Free Math Worksheets

- 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

- When a function gives a total amount, how do you find the amount at a particular time? - Recall the difference between average rate of change over an interval and instantaneous rate of change at one time. - Which function represents the instantaneous production rate? - How can you find the maximum value of that rate function?

1. Evaluate the total-production function at \(t = 20\): \(G(20) = -0.05 \cdot 20^3 + 1.5 \cdot 20^2 + 10 \cdot 20 = -400 + 600 + 200 = 400\). Therefore, \(400\,\text{m}^3\) of gas is produced. 2. Compute the average rate of change on \([4, 12]\). Since \(G(12) = 249.6\) and \(G(4) = 60.8\), the average 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 \(G'(t) = -0.15t^2 + 3t + 10\). Its derivative is \(G''(t) = -0.3t + 3\). Solving \(G''(t)=0\) gives \(t=10\). Because \(G'\) is a downward-opening quadratic, its maximum on \([0, 20]\) occurs at \(t=10\).

a) \(400\,\text{m}^3\) b) \(23.6\,\frac{\text{m}^3}{\text{h}}\) c) \(t = 10\,\text{h}\)

- Decide whether an average rate requires a difference quotient or a derivative. - An instantaneous rate of change is the slope of a tangent line. - Check critical values and interval endpoints when finding an absolute maximum. - What does the second derivative say about how the first derivative changes? - Interpret the inflection point in terms of the net rate of volume change.

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 derivative is \(V'(t)=-3t^2+30t\). Thus, \(V'(4)=-3\cdot16+30\cdot4=72\,\frac{\text{m}^3}{\text{h}}\). 3. Solve \(V'(t)=-3t(t-10)=0\). The relevant interior critical value is \(t=10\). Since \(V''(10)=-30<0\), this is a local maximum. Also, \(V(0)=200\), \(V(10)=700\), and \(V(12)=632\), so the absolute maximum on the interval is \(700\,\text{m}^3\) at \(t=10\). 4. Since \(V''(t)=-6t+30\), the inflection value is \(t=5\). The point is \((5, 450)\). The net rate of change \(V'(t)\) reaches its maximum there, and \(V'(5)=75\,\frac{\text{m}^3}{\text{h}}\). Thus, after \(5\) hours, the volume is increasing at its greatest modeled rate.

a) \(66\,\frac{\text{m}^3}{\text{h}}\) b) \(72\,\frac{\text{m}^3}{\text{h}}\) c) After \(10\) hours; \(700\,\text{m}^3\) d) Inflection point: \((5, 450)\). At \(t=5\), the volume is increasing at its maximum modeled rate of \(75\,\frac{\text{m}^3}{\text{h}}\).

- Distinguish between average and instantaneous velocity. - Find where the first derivative equals zero to locate a possible maximum height. - The inflection point can mark where a rate reaches a maximum. - Include units when interpreting each rate.

1. Since \(h(0)=20\) and \(h(5)=70\), the average upward velocity is \(\frac{70-20}{5-0}=10\,\frac{\text{m}}{\text{min}}\). 2. The derivative is \(h'(t)=-0.6t^2+6t\). Thus, \(h'(3)=-0.6\cdot9+6\cdot3=12.6\,\frac{\text{m}}{\text{min}}\). 3. Solve \(h'(t)=-0.6t(t-10)=0\). The relevant interior critical value is \(t=10\). Since \(h''(10)=-6<0\), this is a local maximum. Checking the interval endpoints gives \(h(0)=20\), \(h(10)=120\), and \(h(15)=20\), so the maximum height is \(120\,\text{m}\) after \(10\) minutes. 4. Since \(h''(t)=-1.2t+6\), the inflection value is \(t=5\). The slope there is \(h'(5)=15\,\frac{\text{m}}{\text{min}}\). Therefore, after \(5\) minutes, the balloon reaches its greatest upward velocity; after that time, its upward velocity begins to decrease.

a) \(10\,\frac{\text{m}}{\text{min}}\) b) \(12.6\,\frac{\text{m}}{\text{min}}\) c) After \(10\) minutes; \(120\,\text{m}\) d) The inflection time is \(t=5\), and the slope there is \(15\,\frac{\text{m}}{\text{min}}\). This is the balloon's maximum upward velocity.

- 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 the power output changes. - Check critical values and endpoints for an absolute maximum. - “Increasing most rapidly” asks for the maximum value of the first derivative.

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 in \(\frac{\text{W}}{\text{h}}\). The second derivative is \(P''(t)=-6t+24\), which is the rate at which \(P'(t)\) changes, in \(\frac{\text{W}}{\text{h}^2}\). 3. Solve \(P'(t)=0\): the solutions are \(t=-2\) and \(t=10\), so only \(t=10\) is in the interval. Since \(P''(10)=-36<0\), this is a local maximum. Comparing \(P(0)=0\), \(P(10)=800\), and \(P(12)=720\) shows that the absolute maximum is \(800\,\text{W}\) at \(t=10\), which is \(4\) p.m. 4. The power output increases most rapidly when \(P'(t)\) is largest. Since \(P'\) is a downward-opening quadratic, its maximum occurs where \(P''(t)=0\): \(t=4\). Therefore, the output is increasing most rapidly at \(10\) a.m.

a) \(368\,\text{W}\) b) \(P'(t)=-3t^2+24t+60\) is the rate of change of power; \(P''(t)=-6t+24\) is the rate of change of that rate. c) At \(4\) p.m.; \(800\,\text{W}\) d) At \(10\) a.m.

- 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

- What does the derivative of a population function represent? - How does an average rate over an interval differ from an instantaneous rate? - To find the greatest growth rate, maximize the derivative function.

1. The average growth rate on \([0, 5]\) is \(\frac{B(5)-B(0)}{5-0}\). Since \(B(0)=100\) and \(B(5)=475\), the average rate is \(\frac{475-100}{5}=75\) bacteria per hour. 2. The instantaneous growth rate is \(B'(t)=40t-3t^2\). Thus, \(B'(5)=200-75=125\) bacteria per hour. 3. To maximize \(B'\), solve \(B''(t)=40-6t=0\). This gives \(t=\frac{20}{3}\). Because \(B'(t)\) is a downward-opening quadratic, this critical value gives its maximum on \([0, 10]\).

a) \(75\) bacteria per hour b) \(125\) bacteria per hour c) \(t=\frac{20}{3}\,\text{h}\approx6.67\,\text{h}\)

- 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

- Use the sign of the first derivative to determine monotonicity. - Examine the exponential term as \(t\) becomes large. - Substitute the expression for \(T(t)\) into the right side of the proposed equation.

1. Differentiate: \(T'(t)=1.4e^{-0.1t}\). Since \(e^{-0.1t}>0\), \(T'(t)>0\) for all \(t\ge 0\). Therefore, \(T\) is strictly increasing. 2. As \(t\to\infty\), \(e^{-0.1t}\to 0\), so \(T(t)\to 22\). For every finite \(t\), \(14e^{-0.1t}>0\), so \(T(t)=22-14e^{-0.1t}<22\). 3. Since \(22-T(t)=14e^{-0.1t}\), \(0.1(22-T(t))=1.4e^{-0.1t}=T'(t)\).

1. \(T'(t)=1.4e^{-0.1t}>0\), so \(T\) is strictly increasing. 2. \(\lim_{t\to\infty}T(t)=22\), and \(T(t)<22\) for every finite \(t\). 3. \(T'(t)=1.4e^{-0.1t}=0.1(22-T(t))\)

- Substitute \(t=0\) and \(t=20\) into the model. - Differentiate and compare the result with \(400-B(t)\). - Isolate the exponential expression before taking a logarithm.

1. \(B(0)=400-320=80\). Also, \(B(20)=400-320e^{-1}\approx 282.28\), so the model predicts about \(282\) birds. 2. Differentiate: \(B'(t)=16e^{-0.05t}\). Also, \(400-B(t)=320e^{-0.05t}\). Therefore, \(B'(t)=0.05(400-B(t))\), and the constant of proportionality is \(0.05\). 3. Set \(B(t)=350\): \(350=400-320e^{-0.05t}\). Then \(e^{-0.05t}=\frac{5}{32}\), so \(t=\frac{\ln(5/32)}{-0.05}\approx 37.13\) years.

1. Initial population: \(80\) birds; after \(20\) years: about \(282\) birds 2. \(B'(t)=0.05(400-B(t))\); constant of proportionality: \(0.05\) 3. About \(37.13\,\text{years}\)

- 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) \(m'(0)\approx -11.77\,\text{mg/day}\); \(m'(20)\approx -1.66\,\text{mg/day}\)

- Substitute the specified times directly into the model for part a). - For an absolute maximum on a closed interval, check critical points and endpoints. - The second derivative describes how the first derivative changes. - To maximize a rate of change, analyze the derivative as its own function. - Set the derivative of that rate equal to zero.

1. Evaluate the model: \(c(0)=5\) and \(c(12)=-0.1(12)^3+1.2(12)^2+5=5\). Both concentrations are \(5\,\text{mg/L}\). 2. The first derivative is \(c'(t)=-0.3t^2+2.4t=0.3t(8-t)\). Its zeros are \(t=0\) and \(t=8\). 3. The second derivative is \(c''(t)=-0.6t+2.4\). Since \(c''(8)=-2.4<0\), \(t=8\) is a local maximum. Comparing with the endpoints gives \(c(0)=5\), \(c(8)=30.6\), and \(c(12)=5\), so the absolute maximum is \(30.6\,\text{mg/L}\) at \(t=8\). 4. At \(t=2\), \(c''(2)=1.2\,\frac{\text{mg}}{\text{L}\cdot\text{min}^2}\). The positive value means the concentration's rate of change is increasing at that time. 5. The concentration increases most rapidly when \(c'(t)\) is maximized. Set its derivative equal to zero: \(c''(t)=0\), giving \(t=4\). 6. The maximum rate is \(c'(4)=-0.3(4)^2+2.4(4)=4.8\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\).

a) \(c(0)=5\,\text{mg/L}\) and \(c(12)=5\,\text{mg/L}\) b) The maximum is \(30.6\,\text{mg/L}\) at \(t=8\) minutes. c) \(c''(2)=1.2\,\frac{\text{mg}}{\text{L}\cdot\text{min}^2}\). The concentration's rate of change is increasing. d) The concentration increases most rapidly at \(t=4\) minutes, at a rate of \(4.8\,\frac{\text{mg}}{\text{L}\cdot\text{min}}\).

- 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) \(h'(9)=\frac{1}{3}\,\text{cm}/\text{s}\approx0.33\,\text{cm}/\text{s}\)

- Interpret a function value using the stated cost units. - Marginal cost is the derivative of total cost. - Rewrite radicals as powers before differentiating if useful. - To prove a function decreases, determine the sign of its derivative. - Analyze each term in the derivative of the average-cost function.

1. a) \(K(64)=5\sqrt{64}+20=60\), so the total cost is \(\$60{,}000\). 2. b) \(K'(x)=\frac{2.5}{\sqrt{x}}\). Set \(\frac{2.5}{\sqrt{x}}=0.25\), giving \(\sqrt{x}=10\) and \(x=100\). The marginal cost there is \(\$250\) per ton. 3. c) Rewrite \(k(x)=\frac{5}{\sqrt{x}}+\frac{20}{x}\). Then \(k'(x)=-\frac{2.5}{x\sqrt{x}}-\frac{20}{x^2}\). 4. For every \(0<x\le500\), both terms are negative, so \(k'(x)<0\). Therefore, average cost is strictly decreasing over the model domain.

a) \(\$60{,}000\) b) \(100\) tons c) Since \(k'(x)=-\frac{2.5}{x\sqrt{x}}-\frac{20}{x^2}<0\) for \(0<x\le500\), average cost is strictly decreasing.

- 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) \(3.6\,\text{m}\)

- The instantaneous rate is the derivative of cumulative use. - Divide cumulative use by time to obtain the average rate. - Compare the resulting expressions for all \(t>0\). - A constant function has derivative \(0\).

1. Model 1: \(E'(t)=0.04t\) and \(d(t)=0.02t\), so \(E'(t)=2d(t)>d(t)\). 2. Model 2: \(E'(t)=1.5\) and \(d(t)=1.5\), so the rates are equal. 3. Model 3: \(E'(t)=\frac{2}{\sqrt{t}}\) and \(d(t)=\frac{4}{\sqrt{t}}\), so \(E'(t)=\frac{1}{2}d(t)<d(t)\). 4. Therefore, model 3 is regressive. 5. Model 2 has \(d(t)=1.5\), so \(d'(t)=0\) and its average rate 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: \(E'(t)=\frac{2}{\sqrt{t}}\), \(d(t)=\frac{4}{\sqrt{t}}\) 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 \(t=8\).

- Planting corresponds to \(t=0\). - To solve for time, isolate the exponential expression and use a logarithm. - A horizontal line at the target height helps identify the corresponding time on the graph. - Instantaneous growth rate is given by the first derivative.

1. \(h(0)=12-10e^0=2\), so the initial height is \(2\,\text{m}\). Also, \(h(20)=12-10e^{-1}\approx 8.32\,\text{m}\). 2. Solve \(12-10e^{-0.05t}=10\). Then \(e^{-0.05t}=0.2\), so \(t=\frac{\ln(0.2)}{-0.05}\approx 32.2\) years. On the graph, locate \(10\) on the vertical axis, move horizontally to the curve, and then move vertically to the time axis. 3. Differentiate: \(h'(t)=0.5e^{-0.05t}\). Therefore, \(h'(10)=0.5e^{-0.5}\approx 0.30\,\text{m/year}\).

a) Initial height: \(2\,\text{m}\); height after \(20\) years: about \(8.32\,\text{m}\) b) About \(32.2\) years; graphically, find the time-coordinate where the curve has height \(10\) c) About \(0.30\,\text{m/year}\)

- 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. Since \(te^{-0.1t}\to0\) as \(t\to\infty\), \(f_k(t)\to100\) signups per day. 2. 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. 3. For \(k=150\), \(f_{150}(10)=1500e^{-1}+100\approx651.82\), or about \(652\) signups per day. 4. 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) \(100\) signups per day b) Day \(10\) c) \(652\) signups per day d) 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 \(f_k'(50)=\frac45\). If \(\theta\) is the angle of inclination, then \(\tan(\theta)=\frac45\), so \(\theta=\arctan\left(\frac45\right)\approx38.66^\circ\). 3. 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.

- A logistic growth curve has its greatest growth rate at its inflection point. - Think about which resources in a petri dish are limited.

1. The area grows most rapidly at the inflection point of the logistic curve. From the model, the inflection point occurs when \(5-0.5t=0\), so \(t=10\). Thus, the maximum growth rate occurs after \(10\) hours. 2. The limiting area of \(500\,\text{mm}^2\) could result from limited space in the petri dish, limited nutrients, or other environmental constraints.

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

- Use the product rule and chain rule. - Find the zero of the first derivative. - Substitute the maximum time into the model. - The most rapid decrease occurs where the slope is minimal.

1. 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 \(k\). 2. The maximum value is \(c_k(4)=\frac{4k}{e}\). Setting this equal to \(4\) gives \(k=e\). 3. Since exponential decay dominates linear growth, \(\lim_{t\to\infty}c_k(t)=0\). 4. 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) \(4\) hours b) \(k=e\) c) \(0\,\text{mg/L}\) d) \(8\) 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. - Compare linear growth with exponential decay. - Use the product and chain rules to locate the extremum. - A positive rate means the quantity 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 at the beginning of the observation period.

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

- The carrying capacity is the value approached as time increases. - Substitute the initial population into the model. - Instantaneous growth rate is represented by the first derivative. - A bounded model of this form requires a negative exponent coefficient.

1. For the population to approach a finite value, \(k<0\), so \(e^{kt}\to0\) as \(t\to\infty\). Therefore, the limiting value is \(c=2000\). 2. Use \(N(0)=200\): \(b+2000=200\), so \(b=-1800\). 3. Differentiate: \(N'(t)=bke^{kt}=-1800ke^{kt}\). Use \(N'(0)=100\): \(-1800k=100\), so \(k=-\frac{1}{18}\). 4. The model is \(N(t)=-1800e^{-t/18}+2000\). The graph begins at \(N(0)=200\) and approaches the horizontal asymptote \(N=2000\).

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

- 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) \(\theta\approx76.0^\circ\)

- 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 \(t=6\). - Compare the long-term effects of the factor \(t\) and the decaying exponential factor.

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. Because exponential decay dominates linear growth, \(\lim_{t\to\infty}8te^{-0.25t}=0\). 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) 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.

- Use the volume formula for a square pyramid. - Similarity relates the side length of the liquid surface to the liquid depth. - Set the geometric volume equal to inflow rate multiplied by time. - Differentiate the depth function to find the rising rate. - Rewrite \(\sqrt[3]{t}\) as \(t^{\frac{1}{3}}\) before applying the power rule.

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: \(h'(t)=t^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{t^2}}\). Thus, \(h'(8)=\frac{1}{4}=0.25\,\text{dm}/\text{min}\). At the time the container becomes full, this is interpreted as the left-hand rate immediately before filling ends.

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

- 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

- How does container width affect the rise rate when inflow is constant? - What does the slope of the height graph represent? - Where does the graph change concavity? - Use a secant slope for the average rate.

1. The slope increases until \(t=6\,\text{s}\) and decreases afterward. Therefore, the water level rises fastest at \(t=6\,\text{s}\). 2. With constant inflow, a larger rise rate corresponds to a smaller horizontal cross-sectional area. The container is wide near the bottom, narrows toward the middle, and widens again near the top. 3. The model gives \(h(0)=\frac{15}{1+e^{4.8}}\approx0.12\,\text{cm}\) and \(h(12)=\frac{15}{1+e^{-4.8}}\approx14.88\,\text{cm}\). The average rate is \(\frac{h(12)-h(0)}{12}\approx1.23\,\text{cm/s}\).

a) The water level rises fastest at \(t=6\,\text{s}\). b) The container is wide at the bottom, narrowest in the middle, and wider again at the top. c) The height is approximately \(14.88\,\text{cm}\), and the average rate of rise is approximately \(1.23\,\text{cm/s}\).

- 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: \(a=100\).