Free math worksheets

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

- The ratio \(m(t)/m_0\) is the fraction remaining. - At the half-life, that ratio is \(0.5\). - Differentiate the exponential model and interpret the derivative's sign.

1. The remaining fraction after \(25\) years is \(e^{-0.0231(25)}\approx 0.5613\). Therefore, about \(56.13\%\) remains. 2. The half-life satisfies \(e^{-0.0231T_H}=0.5\), so \(T_H=\frac{\ln 2}{0.0231}\approx 30.01\) years. 3. Differentiate: \(m'(t)=-0.0231m_0e^{-0.0231t}\). With \(m_0=500\), \(m'(50)\approx -3.64\,\text{mg/year}\). The negative sign indicates that the mass is decreasing.

a) About \(56.13\%\) b) \(T_H\approx 30.01\,\text{years}\) c) \(m'(50)\approx -3.64\,\text{mg/year}\); the negative sign indicates a decrease

- 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.

- Average vertical velocity uses the change in height over the change in time. - The sign of \(v(t)=h'(t)\) tells whether the balloon is rising or falling. - A change from positive velocity to negative velocity identifies when upward motion ends. - Acceleration describes how velocity is changing; interpret its sign in terms of velocity, not height.

1. Since \(h(0)=20\) and \(h(5)=70\), the average vertical velocity on \([0,5]\) is \(\frac{70-20}{5}=10\,\frac{\text{m}}{\text{min}}\). 2. The velocity is \(v(t)=h'(t)=-0.6t^2+6t=-0.6t(t-10)\). Thus \(v(3)=12.6\,\frac{\text{m}}{\text{min}}\). 3. On \(0<t<10\), \(v(t)>0\), and on \(10<t\le 15\), \(v(t)<0\). Therefore, the balloon stops rising and begins falling at \(t=10\,\text{min}\). Its height then is \(h(10)=120\,\text{m}\). 4. The acceleration is \(a(t)=v'(t)=-1.2t+6\). Thus \(a(3)=2.4\,\frac{\text{m}}{\text{min}^2}\) and \(a(7)=-2.4\,\frac{\text{m}}{\text{min}^2}\). At \(t=3\), the positive acceleration means the vertical velocity is increasing; at \(t=7\), the negative acceleration means the vertical velocity is decreasing.

a) \(10\,\frac{\text{m}}{\text{min}}\) b) \(12.6\,\frac{\text{m}}{\text{min}}\) c) The balloon stops rising and begins falling at \(t=10\,\text{min}\), when its height is \(120\,\text{m}\). d) \(a(3)=2.4\,\frac{\text{m}}{\text{min}^2}\), so the vertical velocity is increasing then. \(a(7)=-2.4\,\frac{\text{m}}{\text{min}^2}\), so the vertical velocity is decreasing then.

- The beginning of the observation period corresponds to \(t=0\). - Use the sign of the first derivative to find where the water level decreases. - Solve \(h'(t)=0\) to locate interior candidates for extrema of the water level. - Check both critical values and endpoints for an absolute maximum. - To find the fastest decrease, minimize the first derivative on the time interval.

1. Evaluate the function: \(h(0)=200\) and \(h(5)=-125+375-240+200=210\). The levels are \(200\,\text{cm}\) and \(210\,\text{cm}\). 2. The derivative is \(h'(t)=-3t^2+30t-48=-3(t-2)(t-8)\). It is negative on \([0, 2)\) and \((8, 12]\). Therefore, the water level decreases from \(t=0\) to \(t=2\) and from \(t=8\) to \(t=12\). 3. Check the critical values and endpoints: \(h(0)=200\), \(h(2)=156\), \(h(8)=264\), and \(h(12)=56\). The absolute maximum is \(264\,\text{cm}\) at \(t=8\). 4. The fastest decrease occurs where \(h'(t)\) is smallest on \([0, 12]\). The derivative is a downward-opening parabola, so its minimum on the closed interval occurs at an endpoint. Since \(h'(0)=-48\) and \(h'(12)=-120\), the water level is decreasing most rapidly at \(t=12\).

a) \(200\,\text{cm}\) initially and \(210\,\text{cm}\) after \(5\) hours b) From \(t=0\) to \(t=2\), and from \(t=8\) to \(t=12\) c) After \(8\) hours; \(264\,\text{cm}\) d) At \(t=12\) hours

- 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.

- A maximum concentration occurs at a critical value of the concentration function. - The first derivative is the concentration's rate of change. - “Increasing most rapidly” asks when the first derivative is greatest. - Use only times in the stated interval.

1. The first derivative is \(k'(t)=-3t^2+18t+48=-3(t-8)(t+2)\). The only critical value in \([0, 12]\) is \(t=8\). 2. Since \(k''(t)=-6t+18\) and \(k''(8)=-30<0\), \(t=8\) gives a local maximum. Comparing with the endpoints confirms the absolute maximum: \(k(0)=50\), \(k(8)=498\), and \(k(12)=194\). Thus, the maximum concentration is \(498\,\text{mg/L}\) after \(8\) hours. 3. The concentration increases most rapidly when \(k'(t)\) is largest. Solve \(k''(t)=0\): \(t=3\). Because \(k'''(t)=-6<0\), the first derivative has a maximum there. The rate is \(k'(3)=75\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\).

a) After \(8\) hours; \(498\,\text{mg/L}\) b) After \(3\) hours; \(75\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\)

- Find the critical values of the height function and check the interval endpoints. - The first derivative represents the plant's growth rate. - The fastest growth occurs when the first derivative is greatest. - Keep only times in the stated interval.

1. The first derivative is \(h'(t)=-0.06t^2+1.2t=-0.06t(t-20)\). The critical values are \(t=0\) and \(t=20\). 2. Since \(h''(t)=-0.12t+1.2\) and \(h''(20)=-1.2<0\), \(t=20\) gives a local maximum. Comparing \(h(0)=20\), \(h(20)=100\), and \(h(25)=82.5\) confirms that the maximum modeled height occurs after \(20\) days. 3. The plant grows most rapidly when \(h'(t)\) is largest. Solve \(h''(t)=0\): \(t=10\). Since \(h'''(t)=-0.12<0\), the first derivative has a maximum there. The growth rate is \(h'(10)=6\,\frac{\text{cm}}{\text{day}}\).

a) After \(20\) days b) After \(10\) days; \(6\,\frac{\text{cm}}{\text{day}}\)

- Acceleration is the derivative of velocity. - Find the absolute maximum of velocity by checking critical values and endpoints. - At an inflection point of velocity, interpret the tangent slope as acceleration with the correct units.

1. The first two derivatives are \(v'(t)=-0.3t^2+3.6t\) and \(v''(t)=-0.6t+3.6\). 2. Solve \(v'(t)=0\): \(t=0\) or \(t=12\). Since \(v''(12)=-3.6<0\), \(t=12\) gives a local maximum. Comparing \(v(0)=5\), \(v(12)=91.4\), and \(v(15)=72.5\) confirms that the maximum velocity occurs after \(12\) seconds. 3. Solve \(v''(t)=0\): \(t=6\). Because \(v'''(t)=-0.6<0\), the acceleration \(v'(t)\) reaches its maximum there. The slope is \(v'(6)=10.8\,\frac{\text{m}}{\text{s}^2}\). Thus, after \(6\) seconds, the vehicle has its greatest acceleration.

a) After \(12\) seconds b) The inflection time is \(t=6\) seconds. The tangent slope is \(10.8\,\frac{\text{m}}{\text{s}^2}\), the vehicle's maximum acceleration.

- 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 \(t=8\).

- Convert the percent increase to a growth factor. - Use \(a=e^k\) to convert bases. - The derivative of the area model is the instantaneous area-growth rate.

1. The initial value is \(b=1500\), and the hourly growth factor is \(a=1.035\). Thus, \(A(t)=1500(1.035)^t\). 2. Since \(1.035=e^k\), \(k=\ln(1.035)\approx0.0344\). Therefore, \(A(t)=1500e^{kt}\), where \(k=\ln(1.035)\). 3. The derivative is \(A'(t)=1500ke^{kt}\). Thus, \(A'(10)=1500\ln(1.035)(1.035)^{10}\approx72.79\,\text{mm}^2/\text{h}\).

1. \(A(t)=1500(1.035)^t\) 2. \(k=\ln(1.035)\approx0.0344\) 3. \(A'(10)\approx72.79\,\text{mm}^2/\text{h}\)

- Use the two data points to write two equations involving \(m_0\) and \(a\). - Relate the daily factor to the meaning of half-life. - Use the identity \(a^t=e^{\ln(a)t}\). - The instantaneous rate of change is the value of the derivative at the specified time.

1. The two data points give \(m_0a^2=20\) and \(m_0a^5=2.5\). 2. Dividing the second equation by the first gives \(a^3=\frac{2.5}{20}=0.125\), so \(a=0.5\). 3. Substituting into \(m_0a^2=20\) gives \(m_0(0.5)^2=20\), so \(m_0=80\). Thus, \(m(t)=80(0.5)^t\). 4. Since the mass is multiplied by \(0.5\) each day, the half-life is \(1\) day. 5. Because \((0.5)^t=e^{\ln(0.5)t}\), the natural exponential form is \(m(t)=80e^{\ln(0.5)t}\). 6. Differentiate: \(m'(t)=80\ln(0.5)e^{\ln(0.5)t}\). Therefore, \(m'(1)=40\ln(0.5)\approx -27.73\,\text{mg/day}\).

a) \(a=0.5\); \(m_0=80\,\text{mg}\) b) \(1\,\text{day}\) c) \(m(t)=80e^{\ln(0.5)t}\); \(m'(1)=40\ln(0.5)\approx -27.73\,\text{mg/day}\)

- Compare ratios of consecutive values over equal time intervals. - Relate a growth factor over two days to a growth factor over one day. - Use \(a^t=e^{\ln(a)t}\) to convert between exponential forms. - The growth rate is the first derivative of the area function. - At the doubling time, the function value is twice its initial value.

1. For consecutive \(2\)-day intervals, the ratios are \(\frac{6.1}{5.0}=1.22\), \(\frac{7.4}{6.1}\approx 1.213\), \(\frac{9.0}{7.4}\approx 1.216\), and \(\frac{11.0}{9.0}\approx 1.222\). These ratios are nearly constant, so an exponential model is reasonable. 2. Using the first ratio, the \(2\)-day growth factor is \(1.22\). The daily growth factor is \(\sqrt{1.22}\approx 1.1045\), which corresponds to an increase of about \(10.45\%\) per day. 3. Since \(A_0=5.0\) and \(e^{2k}=1.22\), \(k=\frac{\ln(1.22)}{2}\approx 0.0994\). A model is \(A(t)=5.0e^{0.0994t}\). 4. Differentiate: \(A'(t)=5.0(0.0994)e^{0.0994t}\approx 0.497e^{0.0994t}\). Thus, \(A'(5)\approx 0.817\,\text{m}^2/\text{day}\). 5. For the doubling time \(T\), solve \(e^{kT}=2\). Then \(T=\frac{\ln 2}{k}\approx 6.97\) days.

a) The ratios for consecutive equal time intervals are all approximately \(1.22\). b) \(2\)-day factor: \(1.22\); daily factor: about \(1.1045\); daily increase: about \(10.45\%\) c) \(A(t)=5.0e^{0.0994t}\) d) \(A'(t)\approx 0.497e^{0.0994t}\); \(A'(5)\approx 0.817\,\text{m}^2/\text{day}\) e) About \(6.97\,\text{days}\)

- Compare ratios of consecutive values over equal time intervals. - Relate a decay factor over three hours to a decay factor over one hour. - Use the natural logarithm to convert the decay factor to the base-\(e\) form. - What does the sign of the derivative tell you about the concentration? - At the half-life, half of the initial concentration remains.

1. The ratios for consecutive \(3\)-hour intervals are \(\frac{84.0}{120}=0.7\), \(\frac{58.8}{84.0}=0.7\), \(\frac{41.2}{58.8}\approx 0.701\), and \(\frac{28.8}{41.2}\approx 0.699\). The ratios are nearly constant, so exponential decay is a reasonable model. 2. The \(3\)-hour decay factor is \(0.7\). The hourly factor is \(0.7^{1/3}\approx 0.8879\), so the concentration decreases by about \(11.21\%\) per hour. 3. Since \(C_0=120\) and \(e^{3k}=0.7\), \(k=\frac{\ln(0.7)}{3}\approx -0.1189\). Thus, \(C(t)=120e^{-0.1189t}\). 4. Differentiate: \(C'(t)=120(-0.1189)e^{-0.1189t}\approx -14.267e^{-0.1189t}\). Therefore, \(C'(6)\approx -6.99\,\text{mg}/(\text{L}\cdot\text{h})\). 5. For the half-life \(T\), solve \(e^{kT}=0.5\). Then \(T=\frac{\ln(0.5)}{k}\approx 5.83\) hours.

a) The ratios for consecutive equal time intervals are all approximately \(0.7\). b) \(3\)-hour factor: \(0.7\); hourly factor: about \(0.8879\); hourly decrease: about \(11.21\%\) c) \(C(t)=120e^{-0.1189t}\) d) \(C'(6)\approx -6.99\,\text{mg}/(\text{L}\cdot\text{h})\) e) About \(5.83\,\text{hours}\)

- Use the half-life to write an equation for \(k\). - The parameter \(b\) is the value at \(t=0\). - Use logarithms to solve the threshold inequality. - The derivative gives the instantaneous change in activity per hour.

1. The initial value is \(b=450\). A half-life of \(6\) hours gives \(e^{6k}=0.5\), so \(k=\frac{\ln(0.5)}{6}\approx -0.1155\). Thus, \(A(t)=450e^{(\ln(0.5)/6)t}\approx 450e^{-0.1155t}\). 2. The parameter \(b=450\) is the initial activity in megabecquerels. The hourly factor is \(e^k\approx 0.8909\), so the activity decreases by about \(10.91\%\) each hour. 3. Solve \(450e^{kt}<20\). Since \(k<0\), \(t>\frac{\ln(20/450)}{k}\approx 26.95\). The activity is below \(20\,\text{MBq}\) after about \(26.95\) hours. 4. Differentiate: \(A'(t)=450ke^{kt}\). Then \(A'(0)=450k\approx -51.99\,\text{MBq/h}\), and \(A'(12)=450ke^{12k}\approx -13.00\,\text{MBq/h}\). The negative values mean the activity is decreasing at those instantaneous rates.

a) \(A(t)=450e^{(\ln(0.5)/6)t}\approx 450e^{-0.1155t}\) b) \(b=450\,\text{MBq}\) is the initial activity; hourly decrease: about \(10.91\%\) c) For \(t>26.95\,\text{hours}\) d) \(A'(t)=450ke^{kt}\), where \(k=\frac{\ln(0.5)}{6}\); \(A'(0)\approx -51.99\,\text{MBq/h}\), \(A'(12)\approx -13.00\,\text{MBq/h}\)

- Use the value after \(10\) minutes to solve for \(k\). - At the half-life, the concentration is half the initial concentration. - Use a logarithm to solve for the time at which a threshold is reached. - Include concentration per minute in the derivative's units.

1. Using \(c(10)=200\), \(320e^{10k}=200\). Thus, \(e^{10k}=0.625\), so \(k=\frac{\ln(0.625)}{10}\approx -0.0470\). The model is \(c(t)=320e^{(\ln(0.625)/10)t}\approx 320e^{-0.0470t}\). 2. For the half-life \(T\), solve \(e^{kT}=0.5\). Thus, \(T=\frac{\ln(0.5)}{k}\approx 14.75\) minutes. 3. Solve \(320e^{kt}<10\). Since \(k<0\), \(t>\frac{\ln(10/320)}{k}\approx 73.74\). The concentration is below the limit after about \(73.74\) minutes. 4. Differentiate: \(c'(t)=320ke^{kt}\). At \(t=20\), \(c'(20)\approx -5.88\,\text{mg}/(\text{L}\cdot\text{min})\). This means the concentration is decreasing at about \(5.88\,\text{mg/L}\) per minute at that instant.

a) \(c(t)=320e^{(\ln(0.625)/10)t}\approx 320e^{-0.0470t}\) b) About \(14.75\,\text{minutes}\) c) For \(t>73.74\,\text{minutes}\) d) \(c'(t)=320ke^{kt}\), where \(k=\frac{\ln(0.625)}{10}\); \(c'(20)\approx -5.88\,\text{mg}/(\text{L}\cdot\text{min})\)

- Convert each hourly percent change to a multiplication factor, then use \(k=\ln(q)\). - To find when two populations are equal, set their functions equal. - Differentiate each exponential model to find its instantaneous rate of change. - Interpret a negative rate as a decrease in cells per hour.

1. For culture A, the hourly factor is \(1.042\), so \(k_A=\ln(1.042)\approx 0.0411\). Thus, \(f_A(t)=1200e^{\ln(1.042)t}\approx 1200e^{0.0411t}\). 2. For culture B, the hourly factor is \(0.965\), so \(k_B=\ln(0.965)\approx -0.0356\). Thus, \(f_B(t)=5000e^{\ln(0.965)t}\approx 5000e^{-0.0356t}\). 3. For culture A to reach \(2000\) cells, solve \(1200e^{k_At}=2000\). Then \(t=\frac{\ln(2000/1200)}{k_A}\approx 12.42\) hours. 4. Set the models equal: \(1200e^{k_At}=5000e^{k_Bt}\). Therefore, \(t=\frac{\ln(5000/1200)}{k_A-k_B}\approx 18.59\) hours. 5. The derivatives are \(f_A'(t)=1200k_Ae^{k_At}\) and \(f_B'(t)=5000k_Be^{k_Bt}\). For culture B, \(f_B'(0)\approx -178.14\) cells per hour and \(f_B'(10)\approx -124.75\) cells per hour. Culture B is decreasing at those instantaneous rates.

a) \(f_A(t)=1200e^{\ln(1.042)t}\); \(f_B(t)=5000e^{\ln(0.965)t}\) b) About \(12.42\,\text{hours}\) c) About \(18.59\,\text{hours}\) d) \(f_A'(t)=1200\ln(1.042)e^{\ln(1.042)t}\); \(f_B'(t)=5000\ln(0.965)e^{\ln(0.965)t}\); \(f_B'(0)\approx -178.14\) cells per hour, \(f_B'(10)\approx -124.75\) cells per hour

- 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 \(41.7\,\text{minutes}\)

- The limiting temperature determines \(A\). - Use the initial condition to find \(B\). - Substitute the known point and use logarithms to find \(k\). - Differentiate the model and set the derivative equal to the given rate.

1. The limiting temperature is the room temperature, so \(A=25\). Since \(g(0)=25-B=7\), \(B=18\). 2. Use \(g(20)=16\): \(16=25-18e^{-20k}\). Thus, \(e^{-20k}=0.5\), so \(k=\frac{\ln2}{20}\approx0.0347\,\text{min}^{-1}\). 3. For the second drink, \(g'(t)=18(0.04)e^{-0.04t}=0.72e^{-0.04t}\). Set \(0.72e^{-0.04t}=0.25\). Then \(t=\frac{-\ln(0.25/0.72)}{0.04}\approx26.4\) minutes.

a) \(A=25\), \(B=18\) b) \(k\approx0.0347\,\text{min}^{-1}\) c) About \(26.4\) 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.

- Use the initial value to identify \(a\). - Substitute the value at \(t=5\) to solve for \(k\). - Use the derivative to find the instantaneous growth rate.

1. Since \(f(0)=250\), \(a=250\). Using \(f(5)=600\) gives \(600=250e^{5k}\), so \(e^{5k}=2.4\). Thus, \(k=\frac{\ln(2.4)}{5}\approx 0.1751\), and \(f(t)=250e^{(\ln(2.4)/5)t}\). 2. For the doubling time \(T\), \(e^{kT}=2\). Therefore, \(T=\frac{\ln 2}{k}\approx 3.96\) hours. 3. To reach \(2500\,\text{mm}^2\), solve \(250e^{kt}=2500\). Then \(e^{kt}=10\), so \(t=\frac{\ln 10}{k}\approx 13.15\) hours. 4. Differentiate: \(f'(t)=250ke^{kt}\). Thus, \(f'(8)\approx 177.64\,\text{mm}^2/\text{h}\).

a) \(f(t)=250e^{(\ln(2.4)/5)t}\approx 250e^{0.1751t}\) b) About \(3.96\,\text{hours}\) c) About \(13.15\,\text{hours}\) d) \(f'(8)\approx 177.64\,\text{mm}^2/\text{h}\)

- Use the mass after \(10\) days to solve for \(\lambda\). - At the half-life, half the initial mass remains. - Differentiate the model to find the instantaneous decay rate. - A negative derivative indicates that the mass is decreasing.

1. The initial mass is \(m_0=120\). Using \(m(10)=45\) gives \(45=120e^{-10\lambda}\), so \(e^{-10\lambda}=0.375\). Therefore, \(\lambda=-\frac{\ln(0.375)}{10}\approx 0.0981\), and \(m(t)=120e^{-\lambda t}\). 2. For the half-life \(T\), \(e^{-\lambda T}=0.5\). Thus, \(T=\frac{\ln 2}{\lambda}\approx 7.07\) days. 3. To reach \(10\,\text{mg}\), solve \(120e^{-\lambda t}=10\). Then \(e^{-\lambda t}=\frac{1}{12}\), so \(t=\frac{\ln 12}{\lambda}\approx 25.33\) days. 4. Differentiate: \(m'(t)=-120\lambda e^{-\lambda t}\). Therefore, \(m'(0)\approx -11.77\,\text{mg/day}\), and \(m'(20)\approx -1.66\,\text{mg/day}\).

a) \(m(t)=120e^{-\lambda t}\), where \(\lambda=-\frac{\ln(0.375)}{10}\approx 0.0981\) b) About \(7.07\,\text{days}\) c) About \(25.33\,\text{days}\) d) \(m'(0)\approx -11.77\,\text{mg/day}\); \(m'(20)\approx -1.66\,\text{mg/day}\)

- 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.

- Evaluate the model at \(t=0\) and as \(t\to\infty\). - Consider how the denominator changes as \(t\) increases. - Isolate the exponential expression before taking a logarithm. - The growth rate is the first derivative of the height function.

1. \(h(0)=\frac{3}{6}=0.5\,\text{m}\). As \(t\to\infty\), \(e^{-0.2t}\to 0\), so \(h(t)\to 3\,\text{m}\). 2. The function \(e^{-0.2t}\) strictly decreases. Therefore, the positive denominator \(1+5e^{-0.2t}\) strictly decreases, while the numerator remains positive and constant. Thus, the quotient strictly increases. 3. Set \(h(t)=1.5\). Then \(1+5e^{-0.2t}=2\), so \(e^{0.2t}=5\). Therefore, \(t=5\ln 5\approx 8.05\) weeks. 4. Differentiate: \(h'(t)=\frac{3e^{-0.2t}}{(1+5e^{-0.2t})^2}\). Thus, \(h'(10)\approx 0.144\,\text{m/week}\).

a) Initial height: \(0.5\,\text{m}\); theoretical maximum: \(3\,\text{m}\) b) The positive denominator decreases as \(t\) increases, so the quotient increases. c) \(t=5\ln 5\approx 8.05\,\text{weeks}\) d) \(h'(10)\approx 0.144\,\text{m/week}\)

- 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) \(l\approx4.024\,\text{m}\)

- Rewrite the square root as a rational power. - Interpret a derivative of speed with respect to distance using quotient units. - To show a rate decreases, examine the sign of its derivative. - Simplify the units after dividing speed units by distance units.

1. a) Rewrite \(v(s)=\sqrt{2g}\,s^{\frac{1}{2}}\). Then \(v'(s)=\frac{\sqrt{2g}}{2\sqrt{s}}=\sqrt{\frac{g}{2s}}\), for \(s>0\). 2. b) \(v'(2)=\sqrt{\frac{9.81}{4}}\approx1.566\,\text{s}^{-1}\), and \(v'(8)=\sqrt{\frac{9.81}{16}}\approx0.783\,\text{s}^{-1}\). 3. The derivative measures the increase in speed, in meters per second, per additional meter fallen. Its units simplify to \(\text{s}^{-1}\). 4. c) Differentiate again: \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}\). 5. Since \(v''(s)<0\) for every \(s>0\), \(v'(s)\) is strictly decreasing. Therefore, the speed gained per additional meter becomes smaller as the object falls farther.

a) \(v'(s)=\sqrt{\frac{g}{2s}}\) b) \(v'(2)\approx1.566\,\text{s}^{-1}\); \(v'(8)\approx0.783\,\text{s}^{-1}\). It is the increase in speed per additional meter fallen. c) \(v''(s)=-\frac{\sqrt{2g}}{4s^{\frac{3}{2}}}<0\), so \(v'(s)\) decreases as \(s\) increases.

- 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.

- Use symmetry to locate the right edge from the total width. - Convert a percent grade to a decimal slope. - The derivative at the edge gives a condition on \(a\). - Evaluate the model at the edge to find the depth.

1. Because the bowl is \(12\,\text{m}\) wide and symmetric, the right edge is at \(x=6\). 2. A grade of \(120\%\) corresponds to slope \(1.2\), so \(f'(6)=1.2\). 3. Since \(f'(x)=2ax\), the condition gives \(12a=1.2\), so \(a=0.1\). Thus \(f(x)=0.1x^2\). 4. The depth is the height of the edge above the origin: \(f(6)=0.1(6)^2=3.6\,\text{m}\).

a) \(f(x)=0.1x^2\) b) \(3.6\,\text{m}\)

- 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 \(d'(t)=0\).

- Find the zeros of the first derivative. - Check critical values and endpoints for the maximum. - Use the sign of the first derivative for increasing and decreasing intervals. - The second derivative describes how the growth rate changes.

1. The derivatives are \(f'(t)=-3t(t-8)\) and \(f''(t)=-6t+24\). The critical values in the interval are \(t=0\) and \(t=8\). 2. Since \(f''(8)=-24<0\), the population has a local maximum at \(t=8\). Comparing the endpoints gives \(f(0)=60\), \(f(8)=316\), and \(f(12)=60\), so the maximum population is \(316\) insects after \(8\) weeks. 3. The first derivative is positive for \(0<t<8\) and negative for \(8<t<12\). Therefore, the population increases on \([0, 8]\) and decreases on \([8, 12]\). 4. Solving \(f''(t)=0\) gives \(t=4\). At this inflection time, the growth rate \(f'(t)\) reaches its maximum. Thus, the population is increasing most rapidly after \(4\) weeks.

a) After \(8\) weeks b) \(316\) insects c) Increasing on \([0, 8]\); decreasing on \([8, 12]\) d) At \(t=4\), the population is increasing most rapidly.

- Convert the hourly percent decrease to a remaining factor. - Use \(e^{-k}=0.88\) to find \(k\). - One-fourth remaining means \(N(t)/N_0=0.25\). - Differentiate the exponential model and interpret the sign.

1. The hourly remaining factor is \(0.88\). Since \(e^{-k}=0.88\), \(k=-\ln(0.88)\approx 0.1278\). Therefore, \(N(t)=400e^{-kt}\). 2. One-fourth remains when \((0.88)^t=0.25\). Thus, \(t=\frac{\ln(0.25)}{\ln(0.88)}\approx 10.84\) hours. 3. Differentiate: \(N'(t)=-400ke^{-kt}\). Therefore, \(N'(5)\approx -26.98\,\text{mg/h}\). The negative sign indicates that the amount is decreasing.

1. \(k=-\ln(0.88)\approx 0.1278\); \(N(t)=400e^{-kt}\) 2. About \(10.84\,\text{hours}\) 3. \(N'(5)\approx -26.98\,\text{mg/h}\); the negative sign indicates a decrease

- Use the half-life to set the remaining ratio equal to \(0.5\). - Convert \(65\%\) to \(0.65\) and solve for time. - Differentiate the model and factor out \(N(t)\).

1. The half-life gives \(e^{-5730k}=0.5\). Therefore, \(k=\frac{\ln 2}{5730}\approx 0.00012097\,\text{yr}^{-1}\). 2. Solve \(e^{-kt}=0.65\). Thus, \(t=\frac{\ln(0.65)}{-k}\approx 3561\) years. 3. Differentiate: \(N'(t)=-kN_0e^{-kt}=-kN(t)\). Therefore, the rate of change is proportional to the amount, with constant of proportionality \(-k\).

1. \(k\approx 0.00012097\,\text{yr}^{-1}\) 2. About \(3561\,\text{years}\) 3. \(N'(t)=-kN(t)\); constant of proportionality: \(-k\approx -0.00012097\,\text{yr}^{-1}\)

- Find the vertex of the parabola. - Translate “stay at or above the roadway” into an inequality for the minimum value. - Differentiate to find the slope at the tower. - Use the tangent relationship between slope and angle.

1. The parabola's lowest point occurs halfway between the towers, at \(x=50\). Its height is \(f_a(50)=30-2500a\). 2. To stay at or above the roadway, the minimum height must satisfy \(30-2500a\ge0\). Therefore, \(a\le0.012\). Together with \(a>0\), this gives \(0<a\le0.012\). 3. If the minimum height is \(5\,\text{m}\), then \(30-2500a=5\), so \(a=0.01\). 4. For \(a=0.01\), \(f_a'(x)=0.02x-1\). At \(x=0\), the slope is \(-1\). The acute angle relative to the horizontal is \(\arctan(1)=45^\circ\).

a) \(0<a\le0.012\) b) \(a=0.01\) c) \(45^\circ\)

- Which derivatives locate and verify an inflection point? - What quantity represents the instantaneous rate of change of volume? - Where does the slope reach a maximum when concavity changes from up to down? - Interpret the signs of the first and second derivatives separately.

1. Differentiate: \(V'(t)=-1.5t^2+12t\), \(V''(t)=-3t+12\), and \(V'''(t)=-3\). 2. Solve \(V''(t)=0\): \(-3t+12=0\), so \(t=4\). Since \(V'''(4)=-3\ne0\), the graph has an inflection point there. Also, \(V(4)=84\), so the point is \((4, 84)\). On the graph, this is where the slope stops increasing and begins decreasing. 3. The rate \(V'(t)\) reaches its maximum at \(t=4\). Thus the maximum rate is \(V'(4)=24\,\text{m}^3/\text{h}\). 4. For \(0\le t<4\), \(V''(t)>0\), so the graph is concave up and the rate of change is increasing. For \(4<t\le10\), \(V''(t)<0\), so the graph is concave down and the rate of change is decreasing. Since \(V'(8)=0\), the volume increases through \(t=8\) and decreases after \(t=8\).

a) \((4, 84)\); the volume’s rate of increase is greatest there. b) \(24\,\text{m}^3/\text{h}\) c) Concave up for \(0\le t<4\) and concave down for \(4<t\le10\). The rate of change increases before \(t=4\) and decreases afterward; the volume begins decreasing after \(t=8\).

- 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 where the two displayed curves reach their peaks. - Exponential decay eventually dominates linear growth. - Set the first derivative equal to zero for the maximum. - Check whether the parameter cancels from the critical-point equation. - The most rapid decrease occurs where the slope is smallest.

1. The two displayed curves appear to reach their maximum at the same time, near day \(10\). 2. Since \(te^{-0.1t}\to0\) as \(t\to\infty\), \(f_k(t)\to100\) signups per day. 3. The first derivative is \(f_k'(t)=ke^{-0.1t}(1-0.1t)\). Because \(k>0\) and the exponential factor is positive, the derivative is zero at \(t=10\), independent of \(k\). The second derivative there is negative, so this is the maximum. 4. For \(k=150\), \(f_{150}(10)=1500e^{-1}+100\approx651.82\), or about \(652\) signups per day. 5. The rate decreases most rapidly where the first derivative is minimal. The second derivative is \(f_k''(t)=ke^{-0.1t}(0.01t-0.2)\), which is zero at \(t=20\). The third derivative is positive there, so the first derivative has a minimum at \(t=20\).

a) The maximum occurs at the same time for both displayed curves, near day \(10\). b) \(100\) signups per day c) Day \(10\) d) \(652\) signups per day e) Day \(20\)

- Use the midpoint coordinates in the cable model. - The derivative gives the slope of the cable at the tower. - Relate slope to angle using the tangent function. - For an upward-opening parabola, the minimum occurs at the vertex.

1. At the midpoint, \(x=0\). Set the height equal to \(30\): \(50-2500k=30\). Thus, \(k=\frac{20}{2500}=\frac1{125}\). 2. The right tower is at \(x=50\). Differentiate: \(f_k'(x)=2kx\). For \(k=\frac1{125}\), the slope at the tower is \(f_k'(50)=\frac45\). 3. If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 4. Because \(k>0\), the lowest point is at \(x=0\). The cable touches or passes below the roadway when \(50-2500k\le0\). Therefore, \(k\ge\frac1{50}\). It first touches the roadway when \(k=\frac1{50}\).

a) \(k=\frac1{125}\) b) \(\theta\approx38.66^\circ\) c) \(k\ge\frac1{50}\); it first touches the roadway at \(k=\frac1{50}\).

- Landing at the same height means the function value is zero at the basin. - Use the vertex of the downward-opening parabola. - Convert slope to angle with the inverse tangent.

1. The basin condition is \(h_c(4)=0\): \(-0.5(4)^2+4c=0\). Thus \(-8+4c=0\), so \(c=2\). 2. For \(c=2.5\), the vertex occurs at \(x=2.5\). The maximum height is \(h_{2.5}(2.5)=3.125\,\text{m}\). 3. The initial slope is \(h_c'(0)=c\). For \(c=1.5\), \(\tan(\alpha)=1.5\), so \(\alpha=\arctan(1.5)\approx56.31^\circ\).

a) \(c=2\) b) \(3.125\,\text{m}\) c) \(56.31^\circ\)

- Where is the graph steepest? - What does an increasing slope mean for population growth? - What resource limits might affect a wildlife population? - Relate concavity to whether the growth rate is increasing or decreasing.

1. The population grows fastest at the graph’s inflection point, approximately \(t=7.5\) years. At that time, the population is about \(50\) eagles. 2. Before \(t=7.5\), the graph is concave up: the growth rate is increasing, so the population grows faster each year. After \(t=7.5\), the graph is concave down: the population still grows, but its growth rate decreases. 3. The leveling off suggests a carrying capacity. Limited food, nesting sites, territory, or other habitat resources can prevent unlimited population growth.

a) About \(7.5\) years, with approximately \(50\) eagles. b) The growth rate increases before \(t=7.5\) and decreases afterward. The inflection point marks the maximum growth rate. c) Resource limits and habitat carrying capacity can cause the population to level off.

- Look for where the graph has its greatest positive slope. - Think about which resources in a petri dish are limited.

1. The area grows most rapidly where the displayed curve is steepest. From the graph, the steepest part of the S-shaped curve is centered at about \(t=10\) hours, so the maximum growth rate occurs then. 2. The horizontal line \(h\) is at about \(500\,\text{mm}^2\). The covered area could approach this limit because of limited space, limited nutrients, or other environmental constraints in the petri dish.

a) After \(10\) hours. b) Possible reasons include limited space, limited nutrients, or other limiting environmental conditions.

- Compare the horizontal positions of the two displayed peaks. - Use the product rule and chain rule. - Find the zero of the first derivative and notice which factors can affect it. - Substitute the maximum time into the model. - The most rapid decrease occurs where the slope is minimal.

1. The two displayed curves reach their peaks at the same horizontal position, about \(t=4\) hours. 2. The first derivative is \(c_k'(t)=k(1-0.25t)e^{-0.25t}\). The positive factors do not affect its zeros, so the maximum occurs at \(t=4\) hours, independent of \(k\). 3. The maximum value is \(c_k(4)=\frac{4k}{e}\). Setting this equal to \(4\) gives \(k=e\). 4. Since exponential decay dominates linear growth, \(\lim_{t\to\infty}c_k(t)=0\). 5. The concentration decreases most rapidly where the first derivative is minimal. The second derivative is \(c_k''(t)=k(0.0625t-0.5)e^{-0.25t}\), which is zero at \(t=8\). It changes from negative to positive there, so the slope has a minimum at \(t=8\).

a) The peak time appears unchanged, at about \(4\) hours. b) \(4\) hours c) \(k=e\) d) \(0\,\text{mg/L}\) e) \(8\) hours

- The horizontal asymptote represents the room temperature. - Subtract the room temperature to focus on the temperature difference. - A half-sized difference gives an exponential factor of \(0.5\). - Instantaneous rate of change is found from the first derivative.

1. The horizontal asymptote is \(T=20\), so \(T_R=20\,{}^\circ\text{C}\). The graph gives \(T(0)=90\,{}^\circ\text{C}\). Thus, \(90=a+20\), so \(a=70\). 2. The initial temperature difference is \(70\,{}^\circ\text{C}\). After \(10\) minutes, the difference is \(35\,{}^\circ\text{C}\), so \(70e^{-10k}=35\). Therefore, \(e^{-10k}=0.5\) and \(k=\frac{\ln 2}{10}\). 3. The derivative is \(T'(t)=-70ke^{-kt}\). Thus, \(T'(0)=-70\left(\frac{\ln 2}{10}\right)=-7\ln 2\approx -4.85\,{}^\circ\text{C}/\text{min}\).

a) \(T_R=20\,{}^\circ\text{C}\), \(T(0)=90\,{}^\circ\text{C}\), and \(a=70\) b) \(k=\frac{\ln 2}{10}\) c) \(T'(0)\approx -4.85\,{}^\circ\text{C}/\text{min}\)

- The exponential factor is always positive, so the zero comes from the remaining factor. - Compare linear growth with exponential decay for the long-run behavior. - Use the sign of \(g'(x)\) on either side of its critical point to classify the local extremum. - A positive volume-change rate means the reservoir's volume is increasing.

1. Since \(e^{-0.5x}>0\), \(g(x)=0\) only when \(x+1=0\). Thus, the zero is \(x=-1\). As \(x\to\infty\), exponential decay dominates linear growth, so \(g(x)\to0\). 2. Differentiate: \(g'(x)=e^{-0.5x}(0.5-0.5x)\). The derivative is zero at \(x=1\), and it changes from positive to negative. Therefore, the graph has a local maximum at \(\left(1,2e^{-0.5}\right)\approx(1,1.21)\). 3. The graph gives \(g(0)=1\,\text{m}^3/\text{day}\). Because this rate is positive, the water volume is increasing on the reference date.

a) Zero at \(x=-1\); \(\lim_{x\to\infty}g(x)=0\) b) Local maximum at \(\left(1,2e^{-0.5}\right)\approx(1,1.21)\) c) \(1\,\text{m}^3/\text{day}\); the water volume is increasing

- Interpret a function value as height and a derivative value as slope. - Use vertex form because the highest point is given. - The water reaches the surface at a zero of the function. - The tangent of the launch angle equals the slope at the nozzle.

1. The condition \(f(0)=2\) means the water leaves the nozzle \(2\,\text{ft}\) above the surface. The condition \(f'(4)=0\) means the path has a horizontal tangent at a horizontal distance of \(4\,\text{ft}\), where it reaches its maximum height. 2. Using vertex form with vertex \((4, 10)\), write \(f(x)=a(x-4)^2+10\). Since \(f(0)=2\), \(16a+10=2\), so \(a=-\frac{1}{2}\). Thus, \(f(x)=-\frac{1}{2}(x-4)^2+10=-\frac{1}{2}x^2+4x+2\). 3. Set \(f(x)=0\): \(-\frac{1}{2}x^2+4x+2=0\), or \(x^2-8x-4=0\). The positive solution is \(x=4+2\sqrt{5}\approx8.47\). The stream hits the water about \(8.47\,\text{ft}\) from the nozzle. 4. The initial slope is \(f'(0)\). Since \(f'(x)=-x+4\), \(f'(0)=4\). Therefore, \(\tan(\theta)=4\), so \(\theta=\arctan(4)\approx76.0^\circ\).

a) \(f(0)=2\) gives the nozzle's height, and \(f'(4)=0\) identifies the horizontal tangent at the maximum height. b) \(f(x)=-\frac{1}{2}x^2+4x+2\) c) \(4+2\sqrt{5}\,\text{ft}\approx8.47\,\text{ft}\) d) \(\theta\approx76.0^\circ\)

- 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.

- Which derivative identifies a maximum of the concentration? - How can you locate where the rate of change is most negative? - Use the point-slope form of the tangent line at \(t=6\). - For part d, return to the original exponential model rather than the temporary linear continuation.

1. Differentiate using the product and chain rules: \(c'(t)=8e^{-0.25t}-2te^{-0.25t}=(8-2t)e^{-0.25t}\). 2. Since \(e^{-0.25t}>0\), \(c'(t)=0\) when \(8-2t=0\), so \(t=4\). The derivative changes from positive to negative there, so the concentration is maximal. \(c(4)=32e^{-1}\approx11.77\,\text{mg/L}\). 3. The concentration decreases most rapidly where \(c'\) is smallest, which occurs where \(c''(t)=0\). Differentiating again gives \(c''(t)=(0.5t-4)e^{-0.25t}\). Thus, \(t=8\), and the sign change of \(c''\) confirms a minimum of \(c'\). 4. At \(t=6\), \(c(6)=48e^{-1.5}\) and \(c'(6)=-4e^{-1.5}\). The tangent line is \(L(t)=48e^{-1.5}-4e^{-1.5}(t-6)\). 5. Set \(L(t)=0\): \(48-4(t-6)=0\), so \(t=18\). This is \(12\) additional hours after the intervention. 6. Returning to the original exponential model, exponential decay dominates the linear factor, so \(\lim_{t\to\infty}8te^{-0.25t}=0\). The modeled medication concentration approaches \(0\,\text{mg/L}\) over time.

a) The maximum occurs at \(t=4\) hours and is approximately \(11.77\,\text{mg/L}\). b) The concentration is decreasing most rapidly at \(t=8\) hours. c) It reaches \(0\) after \(12\) additional hours, or \(18\) hours after the medication was taken. d) For the original exponential model, the concentration approaches \(0\,\text{mg/L}\).

- Use the product rule because \(t\) appears both in a power and in the exponent. - Interpret the difference between function values at times \(t\) and \(t+2\). - Which part of the function approaches zero as \(t\) becomes large? - Express the location and value of the maximum in terms of \(a\) and \(b\).

1. Differentiate: \(T'(t)=20te^{-0.5t}-5t^2e^{-0.5t}=5t(4-t)e^{-0.5t}\). 2. The critical times are \(t=0\) and \(t=4\). For \(t>0\), the derivative changes from positive to negative at \(t=4\), so the maximum occurs there. 3. \(T(4)=160e^{-2}+20\approx41.65\,^{\circ}\text{C}\). 4. The equation \(T(t+2)=T(t)-4\) means that, for a time \(t\) satisfying the equation, the component is \(4\,^{\circ}\text{C}\) cooler two hours later. 5. Since \(t^2e^{-0.5t}\to0\) as \(t\to\infty\), \(T(t)\to20\,^{\circ}\text{C}\). 6. For \(T(t)=at^2e^{-bt}+c\) with \(a,b>0\), the positive-time maximum occurs at \(t_{\max}=\frac{2}{b}\). Increasing \(b\) makes the maximum occur earlier. The maximum value is \(c+\frac{4a}{b^2e^2}\), so decreasing \(a\) lowers the maximum; increasing \(b\) also lowers it.

a) The maximum temperature is approximately \(41.65\,^{\circ}\text{C}\), reached after \(4\) hours. b) At any time \(t\) satisfying the equation, the temperature two hours later is \(4\,^{\circ}\text{C}\) lower. c) The temperature approaches \(20\,^{\circ}\text{C}\). d) Increase \(b\) to make the maximum occur earlier, and decrease \(a\) to lower the maximum. Increasing \(b\) also lowers the maximum.

- Translate “increasing most rapidly” into a statement about the first derivative. - Which derivative helps locate a maximum of the first derivative? - Use a sign test or the next derivative to verify the maximum.

1. Differentiate using the product and chain rules: \(v'(t)=(1.5-0.05t)e^{0.05t}\). 2. The sales rate is increasing most rapidly where \(v'\) reaches its maximum. Differentiate again: \(v''(t)=(0.025-0.0025t)e^{0.05t}\). 3. Set \(v''(t)=0\). Since the exponential factor is positive, \(0.025-0.0025t=0\), so \(t=10\). 4. The third derivative is \(v'''(t)=(-0.00125-0.000125t)e^{0.05t}\), and \(v'''(10)<0\). Therefore, \(v'\) has a maximum at \(t=10\).

The sales rate is increasing most rapidly \(10\) days after release.

- Launch corresponds to \(t=0\). - A maximum occurs where the first derivative changes from positive to negative. - Interpret subtraction as a signed difference. - Analyze the limit as time increases. - The most rapid decrease occurs where the first derivative is smallest.

1. At launch, \(S(0)=400+500=900\), so the model gives \(900\) bikes per week. 2. The first derivative is \(S'(t)=(160-20t)e^{-0.1t}\). Since the exponential factor is positive, \(S'(t)=0\) at \(t=8\). The derivative changes from positive to negative there, so sales are maximized after \(8\) weeks. The maximum is \(S(8)=2000e^{-0.8}+500\approx1398.66\), or about \(1399\) bikes per week. 3. The value \(h(t)\) is the difference between weekly sales at week \(t\) and weekly sales at week \(20\). A positive value means sales are higher at week \(t\); a negative value means they are lower. 4. Since \((200t+400)e^{-0.1t}\to0\), \(\lim_{t\to\infty}S(t)=500\). The model approaches \(500\) bikes per week. 5. Sales decrease most rapidly where \(S'(t)\) is smallest. The second derivative is \(S''(t)=(2t-36)e^{-0.1t}\), so \(S''(18)=0\). The third derivative is \(S'''(t)=(5.6-0.2t)e^{-0.1t}\), and \(S'''(18)>0\). Thus \(S'\) has a local minimum at \(t=18\), so sales are decreasing most rapidly then.

a) \(900\) bikes per week. b) After \(8\) weeks; approximately \(1399\) bikes per week. c) \(h(t)\) is weekly sales at week \(t\) minus weekly sales at week \(20\). d) \(500\) bikes per week. e) \(S''(18)=0\) and \(S'''(18)>0\), so \(S'\) has a minimum at \(t=18\). Therefore, weekly sales are decreasing most rapidly after \(18\) weeks.

- The initial value occurs at \(t=0\). - Find where the first derivative changes from positive to negative. - Analyze the exponential term as \(t\to\infty\). - The most rapid decrease occurs where the first derivative reaches its minimum.

1. At \(t=0\), \(c(0)=5\), so the initial concentration is \(5\,\text{mg/L}\). 2. The first derivative is \(c'(t)=(1-0.5t)e^{-0.25t}\). It is zero at \(t=2\) and changes from positive to negative there. Thus the maximum occurs after \(2\) hours. Its value is \(c(2)=8e^{-0.5}+1\approx5.85\,\text{mg/L}\). 3. Since \((2t+4)e^{-0.25t}\to0\), \(\lim_{t\to\infty}c(t)=1\). The model approaches \(1\,\text{mg/L}\). 4. The concentration decreases most rapidly where \(c'(t)\) is smallest. The second derivative is \(c''(t)=(0.125t-0.75)e^{-0.25t}\), which is zero at \(t=6\). The third derivative is positive at \(t=6\), so \(c'\) has a local minimum there. Therefore, the concentration is decreasing most rapidly after \(6\) hours.

a) \(5\,\text{mg/L}\). b) After \(2\) hours; approximately \(5.85\,\text{mg/L}\). c) \(1\,\text{mg/L}\). d) After \(6\) hours.

- The upper edges lie on the x-axis, so use their coordinates in the model. - A \(50\%\) grade corresponds to derivative value \(0.5\). - The parabola's lowest point occurs at its vertex. - Report depth as a positive distance.

1. The top edges are at \(x=\pm10\), and \(f(10)=0\). 2. Since \(f'(x)=2ax\) and \(f'(10)=0.8\), \(20a=0.8\), so \(a=0.04\). 3. Use \(f(10)=0\): \(0.04(10)^2+c=0\), giving \(c=-4\). Thus \(f(x)=0.04x^2-4\). 4. The lowest point is \(f(0)=-4\), so the maximum depth is \(4\,\text{m}\). 5. A \(50\%\) slope is \(0.5\). Solve \(f'(x)=0.08x=0.5\), giving \(x=6.25\).

a) \(f(x)=0.04x^2-4\) b) \(4\,\text{m}\) c) \(6.25\,\text{m}\) from the center

- 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 \(\approx1.23\,\text{cm/s}\)

- Find the limit of the exponential term as \(t\to\infty\). - Use the first derivative to locate the maximum. - The most rapid decrease occurs where the first derivative is smallest. - Compare the long-term values of the two displayed graphs.

1. Since \((t-20)e^{-0.04t}\to0\) as \(t\to\infty\), \(\lim_{t\to\infty}v_a(t)=20a\). 2. Differentiate: \(v_a'(t)=ae^{-0.04t}(1.8-0.04t)\). The derivative is zero at \(t=45\), positive before \(45\), and negative after \(45\). Therefore, the greatest daily attendance occurs on day \(45\). 3. Differentiate again: \(v_a''(t)=-0.04ae^{-0.04t}(2.8-0.04t)\). This expression changes from negative to positive at \(t=70\), so \(v_a'\) has its minimum there. Thus, attendance is decreasing most rapidly on day \(70\). 4. For \(a=50\), the long-term value is \(1000\). For \(a=100\), it is \(2000\). Graph 1 approaches \(1000\), and graph 2 approaches \(2000\). Therefore, graph 1 corresponds to \(a=50\), and graph 2 corresponds to \(a=100\).

a) \(20a\) visitors per day b) Day \(45\) c) Day \(70\) d) Graph 1: \(a=50\); graph 2: \(a=100\).

- 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 \(c=2000\); \(N(t)=-1800e^{-t/18}+2000\).