- 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 \(I'(6)=18\,\text{cartons/min}\).

- Which part of the price depends on the banner's side length? - What units should a derivative of dollars with respect to feet have?

1. Differentiate cost with respect to side length: \(C'(x)=4x\,\text{dollars/ft}\). 2. For a small increase in side length, the cost increases at about \(4x\) dollars per additional foot when the current side length is \(x\) feet. The fixed \(\$20\) charge does not affect this rate.

\(C'(x)=4x\,\text{dollars/ft}\). This is the instantaneous increase in cost per foot of side length.

- A decreasing water volume has a negative derivative. - Compare how steeply the graph falls at the two times.

1. The volume graph falls at both times, so the instantaneous volume rates are negative. 2. The graph is steeper downward at \(t=3\), meaning the volume is decreasing faster then.

At \(t=3\); the graph's downward slope is steeper there.

- The given rate is a derivative, not a total amount of water. - Differentiate the model before substituting the stated time. - Find the unit that makes \(at^2\) a quantity measured in gallons.

1. The rate of water use is \(W'(t)=2at+6\). 2. Use the rate at \(t=2\): \(4a+6=14\), so \(a=2\). 3. Since \(at^2\) is a volume in gallons, \(a\) has units of gallons per hour squared.

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

- 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.

- A rate in kilowatts comes from differentiating an energy total measured in kilowatt-hours. - A candidate must match both given rates, not just one. - Check the rate after one hour and after three hours separately.

1. Each model's electricity-use rate is its derivative, measured in kilowatts. 2. C gives \(E'(t)=2t+4\), so the rates are \(E'(1)=6\) and \(E'(3)=10\). 3. A gives rates \(6\) and \(14\), and B gives a constant rate of \(8\). Only C matches both times.

Model C, \(E(t)=t^2+4t\).

- The tank is empty when the remaining volume first reaches zero. - Check whether the model still describes a possible water volume at the later time. - A derivative can be calculated outside a model's meaningful time interval without representing a real rate.

1. Solve \(60-7t-t^2=0\), or \((t-5)(t+12)=0\). The tank empties after \(5\) minutes; the negative root is outside the context. 2. The model has \(V(7)=60-49-49=-38\,\text{L}\), which is impossible for a tank's remaining water. The real draining process described ends when the tank is empty, so \(V'(7)\) cannot describe its actual water loss.

a) After \(5\,\text{min}\). b) The model already gives a negative water volume at \(t=7\). It only describes the draining tank for \(0\leq t\leq5\), so \(V'(7)\) has no valid physical interpretation here.

- 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 \(t=8\).

- Evaluate the temperature model at the start and consider its long-term value. - A cooling rate is negative even though the requested cooling speed is positive. - Set the derivative equal to the rate in part c), then solve for time.

1. \(T(0)=21+154=175\,{}^\circ\text{C}\). Since the exponential term tends to zero, the kitchen temperature is \(21\,{}^\circ\text{C}\). 2. \(T(40)=21+154e^{-1.52}\approx54.7\,{}^\circ\text{C}\). 3. \(T'(t)=-5.852e^{-0.038t}\). Cooling at \(1.2\,{}^\circ\text{C}/\text{min}\) means \(T'(t)=-1.2\). Thus \(e^{-0.038t}=\frac{1.2}{5.852}\) and \(t=-\frac{\ln(1.2/5.852)}{0.038}\approx41.7\,\text{min}\).

a) Initial temperature: \(175\,{}^\circ\text{C}\); kitchen temperature: \(21\,{}^\circ\text{C}\). b) Approximately \(54.7\,{}^\circ\text{C}\). c) After approximately \(41.7\,\text{min}\).

- 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.

- Treat \(g\) as constant when differentiating with respect to length. - Compare the two derivative values, keeping the units of seconds per meter. - In part c), solve the equation involving the derivative, not the period itself.

1. Differentiate: \(T'(l)=\frac{2\pi}{\sqrt{g}}\cdot\frac{1}{2\sqrt{l}}=\frac{\pi}{\sqrt{gl}}\). 2. \(T'(0.5)=\frac{\pi}{\sqrt{9.81\cdot0.5}}\approx1.419\,\text{s/m}\) and \(T'(2)=\frac{\pi}{\sqrt{9.81\cdot2}}\approx0.709\,\text{s/m}\). Increasing length raises the period, but the change per extra meter is smaller for longer pendulums. 3. Solve \(\frac{\pi}{\sqrt{9.81l}}=0.5\): \(\sqrt{9.81l}=2\pi\), so \(l=\frac{4\pi^2}{9.81}\approx4.024\,\text{m}\).

a) \(T'(l)=\frac{\pi}{\sqrt{gl}}\). b) \(T'(0.5)\approx1.419\,\text{s/m}\) and \(T'(2.0)\approx0.709\,\text{s/m}\). The period increases with length, more slowly at greater lengths. 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 \(t=30\).

- Compare the curves' heights at the marked time for stored amounts. - Estimate each straight tangent's rise over run for the rates. - Keep the value of a function separate from the slope of its graph.

1. At \(t=4\), \(A(4)=22\,\text{MJ}\) and \(B(4)=24\,\text{MJ}\), so Tank B stores more heat. 2. The tangent to A has a slope of \(4\), giving \(A'(4)\approx4\,\text{MJ/h}\). The tangent to B has a slope of \(3\), giving \(B'(4)\approx3\,\text{MJ/h}\). 3. Tank A's stored heat is increasing faster because its tangent is steeper. Graph height gives the energy stored; tangent slope gives its instantaneous rate of change.

a) Tank B. b) \(A'(4)\approx4\,\text{MJ/h}\); \(B'(4)\approx3\,\text{MJ/h}\). c) Tank A is increasing faster. The higher curve shows more stored heat, not necessarily a faster increase.

- What do the graph's height and the table value describe at the same time? - How could you compare two rates if one is given by a tangent and the other by a model? - What symmetry would a quadratic need to pass through the hill's three measurements? - Could a different smooth curve pass through those points with a different slope at the middle time?

1. At \(t=2\), the meadow has about \(16\,\text{cm}\) of snow, while the hill has \(20\,\text{cm}\). Snow is deeper at the hill. 2. The meadow's tangent slope at \(t=2\) is about \(2\,\text{cm/h}\). 3. The unique quadratic through the hill's three values is \(H(t)=20-(t-2)^2\), so \(H'(t)=-2(t-2)\) and \(H'(2)=0\,\text{cm/h}\). Snow depth is increasing faster in the meadow. 4. Without the quadratic assumption, different smooth models could fit the three table values but have different derivatives at \(t=2\). The symmetric table gives a centered rate estimate of zero, not an exact instantaneous rate.

a) The sledding hill (\(20\,\text{cm}\), compared with about \(16\,\text{cm}\) in the meadow). b) Meadow: about \(2\,\text{cm/h}\); hill: exactly \(0\,\text{cm/h}\) under the quadratic model. Snow depth is increasing faster in the meadow. c) No. Without the quadratic assumption, the three readings do not fix the exact derivative at \(t=2\).