{2\pi} y\,\dot x\,dt=\int_0^{2\pi} r(1-\cos t)\cdot r(1-\cos t)\,dt=r^2\int_0^{2\pi}(1-\cos t)^2dt=r^2\cdot3\pi=3\pi r^2$. F2. $t_{\text{flight}}=\dfrac{2v_0\sin\theta}{g}=\dfrac{60\sin40^\circ}{9.8}\approx3.94$ s; $R=\dfrac{v_0^2\sin80^\circ}{g}\approx\dfrac{900(0.985)}{9.8}\approx90.4$ m. G3. $\theta_1'=e^{-t}(-\cos t-\sin t)$, $\theta_2'=e^{-t}(\cos t-\sin t)$; $L=\int_0^4\sqrt{\theta_1'^2+\theta_2'^2}\,dt=\int_0^4 e^{-t}\sqrt2\,dt=\sqrt2(1-e^{-4})$. At $t=0$: $\dfrac{d\theta_2}{d\theta_1}=\dfrac{\cos0-\sin0}{-\cos0-\sin0}=\dfrac{1}{-1}=-1$. 26.1 $x=3\cos\tfrac\pi4=\tfrac{3\sqrt2}{2}$, $y=3\sin\tfrac\pi4=\tfrac{3\sqrt2}{2}$. Point $\big(\tfrac{3\sqrt2}{2},\tfrac{3\sqrt2}{2}\big)$. 26.3 Negative $r$: $x=-2\cos\tfrac\pi6=-\sqrt3$, $y=-2\sin\tfrac\pi6=-1$. Point $(-\sqrt3,-1)$. 26.5 $(0,-3)$: $r=3$, and the point is on the negative $y$-axis, so $\theta=\tfrac{3\pi}{2}$. Polar $(3,\tfrac{3\pi}{2})$. 26.9 Multiply by $r$: $r^2=2r\cos\theta\Rightarrow x^2+y^2=2x\Rightarrow(x-1)^2+y^2=1$. Circle, center $(1,0)$, radius $1$. 26.15 Rose $\cos(k\theta)$/$\sin(k\theta)$: $k$ odd $\Rightarrow k$ petals, $k$ even $\Rightarrow 2k$ petals. (a) $3$; (b) $8$; (c) $5$; (d) $12$. 26.19 $r=1+\cos\theta$, $y=(1+\cos\theta)\sin\theta$: $\dfrac{dy}{d\theta}=2\cos^2\theta+\cos\theta-1=(2\cos\theta-1)(\cos\theta+1)=0$. So $\cos\theta=\tfrac12$ ($\theta=\tfrac\pi3,\tfrac{5\pi}{3}$) or $\cos\theta=-1$ ($\theta=\pi$, the cusp). Horizontal tangents at $\theta=\tfrac\pi3,\tfrac{5\pi}{3}$. 26.23 $A=\tfrac12\int_0^{2\pi}(1+\cos\theta)^2d\theta=\tfrac12\int_0^{2\pi}\big(1+2\cos\theta+\tfrac{1+\cos2\theta}{2}\big)d\theta=\tfrac12\big(2\pi+0+\pi\big)=\dfrac{3\pi}{2}$. 26.25 One petal of $r=2\sin2\theta$ over $[0,\tfrac\pi2]$: $A=\tfrac12\int_0^{\pi/2}4\sin^2 2\theta\,d\theta=2\cdot\tfrac\pi4=\tfrac\pi2$. Four petals: total $=2\pi$. 26.27 Lemniscate $r^2=2\cos2\theta$, one lobe on $[-\tfrac\pi4,\tfrac\pi4]$: lobe area $=\tfrac12\int_{-\pi/4}^{\pi/4}2\cos2\theta\,d\theta=[\tfrac12\sin2\theta]_{-\pi/4}^{\pi/4}=\tfrac12(1-(-1))=1$. Both lobes: total $=2$. 26.33(b) $L=\dfrac{a}{2}\big[\Theta\sqrt{1+\Theta^2}+\ln(\Theta+\sqrt{1+\Theta^2})\big]$ with $a=2,\Theta=4\pi\approx12.566$: $L\approx 1\cdot[12.566\cdot12.606+\ln(25.17)]\approx158.4+3.23\approx162$ m. A1. (a) ellipse, center $(0,0)$; (b) hyperbola, center $(0,0)$; (c) parabola, vertex $(0,0)$ (opens up); (d) circle, center $(0,0)$, radius $6$. A3. $\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$: $a=3$, $b=4$, vertices $(0,\pm3)$, branches open along the $y$-axis. B1. $a^2=169,b^2=144$, $c=\sqrt{169-144}=5$. Foci $(\pm5,0)$; $e=c/a=5/13$. B3. $y^2=-20x$: $4p=-20\Rightarrow p=-5$. Focus $(-5,0)$, directrix $x=5$, opens left. C1. Foci $(\pm3,0)$, vertex $(5,0)$: $a=5$, $c=3$, $b^2=a^2-c^2=16$. $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$. C3. Hyperbola, foci $(\pm5,0)$, $e=5/3$: $c=5$, $a=c/e=3$, $b^2=c^2-a^2=16$. $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$. C5. Set $x=c$ in $\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1$: $y^2=b^2(1-c^2/a^2)=b^2\cdot\tfrac{b^2}{a^2}$, so $y=\pm\tfrac{b^2}{a}$. Chord length $=\dfrac{2b^2}{a}$. D3. $y^2-2y-4x+13=0\Rightarrow(y-1)^2=4x-12=4(x-3)$. Parabola, vertex $(3,1)$, $4p=4\Rightarrow p=1$: focus $(4,1)$, directrix $x=2$. E5. $x^2=4py\Rightarrow 2x=4p\,y'\Rightarrow y'=\tfrac{x_0}{2p}$. Tangent: $y-y_0=\tfrac{x_0}{2p}(x-x_0)\Rightarrow 2p\,y-2p\,y_0=x_0 x-x_0^2$. Using $x_0^2=4py_0$: $x x_0=2p(y+y_0)$. F1. Mercury: perihelion $a(1-e)=0.387(0.794)\approx0.307$ AU; aphelion $a(1+e)=0.387(1.206)\approx0.467$ AU. A1. $\mathbf r(t)=\langle2+t,\,-1+4t,\,3-2t\rangle$. At $t=2$: $\langle4,7,-1\rangle$. A3. $z=t$ rises from $0$ to $2\pi$ as $t$ goes $0\to2\pi$, so one full turn rises by $2\pi$. B1. $\mathbf r'(t)=\langle3t^2,\,-2\sin2t,\,-e^{-t}\rangle$. B3. $\int_0^1\langle t^2,4t^3,1\rangle dt=\langle\tfrac13,\,1,\,1\rangle$. A definite integral evaluates each component to a number, giving a fixed vector. C1. $\mathbf r'=\langle2t,2,1/t\rangle$, at $t=1$ velocity $\langle2,2,1\rangle$, speed $\sqrt{4+4+1}=3$; $\mathbf r''=\langle2,0,-1/t^2\rangle$, accel $\langle2,0,-1\rangle$. D1. $\mathbf r'=\langle-3\sin t,3\cos t,4\rangle$, speed $\sqrt{9+16}=5$. $L=\int_0^{2\pi}5\,dt=10\pi$. E1. Circle of radius $7$: $\kappa=1/R=1/7$ at every point, confirming the curvature of a circle is the reciprocal of its radius. E7. Twisted cubic $\langle t,t^2,t^3\rangle$: $\mathbf r'=\langle1,2t,3t^2\rangle$, $\mathbf r''=\langle0,2,6t\rangle$, $\mathbf r'\times\mathbf r''=\langle6t^2,-6t,2\rangle$, magnitude $\sqrt{36t^4+36t^2+4}$. $\kappa=\dfrac{\sqrt{36t^4+36t^2+4}}{(1+4t^2+9t^4)^{3/2}}$; at $t=0$, $\kappa=2$, the maximum. F1. $a_T=\dfrac{\mathbf v\cdot\mathbf a}{|\mathbf v|}=\dfrac{2+1+2}{\sqrt{4+1+4}}=\dfrac{5}{3}$. F5. $a_N=v^2/R=400/50=8$ m/s²$=8/9.8\approx0.82\,g$ — far above the $0.1$–$0.15\,g$ comfort range; the turn is too tight for that speed. A1. Need $x^2+y^2-4\ge0$: domain is $\{(x,y):x^2+y^2\ge4\}$, the plane outside (and on) the circle of radius $2$. A3. $\dfrac{x+y}{x-y}$ defined where $x-y\ne0$: all of $\mathbb R^2$ except the line $y=x$. B1. $z=3-x-y$ is a plane (slope $-1$ in both $x$ and $y$), $z$-intercept $3$. C1. Polynomial, continuous everywhere, so just substitute: $1\cdot2+3\cdot1\cdot2-5=2+6-5=3$. No path analysis needed for continuous functions. C3. Along $y=mx$: $\dfrac{x\cdot mx}{x^2+m^2x^2}=\dfrac{m}{1+m^2}$, which depends on $m$ (e.g. $0$ on $y=0$, $\tfrac12$ on $y=x$). Limit does not exist. D1. $f_x=6x-4y$, $f_y=-4x+3y^2$. D5. $f_x=2xy\cos(x^2y)+\dfrac{1}{x+y}$, $f_y=x^2\cos(x^2y)+\dfrac{1}{x+y}$. E1. $f=x^3y^2$: $f_{xx}=6xy^2$, $f_{yy}=2x^3$. F1. $z=x^2+y^2$ at $(1,2)$, $z_0=5$: $f_x=2x=2$, $f_y=2y=4$. Plane $z=5+2(x-1)+4(y-2)$. H1. $Q=10L^{0.4}K^{0.6}$: $Q_L=4L^{-0.6}K^{0.6}$, $Q_K=6L^{0.4}K^{-0.4}$. J3. $u=\tfrac12\ln(x^2+y^2)$: $u_x=\dfrac{x}{x^2+y^2}$, $u_{xx}=\dfrac{(x^2+y^2)-x\cdot2x}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$; by symmetry $u_{yy}=\dfrac{x^2-y^2}{(x^2+y^2)^2}$. Sum $=0$, so $u$ is harmonic. 1. $\dfrac{dz}{dt}=z_x\dot x+z_y\dot y=2x\cdot1+1\cdot3t^2=2t+3t^2$. Check: $z=t^2+t^3\Rightarrow z'=2t+3t^2$. ✓ 3. $\dfrac{dz}{dt}=\cos(xy)\cdot(y\dot x+x\dot y)=\cos(xy)(y\cdot2t+x\cdot1)$; with $x=t^2,y=t$ this is $\cos(t^3)\cdot3t^2$. 7. Tree has $w\to x,y$ and each $\to u,v$. $\dfrac{\partial w}{\partial u}=f_x\dfrac{\partial x}{\partial u}+f_y\dfrac{\partial y}{\partial u}$. 15. $\nabla f=\langle yz,xz,xy\rangle$; at $(1,2,3)$: $\langle6,3,2\rangle$. 17. $f=\dfrac{x}{x^2+y^2}$: $f_x=\dfrac{(x^2+y^2)-x\cdot2x}{(x^2+y^2)^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$, $f_y=\dfrac{-2xy}{(x^2+y^2)^2}$. At $(1,1)$: $\langle0,-\tfrac12\rangle$. 19. $\nabla f=\langle2x-2,\,2y+4\rangle=\mathbf0\Rightarrow x=1,y=-2$. Single critical point $(1,-2)$. 21. $\nabla f=\langle2xy,x^2\rangle=\langle4,4\rangle$ at $(2,1)$; $\hat u=\tfrac15\langle3,4\rangle$. $D_{\mathbf u}f=\tfrac15(12+16)=\dfrac{28}{5}=5.6$. 25. $D_{\langle1,0\rangle}f=f_x=6$. $D_{\mathbf v}f=\tfrac{1}{\sqrt2}(f_x+f_y)=10\Rightarrow f_x+f_y=10\sqrt2\Rightarrow f_y=10\sqrt2-6$. So $\nabla f=\langle6,\,10\sqrt2-6\rangle$. 27. $\nabla T=\langle-2x,-4y\rangle=\langle-4,-4\rangle$ at $(2,1)$; crawl in direction $\langle-4,-4\rangle$ (i.e. $\langle-1,-1\rangle$). Max rate $=|\nabla T|=\sqrt{32}=4\sqrt2\approx5.66$ per unit distance. 31. $F=x^2+y^2+z^2$, $\nabla F=\langle2x,2y,2z\rangle=\langle4,2,4\rangle$ at $(2,1,2)$. Tangent plane $4(x-2)+2(y-1)+4(z-2)=0$, i.e. $2x+y+2z=9$. Normal line $(2,1,2)+s\langle2,1,2\rangle$. A5 $f_x = 3x^2 - 3 = 0$ gives $x = \pm 1$, and $f_y = 2y = 0$ gives $y = 0$. The two critical points are $(1,0)$ and $(-1,0)$. A6 Setting $f_x = 3x^2 - 3y = 0$ and $f_y = 3y^2 - 3x = 0$ gives $y = x^2$ and $x = y^2$, so $x = x^4$, hence $x = 0$ or $x = 1$. The two real critical points are $(0,0)$ and $(1,1)$. B5 With $f_{xx} = 6x$, $f_{yy} = 2$, $f_{xy} = 0$, the discriminant is $D = 12x$. At $(1,0)$, $D = 12 > 0$ and $f_{xx} > 0$, a local minimum; at $(-1,0)$, $D = -12 < 0$, a saddle point. B9 $f = x^4 + y^4$: $f_x = 4x^3$, $f_y = 4y^3$ vanish only at the origin, where $f_{xx} = f_{yy} = f_{xy} = 0$ so $D = 0$ (test inconclusive). Directly, $f(x,y) \ge 0 = f(0,0)$ for all $(x,y)$, so the origin is a (global) minimum. C1 Interior: $\nabla f = \langle 2x-2,\,2y\rangle = \mathbf 0$ at $(1,0)$, where $f = -1$. Boundary $x^2+y^2=4$: $f = 4 - 2x$ with $-2\le x\le 2$, giving max $f=8$ at $(-2,0)$ and min $f=0$ at $(2,0)$. Absolute min $-1$ at $(1,0)$, absolute max $8$ at $(-2,0)$. D1 $\nabla f = \lambda\nabla g$: $2x = \lambda$, $2y = 2\lambda$, so $y = 2x$. The constraint $x + 2y = 5$ gives $x = 1$, $y = 2$, and $f = 1 + 4 = 5$ — the squared distance from the origin to the line, matching $25/(1^2+2^2)=5$. D5 Minimizing $S = xy + 2xz + 2yz$ with $xyz = 32$ yields $x = y = 4$, $z = 2$ (so $z = x/2$, a square base twice as wide as tall). Check: $xyz = 32$ ✓ and $S = 16 + 16 + 16 = 48\,\text{m}^2$. E4 Normal equations from $(0,1),(1,1),(2,4)$ with $n=3$, $\sum x = 3$, $\sum x^2 = 5$, $\sum y = 6$, $\sum xy = 9$: $\;5m + 3c = 9$, $3m + 3c = 6$. Subtracting, $2m = 3$, so $m = \tfrac32$, $c = \tfrac12$. The Hessian $\nabla^2 S$ has positive diagonal and $D = 4(\sum x^2 \cdot n - (\sum x)^2)>0$, so the fit is the global minimum. 1 $\displaystyle\int_0^2\!\int_0^3 (x+y)\,dy\,dx = \int_0^2\!\big(3x + \tfrac92\big)\,dx = 6 + 9 = 15$. 3 $\displaystyle\int_0^1\!\int_0^2 (2x+4y)\,dy\,dx = \int_0^1 (4x + 8)\,dx = 2 + 8 = 10$. 15 Reversing to $0\le x\le 1$, $0\le y\le x$: $\displaystyle\int_0^1 e^{x^2}\!\!\int_0^x dy\,dx = \int_0^1 x\,e^{x^2}\,dx = \tfrac12(e-1)\approx 0.859$. 25 One petal: area $= \tfrac12\displaystyle\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta = \tfrac12\cdot\tfrac{\pi}{4} = \dfrac{\pi}{8}$. 27 Polar: $\displaystyle\int_0^{2\pi}\!\int_0^3 (9 - r^2)\,r\,dr\,d\theta = 2\pi\Big[\tfrac{9r^2}{2} - \tfrac{r^4}{4}\Big]_0^3 = 2\pi\cdot\tfrac{81}{4} = \dfrac{81\pi}{2}$. 29 $\displaystyle\iiint_E z\,dV = \Big(\int_0^1\!dx\Big)\Big(\int_0^2\!dy\Big)\Big(\int_0^3 z\,dz\Big) = 1\cdot 2\cdot\tfrac92 = 9$. 31 Cylindrical: $\displaystyle\int_0^{2\pi}\!\int_0^1 (4 - r^2)\,r\,dr\,d\theta = 2\pi\big(2 - \tfrac14\big) = \dfrac{7\pi}{2}$. 33 Density $\propto$ distance, $\delta = k\rho$: $M = \displaystyle\int_0^{2\pi}\!\int_0^\pi\!\int_0^R k\rho\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = k\cdot\tfrac{R^4}{4}\cdot 2\cdot 2\pi = \pi k R^4$. A1 $\det J = \begin{vmatrix}3 & -1\\ 1 & 2\end{vmatrix} = 6 + 1 = 7$. Area is expanded by a factor of $7$. A3 $\det J = \begin{vmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r$, confirming the polar factor. A5 $\det J = \begin{vmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{vmatrix} = \cos^2\alpha + \sin^2\alpha = 1$. A rotation is a rigid motion, so areas are unchanged — geometrically obvious. B3 With $x = 2u$, $y = 3v$, $\det J = 6$ and the ellipse maps to the unit disk. Area $= \iint_{u^2+v^2\le1} 6\,du\,dv = 6\pi = \pi\,(2)(3) = \pi ab$. ✓ D1 Spherical: $V = \displaystyle\int_0^{2\pi}\!\int_0^\pi\!\int_0^a \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot 2\cdot\tfrac{a^3}{3} = \dfrac{4}{3}\pi a^3$. ✓ D4 The map $x=u,\;y=u+v,\;z=u+v+w$ has lower-triangular $J$ with $1$'s on the diagonal, so $\det J = 1$ and $\text{Vol}(R) = 1\cdot\text{Vol}([0,1]^3) = 1$. F1 Using $r^2 = \rho^2\sin^2\phi$ and $\delta = M/(\tfrac43\pi a^3)$: $I = \delta\!\int_0^{2\pi}\!\int_0^\pi\!\int_0^a \rho^2\sin^2\phi\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = \delta\cdot 2\pi\cdot\tfrac43\cdot\tfrac{a^5}{5} = \tfrac{2}{5}Ma^2$. ✓ 3 $\mathbf F = \langle -y,x\rangle$ gives $(0,1), (-1,0), (0,-1), (1,0)$ at the four points — each arrow is the position vector turned $90°$ CCW, so the circulation is counterclockwise. 7 $\nabla f = \langle 2x, 2y\rangle$: a radial source field pointing outward, with magnitude $2r$ growing with distance from the origin. 17 $\nabla\cdot\langle x^2, y^2, z^2\rangle = 2x + 2y + 2z$; at $(1,-1,2)$ this is $2 - 2 + 4 = 4$. 19 $\nabla\cdot\langle x^2y, -2xy\rangle = 2xy - 2x = 2x(y-1)$. At $(1,3)$: $2(1)(2) = 4 > 0$, a source. At $(2,0)$: $2(2)(-1) = -4 < 0$, a sink. 27 Curl test on $\mathbf F = \langle 2xy, x^2\rangle$: $Q_x = 2x = P_y$ ✓ conservative. A potential is $f = x^2 y$ (since $f_x = 2xy$, $f_y = x^2$). 29 $\mathbf F = \langle e^x\cos y, -e^x\sin y\rangle$: $Q_x = -e^x\sin y = P_y$ ✓ conservative. Potential $f = e^x\cos y$. 31 $\mathbf F = \langle -y, x\rangle$: $Q_x - P_y = 1 - (-1) = 2 \ne 0$, so not conservative. A circulating field has nonzero curl, and a gradient must be curl-free, so it cannot be a gradient. A1 $\mathbf r(t)=\langle t,2t\rangle$, $|\mathbf r'| = \sqrt5$, integrand $3t$: $\displaystyle\int_0^1 3t\sqrt5\,dt = \tfrac{3\sqrt5}{2}$. A3 Helix: $|\mathbf r'| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt2$, so length $= \displaystyle\int_0^{2\pi}\sqrt2\,dt = 2\sqrt2\,\pi$. B3 $\mathbf F = \langle y,-x\rangle$ on the unit circle: $\mathbf F\cdot\mathbf r' = -\sin^2 t - \cos^2 t = -1$, so $\oint = \displaystyle\int_0^{2\pi}(-1)\,dt = -2\pi$. C3 $\mathbf F = \langle 2xy, x^2 + 2y\rangle$ is conservative ($Q_x = 2x = P_y$) with $f = x^2 y + y^2$; thus $\int_C = f(2,3) - f(1,1) = (12+9) - (1+1) = 19$. E3 Green's theorem with $P = y^2 - x$, $Q = 3xy$: $Q_x - P_y = 3y - 2y = y$. Over the triangle, $\displaystyle\int_0^1\!\int_0^x y\,dy\,dx = \int_0^1\tfrac{x^2}{2}\,dx = \tfrac16$. E5 Green's theorem: $Q_x = 7$, $P_y = 3$, so $Q_x - P_y = 4$ and $\oint_C = \iint_D 4\,dA = 4\cdot 9\pi = 36\pi$ (disk of radius $3$); the $e^{\sin x}$ and $\sqrt{y^4+1}$ terms drop out under differentiation. F1 Area formula: $x\,dy - y\,dx = ab\cos^2 t + ab\sin^2 t = ab$, so Area $= \tfrac12\displaystyle\int_0^{2\pi}ab\,dt = \pi ab$. A1 Cylinder: $\mathbf r(\theta,z) = \langle R\cos\theta,\,R\sin\theta,\,z\rangle$, with $0\le\theta<2\pi$, $0\le z\le h$. A3 Plane through the origin spanned by $\mathbf r_u = \langle 2,0,1\rangle$, $\mathbf r_v = \langle 0,3,-1\rangle$; a normal is $\mathbf r_u\times\mathbf r_v = \langle -3, 2, 6\rangle$. B3 $z = 2x + 2y + 1$: $\sqrt{1 + 4 + 4} = 3$, so Area $= \iint_D 3\,dA = 3\cdot\pi(1)^2 = 3\pi$. C1 $\iint_S 1\,dS$ is the hemisphere's area: half of $4\pi(2)^2 = 16\pi$, i.e. $8\pi$. C3 $M = \displaystyle\int_0^{2\pi}\!\int_0^\pi \sigma_0\cos^2\phi\cdot R^2\sin\phi\,d\phi\,d\theta$; inner $\int_0^\pi\cos^2\phi\sin\phi\,d\phi = \tfrac23$, so $M = \sigma_0 R^2\cdot\tfrac23\cdot 2\pi = \dfrac{4\pi\sigma_0 R^2}{3}$. E3 On the sphere $\hat{\mathbf n} = \mathbf r/R$, so $\mathbf F\cdot\hat{\mathbf n} = \mathbf r\cdot\mathbf r/R = R$; flux $= R\cdot 4\pi R^2 = 4\pi R^3$. E5 Upward flux of $\langle -y,x,z\rangle$ through $z = 1 - x^2 - y^2$: $\iint_D(2xy - 2xy + (1-x^2-y^2))\,dA = \iint_{r\le1}(1-r^2)\,dA = 2\pi(\tfrac12 - \tfrac14) = \dfrac{\pi}{2}$. F3 Heat flux $\mathbf q = -k\nabla T = 4\langle x,y,z\rangle = 4\mathbf r$; on the sphere of radius $3$, $\mathbf q\cdot\hat{\mathbf n} = 4\cdot 9/3 = 12$, so flux $= 12\cdot 4\pi(3)^2 = 432\pi$ (outward, away from the hot center). A1 $\nabla\times\langle -y, x, 0\rangle = \langle \partial_y 0 - \partial_z x,\ \partial_z(-y) - \partial_x 0,\ \partial_x x - \partial_y(-y)\rangle = \langle 0, 0, 2\rangle$. A3 $\nabla\cdot\langle x^2, y^2, z^2\rangle = 2x + 2y + 2z$. B1 Curl is $\langle 0,0,2\rangle$, so the flux side is $\iint_D 2\,dA = 2\cdot\pi(2)^2 = 8\pi$. Directly, $\oint_{\partial S}\langle -y,x\rangle\cdot d\mathbf r = \int_0^{2\pi}(2\sin^2 t + 2\cos^2 t)\cdot 2\,dt$... on radius-2 circle gives $8\pi$ as well. Both sides $= 8\pi$. ✓ D1 $\nabla\cdot\langle x,y,z\rangle = 3$, so $\iiint_E 3\,dV = 3\cdot\tfrac43\pi(2)^3 = 32\pi$. Surface side: $\mathbf F\cdot\hat{\mathbf n} = R = 2$ on the sphere, flux $= 2\cdot 4\pi(2)^2 = 32\pi$. ✓ D5 $\nabla\cdot\langle x^3,y^3,z^3\rangle = 3(x^2+y^2+z^2) = 3\rho^2$. Flux $= \displaystyle\int_0^{2\pi}\!\int_0^\pi\!\int_0^1 3\rho^2\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 3\cdot\tfrac15\cdot 2\cdot 2\pi = \dfrac{12\pi}{5}$. E3 $\nabla\cdot\mathbf F = 1 + 1 + 1 = 3$ (the $\sin(yz)$, $e^{xz}$, $xy$ pieces differentiate to $0$ in the relevant variable). Flux $= 3\cdot\tfrac43\pi(3)^3 = 3\cdot 36\pi = 108\pi$. E5 $\nabla\cdot\langle y^2 z,\ xz,\ x^2 y^2\rangle = 0 + 0 + 0 = 0$, so the outward flux through the closed hemispherical surface is $0$ by the Divergence theorem. G1 $\nabla\cdot\langle x,y,z\rangle = 3$, so $\oiint_{\partial E}\langle x,y,z\rangle\cdot d\mathbf S = \iiint_E 3\,dV = 3\,\text{Vol}(E)$, giving $\text{Vol}(E) = \tfrac13\oiint_{\partial E}\langle x,y,z\rangle\cdot d\mathbf S$. For the unit ball the flux is $4\pi$ (from E3-type result with $R=1$), so $\text{Vol} = \tfrac13\cdot 4\pi = \tfrac43\pi$. ✓ A1 (a) $0$-form (a function); (b) $1$-form; (c) $2$-form; (d) $3$-form. The degree equals the number of wedged differentials. A5 The exterior derivative $d$ reproduces (a) the gradient $\nabla f$ on a $0$-form; (b) the curl $\nabla\times\mathbf F$ on a $1$-form; (c) the divergence $\nabla\cdot\mathbf F$ on a $2$-form. A7 $d^2 = 0$: applying $d$ twice always gives zero. It encodes $\nabla\times(\nabla f) = \mathbf 0$ (curl of a gradient) and $\nabla\cdot(\nabla\times\mathbf F) = 0$ (divergence of a curl). B3 $df = 2xy\,dx + x^2\,dy - 3z^2\,dz$; its coefficients $(2xy,\,x^2,\,-3z^2)$ are exactly $\nabla f$. B5 $d\eta = (1 + 1 + 1)\,dx\wedge dy\wedge dz = 3\,dx\wedge dy\wedge dz$, reproducing $\nabla\cdot\langle x,y,z\rangle = 3$. C5 With $P = -y/(x^2+y^2)$, $Q = x/(x^2+y^2)$: both $Q_x$ and $P_y$ equal $\dfrac{y^2 - x^2}{(x^2+y^2)^2}$, so $Q_x - P_y = 0$ and $d\omega = 0$ — the vortex form is closed. D5 Closed: $d\omega = 0$; exact: $\omega = d\eta$. Every exact form is closed because $d\omega = d(d\eta) = 0$ by $d^2 = 0$. The vortex form is closed but not exact on the punctured plane (its loop integral is $2\pi\ne 0$), so the converse fails when a hole is present; $H^k = \{\text{closed}\}/\{\text{exact}\}$ counts those holes (de Rham cohomology). B4 $R_0 = \beta/\gamma$. Example: $\beta = 0.5/\text{day}$, $\gamma = 0.2/\text{day}$ gives $R_0 = 2.5$, meaning one infective produces about $2.5$ secondaries in a fully susceptible population. Since $R_0 > 1$, the outbreak takes off. (Any defensible $\beta,\gamma$ with the correct ratio earns full credit.) B5 $\dfrac{dI}{dt} = \big(\tfrac{\beta S}{N} - \gamma\big)I > 0$ exactly when $\dfrac{\beta S}{\gamma N} > 1$, i.e. $\dfrac{S}{N} > \dfrac{\gamma}{\beta} = \dfrac{1}{R_0}$. The epidemic peaks when the susceptible fraction falls to $1/R_0$ (for $R_0 = 2.5$, that is $0.4$). E4 Eliminating $\lambda$ in the Cobb–Douglas FOCs gives $\dfrac{a}{b}\dfrac{K}{L} = \dfrac{w}{r}$, hence $L^* = \dfrac{a}{a+b}\dfrac{B}{w}$, $K^* = \dfrac{b}{a+b}\dfrac{B}{r}$. With $a=b=\tfrac12$, $w=1$, $r=4$, $B=100$: $L^* = \tfrac12\cdot 100 = 50$, $K^* = \tfrac12\cdot 100/4 = 12.5$. E7 $\dfrac{\partial L^*}{\partial w} = -\dfrac{a}{a+b}\dfrac{B}{w^2} < 0$: a higher wage strictly lowers optimal labor, shifting the firm toward capital (automation) — the derivative quantifying input substitution. P2 Circular speeds $v_{\text{circ}} = \sqrt{\mu/r}$. Model answer (LEO→GEO, $\mu = 3.986\times 10^5\,\text{km}^3/\text{s}^2$, $r_1 = 6678$, $r_2 = 42164$ km): $v_1 \approx 7.73\,\text{km/s}$, $v_2 \approx 3.07\,\text{km/s}$. (Exact figures depend on the student's chosen radii.) P6 Transfer time is half the ellipse's period, $t = \pi\sqrt{a_t^3/\mu}$ with $a_t = (r_1+r_2)/2$. For the LEO→GEO numbers above, $a_t \approx 24421$ km gives $t \approx \pi\sqrt{24421^3/\mu} \approx 1.9\times 10^4\,\text{s}\approx 5.3$ hours. D3 The MSE gradient is $\dfrac{\partial L}{\partial m} = \dfrac{2}{n}\sum_i (mx_i + c - y_i)x_i$ and $\dfrac{\partial L}{\partial c} = \dfrac{2}{n}\sum_i (mx_i + c - y_i)$. Setting both to zero recovers the normal equations; gradient descent walks downhill to that unique minimum because $L$ is convex (positive-semidefinite Hessian). C1 Model answer: a complete portfolio follows the nine-part §39.9 structure — problem statement, defended assumptions, fully symbol-defined equations, a tools-used table citing home chapters, reproducible code, headline results with units, at least one independent cross-check (e.g. SIR's final-size relation $s_\infty = e^{-R_0(1-s_\infty)}$ agreeing with the simulated attack rate), explicit limitations, and a plain-language conclusion that leads with the punchline. The validation cross-check is the load-bearing requirement. 3 FTC: $\int_a^b f'(x)\,dx = f(b) - f(a)$. Every later theorem (line-integral FTC, Green's, Stokes', Divergence) has the form $\int_{\partial M}\omega = \int_M d\omega$ — the integral of a derivative over a region equals the plain values of an antiderivative-like object on the boundary. They are the same theorem in dimensions $1, 2, 3$. 8 $f' = 3x^2$; $\int_0^2 3x^2\,dx = [x^3]_0^2 = 8$, which equals $f(2) - f(0) = 8 - 0 = 8$. ✓ 11 Increasing dimension: plain FTC (Ch. 14, dim 1) → Green's theorem (Ch. 35, dim 2 in the plane) → Divergence theorem (Ch. 37, dim 3) → generalized Stokes' $\int_{\partial M}\omega = \int_M d\omega$ (Ch. 38, all dimensions). Shared idea: the integral of a derivative over a region is determined by boundary data. 13 Net displacement $= \displaystyle\int_0^3 (3t^2 - 4t)\,dt = [t^3 - 2t^2]_0^3 = 27 - 18 = 9$. 15 $f(x) = \sqrt x$ near $a = 100$, $f'(100) = 0.05$, $f''(100) = -\tfrac{1}{4000}$. (a) Linear: $10 + 0.05(1) = 10.05$. (b) Second-order: $10.05 + \tfrac12(-\tfrac{1}{4000})(1)^2 = 10.049875$. (c) The second-order value is closer to the true $10.0498\ldots$. 16 $\pi\displaystyle\int_0^\infty e^{-2x}\,dx = \pi\Big[-\tfrac12 e^{-2x}\Big]_0^\infty = \pi\cdot\tfrac12 = \dfrac{\pi}{2}$. 18 (1) Write $I^2 = \iint_{\mathbb R^2} e^{-(x^2+y^2)}\,dA$ (double integral, Ch. 32). (2) Convert to polar (Ch. 26). (3) The Jacobian factor $r$ appears, $dA = r\,dr\,d\theta$ (Ch. 33). (4) The integral becomes $\int_0^{2\pi}\!\int_0^\infty e^{-r^2} r\,dr\,d\theta = 2\pi\cdot\tfrac12 = \pi$, elementary because $r\,e^{-r^2}$ has the obvious antiderivative; hence $I = \sqrt\pi$. 20 $f'(x) = 2(x-3)$, $\eta = 0.1$, $x_0 = 0$: $x_1 = 0 - 0.1(-6) = 0.6$; $x_2 = 0.6 - 0.1(-4.8) = 1.08$; $x_3 = 1.08 - 0.1(-3.84) = 1.464$. The sequence converges to $x = 3$, the minimizer. 21 (a) At onset $\dfrac{dI}{dt} = (\beta S - \gamma)I > 0$ requires $\beta S > \gamma$. (b) With $S\approx 1$, this is $\beta/\gamma > 1$, i.e. $R_0 = \beta/\gamma$ is the threshold. (c) $\dfrac{dI}{dt}$ is the rate operation (differentiation); accumulating it over time to build the epidemic curve $I(t)$ is the integration operation — the two faces of calculus.Chapter 22 — Convergence Tests
Chapter 23 — Power and Taylor Series
Chapter 24 — Applications of Series
Chapter 25 — Parametric Curves
Chapter 26 — Polar Coordinates
Chapter 27 — Conic Sections
Chapter 28 — Vector-Valued Functions
Chapter 29 — Functions of Several Variables
Chapter 30 — Multivariable Chain Rule and Gradient
Chapter 31 — Optimization in Several Variables
Chapter 32 — Multiple Integrals
Chapter 33 — Change of Variables and Jacobians
Chapter 34 — Vector Fields
Chapter 35 — Line Integrals
Chapter 36 — Surface Integrals
Chapter 37 — Stokes' and Divergence Theorems
Chapter 38 — Generalizing the FTC
Chapter 39 — The Modeling Portfolio (Capstone — model-answer sketches)
Chapter 40 — The Big Picture (synthesis — representative answers)
How to Use These Answers