Calculus: Kinematics (Position, Velocity, Acceleration)

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=s'(t)\\v(t)=3t^2-12t+9\\[6pt]\text{Step 2: Substitute }t=2\\v(2)=3(2)^2-12(2)+9\\[6pt]\text{Step 3: Simplify}\\v(2)=12-24+9\\v(2)=-3\\[6pt]\text{Final Answer: }\boxed{-3}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=3t^2-12t+9\\[6pt]\text{Step 2: Differentiate velocity}\\a(t)=6t-12\\[6pt]\text{Step 3: Substitute }t=2\\a(2)=6(2)-12\\[6pt]\text{Step 4: Simplify}\\a(2)=12-12=0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=3t^2-12t+9\\[6pt]\text{Step 2: Set velocity equal to zero}\\3t^2-12t+9=0\\[6pt]\text{Step 3: Divide by }3\\t^2-4t+3=0\\[6pt]\text{Step 4: Factor}\\(t-1)(t-3)=0\\[6pt]\text{Step 5: Solve}\\t=1,\quad t=3\\[6pt]\text{Final Answer: }\boxed{t=1,\ 3}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity}\\v(t)=3t^2-12t+9\\[6pt]\text{Step 2: Factor}\\v(t)=3(t-1)(t-3)\\[6pt]\text{Step 3: Find zeros}\\t=1,\quad t=3\\[6pt]\text{Step 4: Test velocity signs}\\v(0)=9>0\\v(2)=-3<0\\v(4)=9>0\\[6pt]\text{Step 5: Select intervals where }v>0\\[6pt]\text{Final Answer: }\boxed{[0,1)\cup(3,\infty)}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity}\\v(t)=3t^2-12t+9\\[6pt]\text{Step 2: Factor}\\v(t)=3(t-1)(t-3)\\[6pt]\text{Step 3: Find zeros}\\t=1,\quad t=3\\[6pt]\text{Step 4: Test signs}\\v(0)>0,\quad v(2)<0,\quad v(4)>0\\[6pt]\text{Step 5: Select interval where }v<0\\[6pt]\text{Final Answer: }\boxed{(1,3)}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity}\\v(t)=3(t-1)(t-3)\\[6pt]\text{Step 2: Find acceleration}\\a(t)=6(t-2)\\[6pt]\text{Step 3: Identify boundaries}\\t=1,\quad t=2,\quad t=3\\[6pt]\text{Step 4: Compare signs}\\(0,1):\ v>0,\ a<0\\(1,2):\ v<0,\ a<0\\(2,3):\ v<0,\ a>0\\(3,\infty):\ v>0,\ a>0\\[6pt]\text{Step 5: Choose matching signs}\\[6pt]\text{Final Answer: }\boxed{(1,2)\cup(3,\infty)}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity}\\v(t)=3(t-1)(t-3)\\[6pt]\text{Step 2: Find acceleration}\\a(t)=6(t-2)\\[6pt]\text{Step 3: Identify boundaries}\\t=1,\quad t=2,\quad t=3\\[6pt]\text{Step 4: Compare signs}\\[0,1):\ v>0,\ a<0\\(1,2):\ v<0,\ a<0\\(2,3):\ v<0,\ a>0\\(3,\infty):\ v>0,\ a>0\\[6pt]\text{Step 5: Choose opposite signs}\\[6pt]\text{Final Answer: }\boxed{[0,1)\cup(2,3)}\end{array}$

$\begin{array}{l}\text{Step 1: Write displacement formula}\\\Delta s=s(4)-s(0)\\[6pt]\text{Step 2: Evaluate }s(4)\\s(4)=64-96+36+2=6\\[6pt]\text{Step 3: Evaluate }s(0)\\s(0)=2\\[6pt]\text{Step 4: Subtract}\\\Delta s=6-2=4\\[6pt]\text{Final Answer: }\boxed{4\text{ position units}}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity}\\v(t)=3(t-1)(t-3)\\[6pt]\text{Step 2: Find direction-change times}\\t=1,\quad t=3\\[6pt]\text{Step 3: Evaluate positions}\\s(0)=2,\quad s(1)=6\\s(3)=2,\quad s(4)=6\\[6pt]\text{Step 4: Add absolute position changes}\\D=|6-2|+|2-6|+|6-2|\\[6pt]\text{Step 5: Simplify}\\D=4+4+4=12\\[6pt]\text{Final Answer: }\boxed{12\text{ distance units}}\end{array}$

$\begin{array}{l}\text{Step 1: Write average velocity formula}\\\bar v=\frac{s(4)-s(0)}{4-0}\\[6pt]\text{Step 2: Evaluate positions}\\s(4)=6,\quad s(0)=2\\[6pt]\text{Step 3: Substitute}\\\bar v=\frac{6-2}{4}\\[6pt]\text{Step 4: Simplify}\\\bar v=\frac44=1\\[6pt]\text{Final Answer: }\boxed{1\text{ unit/time unit}}\end{array}$

$\begin{array}{l}\text{Step 1: Find velocity zeros}\\v(t)=3(t-1)(t-3)=0\\t=1,\quad t=3\\[6pt]\text{Step 2: Evaluate positions}\\s(0)=2,\ s(1)=6,\ s(3)=2,\ s(4)=6\\[6pt]\text{Step 3: Find total distance}\\D=|6-2|+|2-6|+|6-2|\\D=12\\[6pt]\text{Step 4: Divide by elapsed time}\\\text{Average speed}=\frac{12}{4}=3\\[6pt]\text{Final Answer: }\boxed{3\text{ units/time unit}}\end{array}$

$\begin{array}{l}\text{Step 1: Set velocity equal to zero}\\t^2-6t+8=0\\[6pt]\text{Step 2: Factor}\\(t-2)(t-4)=0\\[6pt]\text{Step 3: Solve each factor}\\t-2=0,\quad t-4=0\\[6pt]\text{Final Answer: }\boxed{t=2,\ 4}\end{array}$

$\begin{array}{l}\text{Step 1: Factor velocity}\\v(t)=(t-2)(t-4)\\[6pt]\text{Step 2: Find zeros}\\t=2,\quad t=4\\[6pt]\text{Step 3: Test velocity signs}\\v(1)=3>0\\v(3)=-1<0\\v(5)=3>0\\[6pt]\text{Step 4: Identify sign changes}\\+\to-\text{ at }t=2\\-\to+\text{ at }t=4\\[6pt]\text{Final Answer: }\boxed{t=2,\ 4}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate velocity}\\a(t)=v'(t)\\[6pt]\text{Step 2: Apply power rule}\\a(t)=2t-6\\[6pt]\text{Step 3: Substitute }t=5\\a(5)=2(5)-6\\[6pt]\text{Step 4: Simplify}\\a(5)=10-6=4\\[6pt]\text{Final Answer: }\boxed{4}\end{array}$

$\begin{array}{l}\text{Step 1: Identify velocity sign}\\v=-8<0\\[6pt]\text{Step 2: Identify acceleration sign}\\a=3>0\\[6pt]\text{Step 3: Compare signs}\\va=(-8)(3)=-24<0\\[6pt]\text{Step 4: Interpret}\\\text{Opposite signs mean speed decreases}\\[6pt]\text{Final Answer: }\boxed{\text{No; slowing down}}\end{array}$

$\begin{array}{l}\text{Step 1: Identify velocity sign}\\v=-8<0\\[6pt]\text{Step 2: Identify acceleration sign}\\a=-3<0\\[6pt]\text{Step 3: Compare signs}\\va=(-8)(-3)=24>0\\[6pt]\text{Step 4: Interpret}\\\text{Matching signs mean speed increases}\\[6pt]\text{Final Answer: }\boxed{\text{Yes; speeding up}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=s'(t)\\[6pt]\text{Step 2: Apply sine derivative}\\v(t)=5\cos t\\[6pt]\text{Step 3: Differentiate velocity}\\a(t)=v'(t)\\[6pt]\text{Step 4: Apply cosine derivative}\\a(t)=-5\sin t\\[6pt]\text{Final Answer: }\boxed{v(t)=5\cos t,\ a(t)=-5\sin t}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=2e^{2t}\\[6pt]\text{Step 2: Differentiate velocity}\\a(t)=4e^{2t}\\[6pt]\text{Step 3: Evaluate velocity}\\v(1)=2e^{2(1)}=2e^2\\[6pt]\text{Step 4: Evaluate acceleration}\\a(1)=4e^{2(1)}=4e^2\\[6pt]\text{Final Answer: }\boxed{v(1)=2e^2,\ a(1)=4e^2}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate using chain rule}\\P'(t)=1000e^{0.03t}(0.03)\\[6pt]\text{Step 2: Multiply constants}\\P'(t)=30e^{0.03t}\\[6pt]\text{Step 3: Substitute }t=5\\P'(5)=30e^{0.15}\\[6pt]\text{Step 4: Approximate}\\P'(5)\approx34.855\\[6pt]\text{Final Answer: }\boxed{34.855\text{ people/year}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate volume}\\V'(t)=-10t\\[6pt]\text{Step 2: Substitute }t=3\\V'(3)=-10(3)\\[6pt]\text{Step 3: Simplify}\\V'(3)=-30\\[6pt]\text{Step 4: Interpret}\\\text{Volume decreases at }30\text{ L/min}\\[6pt]\text{Final Answer: }\boxed{-30\ \mathrm{L/min}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate using chain rule}\\T'(t)=8\cos(\pi t/12)\frac{\pi}{12}\\[6pt]\text{Step 2: Simplify}\\T'(t)=\frac{2\pi}{3}\cos(\pi t/12)\\[6pt]\text{Step 3: Substitute }t=6\\T'(6)=\frac{2\pi}{3}\cos(\pi/2)\\[6pt]\text{Step 4: Evaluate}\\\cos(\pi/2)=0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate cost}\\C'(x)=12+0.04x\\[6pt]\text{Step 2: Substitute }x=100\\C'(100)=12+0.04(100)\\[6pt]\text{Step 3: Simplify}\\C'(100)=12+4=16\\[6pt]\text{Final Answer: }\boxed{\$16\text{ per unit}}\end{array}$

$\begin{array}{l}\text{Step 1: Write area formula}\\A=\pi r^2\\[6pt]\text{Step 2: Differentiate with respect to time}\\\frac{dA}{dt}=2\pi r\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dA}{dt}=2\pi(10)(0.5)\\[6pt]\text{Step 4: Simplify}\\\frac{dA}{dt}=10\pi\\[6pt]\text{Final Answer: }\boxed{10\pi\ \mathrm{m^2/min}}\end{array}$

$\begin{array}{l}\text{Step 1: Write area formula}\\A=\pi r^2\\[6pt]\text{Step 2: Differentiate}\\\frac{dA}{dt}=2\pi r\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\20\pi=2\pi(5)\frac{dr}{dt}\\[6pt]\text{Step 4: Simplify}\\20\pi=10\pi\frac{dr}{dt}\\[6pt]\text{Step 5: Divide}\\\frac{dr}{dt}=\frac{20\pi}{10\pi}=2\\[6pt]\text{Final Answer: }\boxed{2\ \mathrm{cm/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=\frac43\pi r^3\\[6pt]\text{Step 2: Differentiate}\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dV}{dt}=4\pi(3)^2(2)\\[6pt]\text{Step 4: Square the radius}\\\frac{dV}{dt}=4\pi(9)(2)\\[6pt]\text{Step 5: Multiply}\\\frac{dV}{dt}=72\pi\\[6pt]\text{Final Answer: }\boxed{72\pi\ \mathrm{cm^3/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=\frac43\pi r^3\\[6pt]\text{Step 2: Differentiate}\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\100\pi=4\pi(5)^2\frac{dr}{dt}\\[6pt]\text{Step 4: Simplify}\\100\pi=100\pi\frac{dr}{dt}\\[6pt]\text{Step 5: Divide}\\\frac{dr}{dt}=1\\[6pt]\text{Final Answer: }\boxed{1\ \mathrm{cm/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=s^3\\[6pt]\text{Step 2: Differentiate}\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dV}{dt}=3(4)^2(3)\\[6pt]\text{Step 4: Simplify}\\\frac{dV}{dt}=3(16)(3)=144\\[6pt]\text{Final Answer: }\boxed{144\ \mathrm{cm^3/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write area formula}\\A=s^2\\[6pt]\text{Step 2: Differentiate}\\\frac{dA}{dt}=2s\frac{ds}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dA}{dt}=2(6)(-0.2)\\[6pt]\text{Step 4: Simplify}\\\frac{dA}{dt}=-2.4\\[6pt]\text{Final Answer: }\boxed{-2.4\ \mathrm{m^2/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write area formula}\\A=LW\\[6pt]\text{Step 2: Apply product rule}\\\frac{dA}{dt}=W\frac{dL}{dt}+L\frac{dW}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dA}{dt}=5(2)+10(-1)\\[6pt]\text{Step 4: Simplify}\\\frac{dA}{dt}=10-10=0\\[6pt]\text{Final Answer: }\boxed{0\ \mathrm{m^2/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=\pi r^2h\\[6pt]\text{Step 2: Note radius is constant}\\\frac{dr}{dt}=0\\[6pt]\text{Step 3: Differentiate}\\\frac{dV}{dt}=\pi r^2\frac{dh}{dt}\\[6pt]\text{Step 4: Substitute}\\\frac{dV}{dt}=\pi(3)^2(2)\\[6pt]\text{Step 5: Simplify}\\\frac{dV}{dt}=18\pi\\[6pt]\text{Final Answer: }\boxed{18\pi\ \mathrm{cm^3/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=\pi r^2h\\[6pt]\text{Step 2: Differentiate using product rule}\\\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dV}{dt}=2\pi(2)(5)(1)+\pi(2)^2(-2)\\[6pt]\text{Step 4: Simplify}\\\frac{dV}{dt}=20\pi-8\pi=12\pi\\[6pt]\text{Final Answer: }\boxed{12\pi\ \mathrm{cm^3/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write Pythagorean equation}\\x^2+y^2=100\\[6pt]\text{Step 2: Find height}\\6^2+y^2=100\\y^2=64\\y=8\\[6pt]\text{Step 3: Differentiate}\\2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\[6pt]\text{Step 4: Substitute}\\2(6)(2)+2(8)\frac{dy}{dt}=0\\[6pt]\text{Step 5: Isolate derivative}\\24+16\frac{dy}{dt}=0\\16\frac{dy}{dt}=-24\\[6pt]\text{Step 6: Divide}\\\frac{dy}{dt}=-\frac{24}{16}=-1.5\\[6pt]\text{Final Answer: }\boxed{-1.5\ \mathrm{ft/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write Pythagorean equation}\\x^2+y^2=169\\[6pt]\text{Step 2: Find height}\\5^2+y^2=169\\y^2=144\\y=12\\[6pt]\text{Step 3: Differentiate}\\2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\\[6pt]\text{Step 4: Substitute}\\2(5)(3)+2(12)\frac{dy}{dt}=0\\[6pt]\text{Step 5: Simplify}\\30+24\frac{dy}{dt}=0\\24\frac{dy}{dt}=-30\\[6pt]\text{Step 6: Divide}\\\frac{dy}{dt}=-\frac{30}{24}=-\frac54\\[6pt]\text{Final Answer: }\boxed{-\frac54\ \mathrm{ft/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write distance equation}\\z^2=x^2+y^2\\[6pt]\text{Step 2: Find current separation}\\z^2=3^2+4^2=25\\z=5\\[6pt]\text{Step 3: Differentiate}\\2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\\[6pt]\text{Step 4: Substitute}\\2(5)\frac{dz}{dt}=2(3)(4)+2(4)(3)\\[6pt]\text{Step 5: Simplify}\\10\frac{dz}{dt}=24+24=48\\[6pt]\text{Step 6: Divide}\\\frac{dz}{dt}=\frac{48}{10}=\frac{24}{5}\\[6pt]\text{Final Answer: }\boxed{\frac{24}{5}\ \mathrm{km/h}}\end{array}$

$\begin{array}{l}\text{Step 1: Define variables}\\x=\text{distance from pole}\\s=\text{shadow length}\\[6pt]\text{Step 2: Use similar triangles}\\\frac{15}{x+s}=\frac6s\\[6pt]\text{Step 3: Cross-multiply}\\15s=6(x+s)\\[6pt]\text{Step 4: Collect terms}\\15s=6x+6s\\9s=6x\\[6pt]\text{Step 5: Differentiate}\\9\frac{ds}{dt}=6\frac{dx}{dt}\\[6pt]\text{Step 6: Substitute}\\9\frac{ds}{dt}=6(4)=24\\[6pt]\text{Step 7: Divide}\\\frac{ds}{dt}=\frac{24}{9}=\frac83\\[6pt]\text{Final Answer: }\boxed{\frac83\ \mathrm{ft/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Use similar triangles}\\\frac{15}{x+s}=\frac6s\\[6pt]\text{Step 2: Simplify relationship}\\15s=6x+6s\\9s=6x\\[6pt]\text{Step 3: Differentiate}\\9\frac{ds}{dt}=6\frac{dx}{dt}\\[6pt]\text{Step 4: Find shadow growth rate}\\9\frac{ds}{dt}=6(4)\\\frac{ds}{dt}=\frac83\\[6pt]\text{Step 5: Define shadow-tip position}\\p=x+s\\[6pt]\text{Step 6: Differentiate}\\\frac{dp}{dt}=\frac{dx}{dt}+\frac{ds}{dt}\\[6pt]\text{Step 7: Substitute and simplify}\\\frac{dp}{dt}=4+\frac83=\frac{20}{3}\\[6pt]\text{Final Answer: }\boxed{\frac{20}{3}\ \mathrm{ft/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write cone volume formula}\\V=\frac13\pi r^2h\\[6pt]\text{Step 2: Use similar triangles}\\\frac rh=\frac4{12}=\frac13\\r=\frac h3\\[6pt]\text{Step 3: Substitute for radius}\\V=\frac13\pi(\frac h3)^2h\\V=\frac{\pi h^3}{27}\\[6pt]\text{Step 4: Differentiate}\\\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}\\[6pt]\text{Step 5: Substitute values}\\8\pi=\frac{\pi(6)^2}{9}\frac{dh}{dt}\\[6pt]\text{Step 6: Simplify}\\8\pi=4\pi\frac{dh}{dt}\\[6pt]\text{Step 7: Divide}\\\frac{dh}{dt}=2\\[6pt]\text{Final Answer: }\boxed{2\ \mathrm{cm/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write cone volume}\\V=\frac13\pi r^2h\\[6pt]\text{Step 2: Use similar triangles}\\\frac rh=\frac5{10}=\frac12\\r=\frac h2\\[6pt]\text{Step 3: Substitute}\\V=\frac13\pi(\frac h2)^2h\\V=\frac{\pi h^3}{12}\\[6pt]\text{Step 4: Differentiate}\\\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}\\[6pt]\text{Step 5: Use negative draining rate}\\\frac{dV}{dt}=-2\pi\\[6pt]\text{Step 6: Substitute }h=4\\-2\pi=\frac{\pi(4)^2}{4}\frac{dh}{dt}\\[6pt]\text{Step 7: Simplify}\\-2\pi=4\pi\frac{dh}{dt}\\[6pt]\text{Step 8: Divide}\\\frac{dh}{dt}=-\frac12\\[6pt]\text{Final Answer: }\boxed{-\frac12\ \mathrm{m/min}}\end{array}$

$\begin{array}{l}\text{Step 1: Write sphere volume}\\V=\frac43\pi r^3\\[6pt]\text{Step 2: Differentiate}\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\36\pi=4\pi(3)^2\frac{dr}{dt}\\[6pt]\text{Step 4: Simplify}\\36\pi=36\pi\frac{dr}{dt}\\[6pt]\text{Step 5: Divide}\\\frac{dr}{dt}=1\\[6pt]\text{Final Answer: }\boxed{1\ \mathrm{cm/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write cable-length equation}\\L^2=x^2+4^2\\L^2=x^2+16\\[6pt]\text{Step 2: Find current cable length}\\L^2=3^2+16=25\\L=5\\[6pt]\text{Step 3: Differentiate}\\2L\frac{dL}{dt}=2x\frac{dx}{dt}\\[6pt]\text{Step 4: Substitute}\\2(5)\frac{dL}{dt}=2(3)(3)\\[6pt]\text{Step 5: Simplify}\\10\frac{dL}{dt}=18\\[6pt]\text{Step 6: Divide}\\\frac{dL}{dt}=\frac{18}{10}=\frac95\\[6pt]\text{Final Answer: }\boxed{\frac95\ \mathrm{m/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Write relationship}\\y=x^2\\[6pt]\text{Step 2: Differentiate with respect to time}\\\frac{dy}{dt}=2x\frac{dx}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dy}{dt}=2(2)(3)\\[6pt]\text{Step 4: Multiply}\\\frac{dy}{dt}=12\\[6pt]\text{Final Answer: }\boxed{12\text{ output units/time unit}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate with respect to time}\\2x\frac{dx}{dt}+y\frac{dx}{dt}+x\frac{dy}{dt}+2y\frac{dy}{dt}=0\\[6pt]\text{Step 2: Group derivative terms}\\(x+2y)\frac{dy}{dt}=-(2x+y)\frac{dx}{dt}\\[6pt]\text{Step 3: Substitute}\\(2+4)\frac{dy}{dt}=-(4+2)(3)\\[6pt]\text{Step 4: Simplify}\\6\frac{dy}{dt}=-18\\[6pt]\text{Step 5: Divide}\\\frac{dy}{dt}=-3\\[6pt]\text{Final Answer: }\boxed{-3}\end{array}$

$\begin{array}{l}\text{Step 1: Write linearization formula}\\L(x)=f(a)+f'(a)(x-a)\\[6pt]\text{Step 2: Evaluate function}\\f(2)=2^3=8\\[6pt]\text{Step 3: Differentiate}\\f'(x)=3x^2\\[6pt]\text{Step 4: Evaluate derivative}\\f'(2)=3(2)^2=12\\[6pt]\text{Step 5: Substitute}\\L(x)=8+12(x-2)\\[6pt]\text{Final Answer: }\boxed{L(x)=8+12(x-2)}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=x^3\\[6pt]\text{Step 2: Evaluate at }a=2\\f(2)=8\\[6pt]\text{Step 3: Find derivative}\\f'(x)=3x^2\\f'(2)=12\\[6pt]\text{Step 4: Write linearization}\\L(x)=8+12(x-2)\\[6pt]\text{Step 5: Substitute}\\L(2.01)=8+12(0.01)\\[6pt]\text{Step 6: Simplify}\\L(2.01)=8+0.12=8.12\\[6pt]\text{Final Answer: }\boxed{8.12}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=\sqrt{x}\\[6pt]\text{Step 2: Evaluate at }a=9\\f(9)=3\\[6pt]\text{Step 3: Differentiate}\\f'(x)=\frac1{2\sqrt{x}}\\[6pt]\text{Step 4: Evaluate derivative}\\f'(9)=\frac16\\[6pt]\text{Step 5: Write linearization}\\L(x)=3+\frac16(x-9)\\[6pt]\text{Step 6: Substitute}\\L(9.2)=3+\frac{0.2}{6}\\[6pt]\text{Step 7: Simplify}\\L(9.2)\approx3.03333\\[6pt]\text{Final Answer: }\boxed{3.03333}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=x^{1/3}\\[6pt]\text{Step 2: Evaluate}\\f(8)=2\\[6pt]\text{Step 3: Differentiate}\\f'(x)=\frac13x^{-2/3}\\[6pt]\text{Step 4: Evaluate derivative}\\f'(8)=\frac1{3(4)}=\frac1{12}\\[6pt]\text{Step 5: Write linearization}\\L(x)=2+\frac1{12}(x-8)\\[6pt]\text{Step 6: Substitute}\\L(8.3)=2+\frac{0.3}{12}\\[6pt]\text{Step 7: Simplify}\\L(8.3)=2.025\\[6pt]\text{Final Answer: }\boxed{2.025}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=\ln x\\[6pt]\text{Step 2: Evaluate}\\f(1)=\ln1=0\\[6pt]\text{Step 3: Differentiate}\\f'(x)=\frac1x\\f'(1)=1\\[6pt]\text{Step 4: Write linearization}\\L(x)=0+1(x-1)=x-1\\[6pt]\text{Step 5: Substitute}\\L(1.05)=1.05-1=0.05\\[6pt]\text{Final Answer: }\boxed{0.05}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=e^x\\[6pt]\text{Step 2: Evaluate}\\f(0)=1\\[6pt]\text{Step 3: Differentiate}\\f'(x)=e^x\\f'(0)=1\\[6pt]\text{Step 4: Write linearization}\\L(x)=1+x\\[6pt]\text{Step 5: Substitute}\\L(0.02)=1+0.02=1.02\\[6pt]\text{Final Answer: }\boxed{1.02}\end{array}$

$\begin{array}{l}\text{Step 1: Define function}\\f(x)=\sin x\\[6pt]\text{Step 2: Evaluate}\\f(0)=0\\[6pt]\text{Step 3: Differentiate}\\f'(x)=\cos x\\f'(0)=1\\[6pt]\text{Step 4: Write linearization}\\L(x)=0+1(x-0)=x\\[6pt]\text{Step 5: Substitute}\\L(0.1)=0.1\\[6pt]\text{Final Answer: }\boxed{0.1}\end{array}$

$\begin{array}{l}\text{Step 1: Find radius change}\\dr=10.1-10=0.1\\[6pt]\text{Step 2: Write area formula}\\A=\pi r^2\\[6pt]\text{Step 3: Differentiate}\\dA=2\pi r\,dr\\[6pt]\text{Step 4: Substitute}\\dA=2\pi(10)(0.1)\\[6pt]\text{Step 5: Simplify}\\dA=2\pi\\[6pt]\text{Final Answer: }\boxed{\Delta A\approx2\pi\ \mathrm{cm^2}}\end{array}$

$\begin{array}{l}\text{Step 1: Find radius change}\\dr=5.02-5=0.02\\[6pt]\text{Step 2: Write volume formula}\\V=\frac43\pi r^3\\[6pt]\text{Step 3: Differentiate}\\dV=4\pi r^2\,dr\\[6pt]\text{Step 4: Substitute}\\dV=4\pi(5)^2(0.02)\\[6pt]\text{Step 5: Simplify}\\dV=100\pi(0.02)=2\pi\\[6pt]\text{Final Answer: }\boxed{\Delta V\approx2\pi\ \mathrm{cm^3}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate}\\f'(x)=\frac12x^{-1/2}\\[6pt]\text{Step 2: Find second derivative}\\f''(x)=-\frac14x^{-3/2}\\[6pt]\text{Step 3: Determine sign}\\f''(x)<0\quad(x>0)\\[6pt]\text{Step 4: Interpret concavity}\\\text{The graph is concave down}\\[6pt]\text{Step 5: Compare tangent and graph}\\\text{The tangent lies above the graph}\\[6pt]\text{Final Answer: }\boxed{\text{Overestimate}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate}\\f'(x)=e^x\\[6pt]\text{Step 2: Find second derivative}\\f''(x)=e^x\\[6pt]\text{Step 3: Determine sign}\\e^x>0\\[6pt]\text{Step 4: Interpret concavity}\\\text{The graph is concave up}\\[6pt]\text{Step 5: Compare tangent and graph}\\\text{The tangent lies below the graph}\\[6pt]\text{Final Answer: }\boxed{\text{Underestimate}}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac{\sin0}{0}=\frac00\\[6pt]\text{Step 2: Apply L'Hopital's Rule}\\\lim_{x\to0}\frac{3\cos(3x)}1\\[6pt]\text{Step 3: Substitute }x=0\\3\cos0\\[6pt]\text{Step 4: Evaluate}\\3(1)=3\\[6pt]\text{Final Answer: }\boxed{3}\end{array}$

$\begin{array}{l}\text{Step 1: Check indeterminate form}\\\frac{1-\cos0}{0^2}=\frac00\\[6pt]\text{Step 2: Differentiate numerator and denominator}\\\lim_{x\to0}\frac{\sin x}{2x}\\[6pt]\text{Step 3: Check again}\\\frac{\sin0}{2(0)}=\frac00\\[6pt]\text{Step 4: Apply L'Hopital again}\\\lim_{x\to0}\frac{\cos x}{2}\\[6pt]\text{Step 5: Substitute and evaluate}\\\frac{\cos0}{2}=\frac12\\[6pt]\text{Final Answer: }\boxed{\frac12}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac{e^0-1}{0}=\frac00\\[6pt]\text{Step 2: Apply L'Hopital's Rule}\\\lim_{x\to0}\frac{2e^{2x}}1\\[6pt]\text{Step 3: Substitute }x=0\\2e^0\\[6pt]\text{Step 4: Evaluate}\\2(1)=2\\[6pt]\text{Final Answer: }\boxed{2}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac{\ln1}{0}=\frac00\\[6pt]\text{Step 2: Differentiate numerator}\\\frac{d}{dx}\ln(1+x)=\frac1{1+x}\\[6pt]\text{Step 3: Differentiate denominator}\\\frac{d}{dx}x=1\\[6pt]\text{Step 4: Write new limit}\\\lim_{x\to0}\frac1{1+x}\\[6pt]\text{Step 5: Substitute}\\\frac1{1+0}=1\\[6pt]\text{Final Answer: }\boxed{1}\end{array}$

$\begin{array}{l}\text{Step 1: Identify form}\\\frac{\infty}{\infty}\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to\infty}\frac{2x}{e^x}\\[6pt]\text{Step 3: Form remains indeterminate}\\\frac{\infty}{\infty}\\[6pt]\text{Step 4: Differentiate again}\\\lim_{x\to\infty}\frac2{e^x}\\[6pt]\text{Step 5: Evaluate}\\e^x\to\infty\\\frac2{e^x}\to0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Identify form}\\\frac{\infty}{\infty}\\[6pt]\text{Step 2: Differentiate numerator}\\\frac{d}{dx}\ln x=\frac1x\\[6pt]\text{Step 3: Differentiate denominator}\\\frac{d}{dx}x=1\\[6pt]\text{Step 4: Write new limit}\\\lim_{x\to\infty}\frac1x\\[6pt]\text{Step 5: Evaluate}\\\frac1x\to0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac00\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to0}\frac{1-\cos x}{3x^2}\\[6pt]\text{Step 3: Still indeterminate}\\\frac00\\[6pt]\text{Step 4: Differentiate twice}\\\lim_{x\to0}\frac{\sin x}{6x}\\[6pt]\text{Step 5: Still indeterminate}\\\frac00\\[6pt]\text{Step 6: Differentiate third time}\\\lim_{x\to0}\frac{\cos x}{6}\\[6pt]\text{Step 7: Evaluate}\\\frac{\cos0}{6}=\frac16\\[6pt]\text{Final Answer: }\boxed{\frac16}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac00\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to0}\frac{\sec^2x-1}{3x^2}\\[6pt]\text{Step 3: Use identity}\\\sec^2x-1=\tan^2x\\[6pt]\text{Step 4: Rewrite}\\\frac13\lim_{x\to0}(\frac{\tan x}{x})^2\\[6pt]\text{Step 5: Evaluate inner limit}\\\lim_{x\to0}\frac{\tan x}{x}\\=\lim_{x\to0}\frac{\sec^2x}{1}=1\\[6pt]\text{Step 6: Simplify}\\\frac13(1)^2=\frac13\\[6pt]\text{Final Answer: }\boxed{\frac13}\end{array}$

$\begin{array}{l}\text{Step 1: Identify form}\\\frac{\infty}{\infty}\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to\infty}\frac{6x+2}{10x}\\[6pt]\text{Step 3: Still indeterminate}\\\frac{\infty}{\infty}\\[6pt]\text{Step 4: Differentiate again}\\\lim_{x\to\infty}\frac6{10}\\[6pt]\text{Step 5: Simplify}\\\frac6{10}=\frac35\\[6pt]\text{Final Answer: }\boxed{\frac35}\end{array}$

$\begin{array}{l}\text{Step 1: Identify indeterminate product}\\0\cdot(-\infty)\\[6pt]\text{Step 2: Rewrite as quotient}\\x\ln x=\frac{\ln x}{1/x}\\[6pt]\text{Step 3: Apply L'Hopital's Rule}\\\lim_{x\to0^+}\frac{1/x}{-1/x^2}\\[6pt]\text{Step 4: Simplify}\\\lim_{x\to0^+}(-x)\\[6pt]\text{Step 5: Evaluate}\\-x\to0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Identify indeterminate form}\\\infty-\infty\\[6pt]\text{Step 2: Multiply by conjugate}\\\frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}\\[6pt]\text{Step 3: Use difference of squares}\\\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}\\[6pt]\text{Step 4: Simplify numerator}\\\frac{x}{\sqrt{x^2+x}+x}\\[6pt]\text{Step 5: Factor }x>0\\\frac{x}{x\sqrt{1+1/x}+x}\\[6pt]\text{Step 6: Cancel }x\\\frac1{\sqrt{1+1/x}+1}\\[6pt]\text{Step 7: Evaluate}\\\frac1{\sqrt1+1}=\frac12\\[6pt]\text{Final Answer: }\boxed{\frac12}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac{\sin0}{\sin0}=\frac00\\[6pt]\text{Step 2: Differentiate numerator}\\5\cos(5x)\\[6pt]\text{Step 3: Differentiate denominator}\\2\cos(2x)\\[6pt]\text{Step 4: Write new limit}\\\lim_{x\to0}\frac{5\cos(5x)}{2\cos(2x)}\\[6pt]\text{Step 5: Evaluate}\\\frac{5(1)}{2(1)}=\frac52\\[6pt]\text{Final Answer: }\boxed{\frac52}\end{array}$

$\begin{array}{l}\text{Step 1: Check direct substitution}\\\frac00\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to0}\frac{e^x-1}{2x}\\[6pt]\text{Step 3: Still indeterminate}\\\frac00\\[6pt]\text{Step 4: Differentiate again}\\\lim_{x\to0}\frac{e^x}{2}\\[6pt]\text{Step 5: Evaluate}\\\frac{e^0}{2}=\frac12\\[6pt]\text{Final Answer: }\boxed{\frac12}\end{array}$

$\begin{array}{l}\text{Step 1: Identify form}\\\frac{\infty}{\infty}\\[6pt]\text{Step 2: Differentiate once}\\\lim_{x\to\infty}\frac{3x^2}{2e^{2x}}\\[6pt]\text{Step 3: Differentiate again}\\\lim_{x\to\infty}\frac{6x}{4e^{2x}}\\[6pt]\text{Step 4: Differentiate third time}\\\lim_{x\to\infty}\frac6{8e^{2x}}\\[6pt]\text{Step 5: Evaluate}\\e^{2x}\to\infty\\\frac6{8e^{2x}}\to0\\[6pt]\text{Final Answer: }\boxed{0}\end{array}$

$\begin{array}{l}\text{Step 1: Write linearization}\\L(x)=f(a)+f'(a)(x-a)\\[6pt]\text{Step 2: Substitute values}\\L(x)=5-3(x-2)\\[6pt]\text{Step 3: Substitute }x=2.1\\L(2.1)=5-3(2.1-2)\\[6pt]\text{Step 4: Simplify}\\L(2.1)=5-3(0.1)\\L(2.1)=5-0.3=4.7\\[6pt]\text{Final Answer: }\boxed{f(2.1)\approx4.7}\end{array}$

$\begin{array}{l}\text{Step 1: Write linearization}\\L(x)=f(4)+f'(4)(x-4)\\[6pt]\text{Step 2: Substitute values}\\L(x)=10+2(x-4)\\[6pt]\text{Step 3: Evaluate}\\L(4.2)=10+2(0.2)=10.4\\[6pt]\text{Step 4: Analyze concavity}\\f''(x)>0\\\text{The graph is concave up}\\[6pt]\text{Step 5: Compare tangent and graph}\\\text{The tangent lies below the graph}\\[6pt]\text{Final Answer: }\boxed{10.4\text{, underestimate}}\end{array}$

$\begin{array}{l}\text{Step 1: Write differential approximation}\\\Delta f\approx f'(a)\Delta x\\[6pt]\text{Step 2: Identify values}\\a=1,\quad \Delta x=0.05\\[6pt]\text{Step 3: Substitute}\\\Delta f\approx4(0.05)\\[6pt]\text{Step 4: Multiply}\\\Delta f\approx0.20\\[6pt]\text{Final Answer: }\boxed{0.20}\end{array}$

$\begin{array}{l}\text{Step 1: Write volume formula}\\V=\frac43\pi r^3\\[6pt]\text{Step 2: Differentiate}\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\[6pt]\text{Step 3: Substitute}\\\frac{dV}{dt}=4\pi(10)^2(0.1)\\[6pt]\text{Step 4: Simplify}\\\frac{dV}{dt}=400\pi(0.1)=40\pi\\[6pt]\text{Step 5: Interpret positive rate}\\\text{Volume is increasing}\\[6pt]\text{Final Answer: }\boxed{40\pi\ \mathrm{cm^3/s}}\end{array}$

$\begin{array}{l}\text{Step 1: Factor velocity}\\v(t)=(t-1)(t-3)\\[6pt]\text{Step 2: Find direction changes}\\t=1,\quad t=3\\[6pt]\text{Step 3: Find position antiderivative}\\s(t)=\frac{t^3}{3}-2t^2+3t+C\\[6pt]\text{Step 4: Evaluate positions}\\s(0)=C\\s(1)=\frac43+C\\s(3)=C\\s(4)=\frac43+C\\[6pt]\text{Step 5: Add absolute changes}\\D=|\frac43|+|-\frac43|+|\frac43|\\[6pt]\text{Step 6: Simplify}\\D=\frac43+\frac43+\frac43=4\\[6pt]\text{Final Answer: }\boxed{4\text{ distance units}}\end{array}$

$\begin{array}{l}\text{Step 1: Write displacement integral}\\\Delta s=\int_0^4v(t)\,dt\\[6pt]\text{Step 2: Substitute velocity}\\\Delta s=\int_0^4(t^2-4t+3)\,dt\\[6pt]\text{Step 3: Integrate}\\\Delta s=[\frac{t^3}{3}-2t^2+3t]_0^4\\[6pt]\text{Step 4: Evaluate upper bound}\\\Delta s=\frac{64}{3}-32+12\\[6pt]\text{Step 5: Simplify}\\\Delta s=\frac{64}{3}-20=\frac43\\[6pt]\text{Final Answer: }\boxed{\frac43\text{ position units}}\end{array}$

$\begin{array}{l}\text{Step 1: Factor velocity}\\v(t)=(t-1)(t-3)\\[6pt]\text{Step 2: Find acceleration}\\a(t)=2t-4=2(t-2)\\[6pt]\text{Step 3: Identify boundaries}\\t=1,\quad t=2,\quad t=3\\[6pt]\text{Step 4: Compare signs}\\(0,1):\ v>0,\ a<0\\(1,2):\ v<0,\ a<0\\(2,3):\ v<0,\ a>0\\(3,\infty):\ v>0,\ a>0\\[6pt]\text{Step 5: Select matching signs}\\[6pt]\text{Final Answer: }\boxed{(1,2)\cup(3,\infty)}\end{array}$

$\begin{array}{l}\text{Step 1: Factor velocity}\\v(t)=(t-1)(t-3)\\[6pt]\text{Step 2: Find acceleration}\\a(t)=2t-4=2(t-2)\\[6pt]\text{Step 3: Identify boundaries}\\t=1,\quad t=2,\quad t=3\\[6pt]\text{Step 4: Compare signs}\\[0,1):\ v>0,\ a<0\\(1,2):\ v<0,\ a<0\\(2,3):\ v<0,\ a>0\\(3,\infty):\ v>0,\ a>0\\[6pt]\text{Step 5: Select opposite signs}\\[6pt]\text{Final Answer: }\boxed{[0,1)\cup(2,3)}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate volume}\\V'(t)=20-4t\\[6pt]\text{Step 2: Set rate equal to zero}\\20-4t=0\\[6pt]\text{Step 3: Solve}\\-4t=-20\\t=5\\[6pt]\text{Step 4: Verify sign change}\\V'(4)=4>0\\V'(6)=-4<0\\[6pt]\text{Final Answer: }\boxed{t=5\text{ time units}}\end{array}$

$\begin{array}{l}\text{Step 1: Differentiate position}\\v(t)=3t^2-6t\\[6pt]\text{Step 2: Factor velocity}\\v(t)=3t(t-2)\\[6pt]\text{Step 3: Solve }v(t)=0\\3t(t-2)=0\\t=0,\quad t=2\\[6pt]\text{Step 4: Select first positive time}\\t=2\\[6pt]\text{Step 5: Differentiate velocity}\\a(t)=6t-6\\[6pt]\text{Step 6: Substitute}\\a(2)=6(2)-6=6\\[6pt]\text{Final Answer: }\boxed{6\ \mathrm{m/s^2}}\end{array}$