- What is the derivative of a linear function with respect to time? - The rate units come from cartons divided by minutes.

1. Differentiate the linear model: \(I'(t)=18\). 2. Because the derivative is constant, \(I'(6)=18\,\text{cartons/min}\). Thus, at \(t=6\) minutes, cartons are being inspected at an instantaneous rate of \(18\) cartons per minute.

\(I'(t)=18\), so \(I'(6)=18\,\text{cartons/min}\).

- Differentiate the supplied contextual model with respect to time. - Evaluate the derivative at the requested time. - Use both the sign and the compound units in your interpretation.

1. Differentiate the model: \(Q'(t)=6-0.4t\). 2. Evaluate at \(t=5\): \(Q'(5)=6-2=4\,\text{kg/min}\). 3. The positive value means that at \(t=5\) minutes, the mass on the conveyor is increasing at an instantaneous rate of \(4\) kilograms per minute.

\(Q'(5)=4\,\text{kg/min}\). At \(t=5\) minutes, the mass on the conveyor is increasing at \(4\) kilograms per minute.

- Average rate of change uses two function values and an elapsed time. - Instantaneous rate comes from the derivative at one time. - Put both rates in the same units before comparing them.

1. \(F(1)=3\) and \(F(3)=33\). The average rate on \([1,3]\) is \(\frac{33-3}{3-1}=15\,\text{gal/h}\). 2. Differentiate: \(F'(t)=3t^2+2\). Thus \(F'(2)=14\,\text{gal/h}\). 3. The instantaneous rate at \(t=2\) is \(1\,\text{gal/h}\) lower than the average processing rate from \(t=1\) to \(t=3\).

a) \(15\,\text{gal/h}\) b) \(14\,\text{gal/h}\) c) At \(t=2\), the system is processing water \(1\,\text{gal/h}\) more slowly than its average pace over \([1,3]\).

- Compute the first derivative to obtain the concentration's instantaneous rate of change. - Differentiate again to determine how that rate itself is changing. - Combine the signs of the first and second derivatives when interpreting the behavior.

1. Differentiate once: \(C'(t)=-5+t\), so \(C'(3)=-2\,\frac{\text{g}}{\text{L}\cdot\text{day}}\). 2. Differentiate again: \(C''(t)=1\), so \(C''(3)=1\,\frac{\text{g}}{\text{L}\cdot\text{day}^2}\). 3. At day \(3\), the concentration is decreasing at \(2\,\frac{\text{g}}{\text{L}\cdot\text{day}}\). Because the first derivative is increasing at \(1\,\frac{\text{g}}{\text{L}\cdot\text{day}^2}\), that decrease is becoming less rapid.

\(C'(3)=-2\,\frac{\text{g}}{\text{L}\cdot\text{day}}\) and \(C''(3)=1\,\frac{\text{g}}{\text{L}\cdot\text{day}^2}\). The concentration is decreasing at day \(3\), but its rate of decrease is becoming less rapid.

- Each average rate uses the change in cumulative amount divided by the corresponding one-hour interval. - The target time lies between the two nearby intervals, so use information from both sides of it. - The estimated instantaneous rate should have units of gallons per hour.

1. On \([2,3]\), the average rate is \(\frac{88-62}{3-2}=26\,\text{gal/h}\). 2. On \([3,4]\), the average rate is \(\frac{118-88}{4-3}=30\,\text{gal/h}\). 3. A symmetric estimate for the rate at the middle time is the average of the nearby one-hour rates: \(\frac{26+30}{2}=28\,\text{gal/h}\). Equivalently, this is \(\frac{C(4)-C(2)}{4-2}\). 4. Thus, at \(t=3\) hours, the system is processing water at approximately \(28\) gallons per hour.

a) \(26\,\text{gal/h}\) on \([2,3]\); \(30\,\text{gal/h}\) on \([3,4]\) b) \(C'(3)\approx28\,\text{gal/h}\). At \(t=3\) hours, the system is processing water at approximately \(28\) gallons per hour.

- Because \(G\) is a total amount, compare its values to find an amount produced over an interval. - Average rate of change uses a difference quotient over two times. - Instantaneous production rate is obtained by differentiating \(G\) and evaluating at the requested time. - Compare the two rates in the same units before interpreting them.

1. Since \(G(0)=0\), the amount produced during the first \(20\) hours is \(G(20)=-0.05(20)^3+1.5(20)^2+10(20)=400\,\text{m}^3\). 2. Since \(G(12)=249.6\) and \(G(4)=60.8\), the average production rate is \(\frac{G(12)-G(4)}{12-4}=\frac{249.6-60.8}{8}=23.6\,\frac{\text{m}^3}{\text{h}}\). 3. The instantaneous production rate is \(G'(t)=-0.15t^2+3t+10\). Thus \(G'(10)=25\,\frac{\text{m}^3}{\text{h}}\). 4. The instantaneous rate at \(t=10\) is \(1.4\,\frac{\text{m}^3}{\text{h}}\) greater than the average rate from \(t=4\) to \(t=12\). This means gas is being produced slightly faster at that instant than the average pace over the interval.

a) \(400\,\text{m}^3\) b) \(23.6\,\frac{\text{m}^3}{\text{h}}\) c) \(25\,\frac{\text{m}^3}{\text{h}}\), which is \(1.4\,\frac{\text{m}^3}{\text{h}}\) greater than the average rate in part b).

- Average rate uses two volume values and the elapsed time between them. - The first derivative gives the instantaneous volume-change rate. - The second derivative describes how that first rate is changing and therefore has one additional time unit in the denominator.

1. Evaluate \(V(2)=252\) and \(V(5)=450\). The average rate of change is \(\frac{450-252}{5-2}=66\,\frac{\text{m}^3}{\text{h}}\). 2. The first derivative is \(V'(t)=-3t^2+30t\). Thus \(V'(4)=72\,\frac{\text{m}^3}{\text{h}}\). At \(t=4\), the basin's water volume is increasing at an instantaneous rate of \(72\,\text{m}^3\) per hour. 3. The second derivative is \(V''(t)=-6t+30\), so \(V''(4)=6\,\frac{\text{m}^3}{\text{h}^2}\). At that instant, the net volume-change rate itself is increasing at \(6\,\frac{\text{m}^3}{\text{h}}\) per hour.

a) \(66\,\frac{\text{m}^3}{\text{h}}\) b) \(72\,\frac{\text{m}^3}{\text{h}}\); at \(t=4\), the water volume is increasing at \(72\,\text{m}^3\) per hour. c) \(V''(4)=6\,\frac{\text{m}^3}{\text{h}^2}\); the net volume-change rate is increasing by \(6\,\frac{\text{m}^3}{\text{h}}\) per hour at that instant.

- Convert each clock time to hours after \(6\) a.m. - The first derivative describes how power output is changing at an instant. - The second derivative describes how the first derivative is changing. - Interpret the signs of the two derivatives separately before combining the statements.

1. At \(10\) a.m., \(t=4\). Thus \(P(4)=-64+192+240=368\,\text{W}\). 2. The first derivative is \(P'(t)=-3t^2+24t+60\), which is the instantaneous rate of change of power output, measured in \(\frac{\text{W}}{\text{h}}\). 3. The second derivative is \(P''(t)=-6t+24\), which is the instantaneous rate of change of \(P'(t)\), measured in \(\frac{\text{W}}{\text{h}^2}\). 4. At \(2\) p.m., \(t=8\). Then \(P'(8)=60\,\frac{\text{W}}{\text{h}}\) and \(P''(8)=-24\,\frac{\text{W}}{\text{h}^2}\). 5. The positive first derivative means power output is increasing at \(2\) p.m. The negative second derivative means that this increase is becoming slower at that instant.

a) \(368\,\text{W}\) b) \(P'(t)=-3t^2+24t+60\), in \(\frac{\text{W}}{\text{h}}\), is the instantaneous rate of change of power output. \(P''(t)=-6t+24\), in \(\frac{\text{W}}{\text{h}^2}\), is the instantaneous rate of change of that rate. c) At \(2\) p.m., \(P'(8)=60\,\frac{\text{W}}{\text{h}}\) and \(P''(8)=-24\,\frac{\text{W}}{\text{h}^2}\). The output is increasing, but its rate of increase is decreasing.

- Substitute the temperature function into the activity function. - Expand the squared binomial carefully. - Instantaneous rate of change is represented by a derivative.

1. Compose the functions: \(A(t)=A(T(t))=-0.01((15+2t)-25)^2+10\). 2. Simplifying gives \(A(t)=-0.01(2t-10)^2+10=-0.04t^2+0.4t+9\). 3. The instantaneous rate of change is the derivative, \(A'(t)=-0.08t+0.4\). 4. At \(t=2\), \(A'(2)=-0.08\cdot2+0.4=0.24\).

a) \(A(t)=-0.04t^2+0.4t+9\) b) \(0.24\) activity units per hour

- Average growth rate uses the change in the population over an interval. - Instantaneous growth rate is the derivative of the population model. - Evaluate the same derivative formula at both requested times before comparing the rates.

1. The average growth rate on \([0,5]\) is \(\frac{B(5)-B(0)}{5}\). Since \(B(0)=100\) and \(B(5)=475\), the average rate is \(75\) bacteria per hour. 2. The instantaneous growth rate is \(B'(t)=40t-3t^2\). Thus \(B'(5)=125\) bacteria per hour. 3. At \(t=8\), \(B'(8)=40(8)-3(8)^2=128\) bacteria per hour. 4. The instantaneous growth rate at \(8\) hours is \(3\) bacteria per hour greater than at \(5\) hours, so the culture is growing slightly faster at \(t=8\) than at \(t=5\).

a) \(75\) bacteria per hour b) \(125\) bacteria per hour c) \(128\) bacteria per hour; this is \(3\) bacteria per hour greater than the rate at \(t=5\), so the culture is growing slightly faster at \(t=8\).

- Substitute \(t=0\) for the initial temperature and examine the long-term limit for the room temperature. - Differentiate the temperature function to obtain the instantaneous cooling rate. - Use a logarithm to solve the derivative equation for \(t\).

1. At \(t=0\), \(f(0)=21+154=175\), so the initial temperature is \(175\,{}^\circ\text{C}\). 2. As \(t\to\infty\), \(e^{-0.038t}\to 0\), so \(f(t)\to 21\). The ambient temperature is \(21\,{}^\circ\text{C}\). 3. \(f(40)=21+154e^{-0.038(40)}\approx 54.68\), so the temperature is about \(54.7\,{}^\circ\text{C}\). 4. Differentiate: \(f'(t)=-5.852e^{-0.038t}\). Set \(f'(t)=-1.2\): \(e^{-0.038t}=\frac{1.2}{5.852}\). Therefore, \(t=\frac{\ln(1.2/5.852)}{-0.038}\approx 41.70\) minutes.

a) Initial temperature: \(175\,{}^\circ\text{C}\); kitchen temperature: \(21\,{}^\circ\text{C}\) b) About \(54.7\,{}^\circ\text{C}\) c) After about \(41.7\,\text{minutes}\)

- Differentiate the exponential term carefully and keep track of the time units. - Compare the derivative with the difference between room temperature and drink temperature at the same instant. - In a relation of the form rate = constant times quantity, the constant must supply the reciprocal-time unit.

1. \(T(0)=22-14=8\,{}^\circ\text{C}\). Differentiating gives \(T'(t)=1.4e^{-0.1t}\), so \(T'(0)=1.4\,{}^\circ\text{C}/\text{min}\). The drink starts at \(8\,{}^\circ\text{C}\) and is warming at \(1.4\,{}^\circ\text{C}\) per minute initially. 2. \(T'(10)=1.4e^{-1}\approx0.515\,{}^\circ\text{C}/\text{min}\). Also, \(22-T(10)=14e^{-1}\approx5.150\,{}^\circ\text{C}\). The warming rate is one tenth of the temperature gap at that instant. 3. Since \(22-T(t)=14e^{-0.1t}\), \(0.1(22-T(t))=1.4e^{-0.1t}=T'(t)\). The constant \(0.1\,\text{min}^{-1}\) means the instantaneous warming rate equals \(10\%\) of the current temperature gap per minute.

1. \(T(0)=8\,{}^\circ\text{C}\) and \(T'(0)=1.4\,{}^\circ\text{C}/\text{min}\). The drink starts at \(8\,{}^\circ\text{C}\) and is initially warming at \(1.4\,{}^\circ\text{C}\) per minute. 2. \(T'(10)\approx0.515\,{}^\circ\text{C}/\text{min}\) and \(22-T(10)\approx5.150\,{}^\circ\text{C}\); the rate is one tenth of the temperature gap. 3. \(T'(t)=0.1(22-T(t))\). The proportionality constant is \(0.1\,\text{min}^{-1}\).

- Evaluate the model and its derivative at the same starting time before interpreting them. - Rewrite \(400-B(t)\) and compare it directly with \(B'(t)\). - A proportionality constant relating a population to a population-per-time rate must carry reciprocal-time units. - Use the size of the remaining gap to explain why the later rate differs from the initial rate.

1. \(B(0)=400-320=80\) birds. Differentiating gives \(B'(t)=16e^{-0.05t}\), so \(B'(0)=16\) birds per year. The model begins with \(80\) birds and an instantaneous growth rate of \(16\) birds per year. 2. Since \(400-B(t)=320e^{-0.05t}\), \(0.05(400-B(t))=16e^{-0.05t}=B'(t)\). The constant of proportionality is \(0.05\,\text{year}^{-1}\). 3. \(B'(20)=16e^{-1}\approx5.89\) birds per year. This is smaller than the initial rate because the population is closer to \(400\) birds after \(20\) years, so the gap \(400-B(t)\) is smaller.

1. \(B(0)=80\) birds and \(B'(0)=16\) birds per year. 2. \(B'(t)=0.05(400-B(t))\); the constant of proportionality is \(0.05\,\text{year}^{-1}\). 3. \(B'(20)=16e^{-1}\approx5.89\) birds per year, smaller than the initial \(16\) birds per year because the gap to \(400\) birds is smaller.

- The function value, first derivative, and second derivative describe three different quantities with different units. - Differentiate once to obtain the instantaneous concentration-change rate. - Differentiate again to describe how that rate is changing. - Keep the additional time unit in the denominator when interpreting the second derivative.

1. Evaluate the model: \(c(0)=5\,\text{mg/L}\) and \(c(12)=5\,\text{mg/L}\). 2. The first derivative is \(c'(t)=-0.3t^2+2.4t\). Thus \(c'(2)=3.6\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\). At \(t=2\), the concentration is increasing at \(3.6\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\). 3. The second derivative is \(c''(t)=-0.6t+2.4\), so \(c''(2)=1.2\,\frac{\text{mg}}{\text{L}\cdot\text{min}^2}\). At that instant, the concentration's rate of change is itself increasing at \(1.2\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\) per minute.

a) \(c(0)=5\,\text{mg/L}\) and \(c(12)=5\,\text{mg/L}\) b) \(c'(2)=3.6\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\); the concentration is increasing at that instantaneous rate. c) \(c''(2)=1.2\,\frac{\text{mg}}{\text{L}\cdot\text{min}^2}\); the concentration's rate of change is increasing at that instant.

- Express the water volume in terms of the water depth; similar triangles relate the surface width to the depth. - With a constant inflow rate, total volume is rate multiplied by time. - Set the two volume expressions equal and solve for \(h\) in terms of \(t\). - An instantaneous rate of change is found with a derivative. - Rewrite the square root as a rational power if that makes differentiation easier.

1. Similar triangles give \(\frac{w}{h}=\frac{40}{40}=1\), so the water surface width is \(w=h\). 2. The water volume is \(V=\frac{1}{2}wh(100)=50h^2\). 3. The inflow gives \(V(t)=200t\). Therefore, \(50h^2=200t\), so \(h(t)=2\sqrt{t}\) for \(0\le t\le400\). 4. Differentiate: \(h'(t)=\frac{1}{\sqrt{t}}\). Thus, \(h'(9)=\frac{1}{3}\,\text{cm}/\text{s}\).

a) \(h(t)=2\sqrt{t}\), for \(0\le t\le400\) b) \(h'(9)=\frac{1}{3}\,\text{cm}/\text{s}\approx0.33\,\text{cm}/\text{s}\)

- Use the volume formula for a square pyramid and similarity to relate the liquid-surface side length to its depth. - Set the geometric liquid volume equal to the inflow rate multiplied by time. - Differentiate the depth function only on the interval while the container is still filling. - The requested rate is the left-hand rate as the fill time is approached.

1. Similar pyramids give \(\frac{a}{h}=\frac{6}{6}=1\), so the side length of the liquid surface is \(a=h\). 2. The liquid volume is \(V=\frac{1}{3}a^2h=\frac{1}{3}h^3\). 3. Since \(V(t)=9t\), \(\frac{1}{3}h^3=9t\). Therefore, \(h(t)=3\sqrt[3]{t}\), for \(0\le t\le8\). 4. Differentiate for \(0<t<8\): \(h'(t)=t^{-\frac{2}{3}}\). The left-hand rate as \(t\to8^-\) is \(h'(8)=\frac{1}{4}=0.25\,\text{dm}/\text{min}\).

a) \(h(t)=3\sqrt[3]{t}\), for \(0\le t\le8\) b) \(0.25\,\text{dm}/\text{min}\), the instantaneous rising rate immediately before the container becomes full

- Interpret the units of \(K\) carefully: its values are in thousands of dollars. - Marginal cost is the derivative of total cost with respect to production level. - Evaluate the same marginal-cost function at both production levels before comparing them. - Convert thousand dollars per ton to dollars per ton consistently.

1. \(K(64)=5\sqrt{64}+20=60\), so the total cost is \(\$60{,}000\). 2. \(K'(x)=\frac{2.5}{\sqrt{x}}\). Set \(\frac{2.5}{\sqrt{x}}=0.25\), giving \(\sqrt{x}=10\) and \(x=100\) tons. 3. \(K'(64)=\frac{2.5}{8}=0.3125\) thousand dollars per ton, or \(\$312.50\) per ton. Also, \(K'(100)=0.25\) thousand dollars per ton, or \(\$250\) per ton. 4. The model's marginal cost is lower at \(100\) tons than at \(64\) tons, so an additional ton is modeled as adding less to total cost near \(100\) tons than near \(64\) tons.

a) \(\$60{,}000\) b) \(100\) tons c) At \(64\) tons, the marginal cost is \(\$312.50\) per ton; at \(100\) tons, it is \(\$250\) per ton. The modeled cost of an additional ton is lower near \(100\) tons.

- Rewrite the square root as a power and apply the power rule. - A derivative value gives the instantaneous change in period per unit change in length. - Keep track of the derivative's units. - Isolate the square root before squaring when solving for \(l\).

1. a) Apply the power rule: \(T'(l)=\frac{2\pi}{\sqrt{g}}\cdot\frac{1}{2}l^{-\frac{1}{2}}=\frac{\pi}{\sqrt{gl}}\). 2. b) \(T'(0.5)=\frac{\pi}{\sqrt{9.81(0.5)}}\approx1.419\,\text{s}/\text{m}\), and \(T'(2.0)=\frac{\pi}{\sqrt{9.81(2.0)}}\approx0.709\,\text{s}/\text{m}\). 3. Both rates are positive, so a longer pendulum has a longer period. The smaller rate at \(2.0\,\text{m}\) shows that each additional meter changes the period less for a longer pendulum. 4. c) Solve \(\frac{\pi}{\sqrt{9.81l}}=0.5\). Then \(\sqrt{9.81l}=2\pi\), so \(l=\frac{4\pi^2}{9.81}\approx4.024\,\text{m}\).

a) \(T'(l)=\frac{\pi}{\sqrt{gl}}\) b) \(T'(0.5)\approx1.419\,\text{s}/\text{m}\); \(T'(2.0)\approx0.709\,\text{s}/\text{m}\). The period increases with length, but at a decreasing rate. c) \(l\approx4.024\,\text{m}\)

- Solve the circle-area formula for the radius. - Apply the chain rule to the square-root function. - The first derivative gives the instantaneous radial growth rate. - A negative second derivative means the first derivative is decreasing.

1. a) Since \(A=\pi r^2\), \(r=\sqrt{\frac{A}{\pi}}\). Therefore, \(r(t)=\sqrt{4t+9}\). 2. b) By the chain rule, \(r'(t)=\frac{2}{\sqrt{4t+9}}\). Thus, \(r'(4)=\frac{2}{5}=0.4\,\text{m}/\text{h}\). 3. c) Differentiate again: \(r''(t)=-\frac{4}{(4t+9)^{\frac{3}{2}}}\). Therefore, \(r''(4)=-\frac{4}{125}=-0.032\,\text{m}/\text{h}^2\). 4. The negative second derivative means the radius is still increasing, but its rate of increase is decreasing.

a) \(r(t)=\sqrt{4t+9}\) b) \(r'(4)=0.4\,\text{m}/\text{h}\) c) \(r''(4)=-0.032\,\text{m}/\text{h}^2\); the radius grows at a decreasing rate.

- Substitute \(E(t)\) into the tax function before simplifying \(R(t)\). - Factor the common growth factor from the numerator before canceling. - Interpret the sign and units of the derivative as a change in real after-tax income over time.

1. Substitute the income function into the tax function: \(T(E(t))=0.000002[50{,}000(1+0.02t)]^2=5000(1+0.02t)^2\). 2. Then \(R(t)=\frac{50{,}000(1+0.02t)-5000(1+0.02t)^2}{1+0.02t}=50{,}000-5000(1+0.02t)\). 3. Simplify: \(R(t)=45{,}000-100t\). Therefore, \(R'(t)=-100\,\frac{\$}{\text{year}}\). 4. The negative derivative means the model's real after-tax income decreases by \(\$100\) per year. The nonlinear tax formula makes the tax share \(\frac{T(E(t))}{E(t)}=0.1(1+0.02t)\) increase with time, so after-tax purchasing power falls even though gross income and the modeled price level use the same growth factor.

1. \(R(t)=50{,}000-5000(1+0.02t)\) 2. \(R'(t)=-100\,\frac{\$}{\text{year}}\) 3. The model's real after-tax income decreases by \(\$100\) per year because the nonlinear tax formula takes an increasing share of gross income.

- Differentiate cumulative energy use to obtain instantaneous use per hour. - Divide cumulative energy by elapsed time to obtain average use per hour. - Compare the two rate expressions for each model for all positive times. - A constant average-rate function has derivative zero.

1. Model 1: \(E'(t)=0.04t\) and \(d(t)=0.02t\), both in kilowatt-hours per hour, so \(E'(t)=2d(t)>d(t)\). 2. Model 2: \(E'(t)=1.5\) and \(d(t)=1.5\), both in kilowatt-hours per hour, so the rates are equal. 3. Model 3: \(E'(t)=\frac{2}{\sqrt{t}}\) and \(d(t)=\frac{4}{\sqrt{t}}\), both in kilowatt-hours per hour, so \(E'(t)=\frac{1}{2}d(t)<d(t)\). 4. Therefore, Model 3 has instantaneous usage rate less than average use per hour for every \(t>0\). 5. Model 2 has \(d(t)=1.5\), so \(d'(t)=0\) and its average use per hour is constant.

a) Model 1: \(E'(t)=0.04t\), \(d(t)=0.02t\); Model 2: \(E'(t)=1.5\), \(d(t)=1.5\); Model 3: \(E'(t)=\frac{2}{\sqrt{t}}\), \(d(t)=\frac{4}{\sqrt{t}}\). All rates are in kilowatt-hours per hour. b) Model 3 c) Model 2, because \(d'(t)=0\).

- Function values describe height, while derivative values describe height change per year. - Differentiate the supplied model once and evaluate the same derivative at both requested times. - On the graph, compare how steep the curve looks near \(t=10\) and near \(t=30\).

1. \(h(0)=12-10=2\,\text{m}\). Also, \(h(20)=12-10e^{-1}\approx8.32\,\text{m}\). 2. Differentiate: \(h'(t)=0.5e^{-0.05t}\). Thus \(h'(10)=0.5e^{-0.5}\approx0.30\,\text{m/year}\). After \(10\) years, the tree's height is increasing at about \(0.30\,\text{m/year}\). 3. \(h'(30)=0.5e^{-1.5}\approx0.11\,\text{m/year}\). The modeled growth rate is smaller at \(30\) years than at \(10\) years. The graph supports this because the curve is visibly flatter at later times.

a) Initial height: \(2\,\text{m}\); height after \(20\) years: approximately \(8.32\,\text{m}\) b) \(h'(10)\approx0.30\,\text{m/year}\); the tree's height is increasing at about \(0.30\,\text{m}\) per year then. c) \(h'(30)\approx0.11\,\text{m/year}\), which is smaller than the rate after \(10\) years; the graph is flatter near \(t=30\).

- Differentiate the cumulative water-use model before using the instantaneous-rate information. - Substitute the stated time and rate into the derivative equation. - Determine the parameter's units from the term \(at^2\).

1. Differentiate the cumulative-use model: \(W'(t)=2at+6\). 2. Use the given instantaneous rate at \(t=2\): \(4a+6=14\). 3. Solving gives \(a=2\). Because \(at^2\) must have units of gallons, \(a\) has units of gallons per hour squared.

\(a=2\,\text{gal/h}^2\).

- Separate the question about graph height from the question about tangent slope. - Estimate each derivative from rise over run on its tangent segment. - A larger function value does not imply a larger derivative value.

1. At \(t=4\), the graph shows \(A(4)\approx22\,\text{MJ}\) and \(B(4)\approx24\,\text{MJ}\), so system B contains more stored energy. 2. The tangent to A rises from about \(18\) to \(26\) megajoules while time increases from \(3\) to \(5\) hours, giving \(A'(4)\approx\frac{26-18}{5-3}=4\,\text{MJ/h}\). 3. The tangent to B rises from about \(21\) to \(27\) megajoules over the same two-hour span, giving \(B'(4)\approx3\,\text{MJ/h}\). 4. System A is increasing faster because its tangent slope is greater, even though system B has the larger stored-energy value at that instant.

a) System B b) \(A'(4)\approx4\,\text{MJ/h}\) and \(B'(4)\approx3\,\text{MJ/h}\) c) System A is increasing faster. Graph height gives the quantity value; tangent slope gives the instantaneous rate.

- For part a), compare the heights of the two curves at \(t=4\). - For part b), estimate rise over run along each straight tangent segment. - For part c), distinguish the value of a function from the slope of its graph.

1. At \(t=4\), the graph shows \(A(4)=22\,\text{MJ}\) and \(B(4)=24\,\text{MJ}\), so system \(B\) contains more stored energy. 2. The tangent to \(A\) has slope \(4\), so \(A'(4)\approx4\,\text{MJ/h}\). 3. The tangent to \(B\) has slope \(3\), so \(B'(4)\approx3\,\text{MJ/h}\). 4. Because \(4>3\), system \(A\) is increasing faster at \(t=4\). 5. Graph height represents stored energy, while tangent slope represents instantaneous rate of change. A system can therefore contain more energy while increasing more slowly.

a) System \(B\). b) \(A'(4)\approx4\,\text{MJ/h}\) and \(B'(4)\approx3\,\text{MJ/h}\). c) System \(A\) is increasing faster. The higher graph has the greater stored-energy value, while the steeper tangent has the greater instantaneous rate of change.

- The slope of the height graph represents the rate at which the water level rises. - With constant inflow, a steeper height graph corresponds to a smaller horizontal cross-sectional area. - Use the supplied formula, not visual measurement, for the requested two-decimal numerical values. - Average rate of rise is a secant slope over the stated interval.

1. The graph's slope increases until about \(t=6\,\text{s}\) and decreases afterward. Therefore, the water level rises fastest at about \(t=6\,\text{s}\). 2. With constant inflow, a larger rise rate corresponds to a smaller horizontal cross-sectional area. The container is relatively wide near the bottom, narrows toward the middle, and widens again near the top. 3. Using the supplied formula, \(h(0)=\frac{15}{1+e^{4.8}}\approx0.12\,\text{cm}\) and \(h(12)=\frac{15}{1+e^{-4.8}}\approx14.88\,\text{cm}\). 4. The average rate of rise is \(\frac{h(12)-h(0)}{12}\approx1.23\,\text{cm/s}\).

a) The water level rises fastest at about \(t=6\,\text{s}\); its rate of rise increases before then and decreases afterward. b) The container is wide near the bottom, narrowest around the middle, and wider again near the top. c) \(h(12)\approx14.88\,\text{cm}\); average rate of rise \(\approx1.23\,\text{cm/s}\)

- What value must the model approach as \(t\) becomes large? - Use the initial population to relate \(b\) and \(c\). - The initial growth rate is a value of the first derivative, not a population value. - Check that the sign of \(k\) is consistent with a bounded population approaching its carrying capacity.

1. A finite carrying capacity requires \(k<0\), so \(e^{kt}\to0\) as \(t\to\infty\). Therefore, \(c=2000\). 2. Use \(N(0)=200\): \(b+2000=200\), so \(b=-1800\). 3. Differentiate: \(N'(t)=bke^{kt}\). The initial growth-rate condition gives \(N'(0)=bk=100\). Thus \((-1800)k=100\), so \(k=-\frac{1}{18}\). 4. Hence the population model is \(N(t)=-1800e^{-t/18}+2000\).

\(b=-1800\), \(k=-\frac{1}{18}\), and \(c=2000\); \(N(t)=-1800e^{-t/18}+2000\).