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Exponential models

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55604812
A quantity \(N\) satisfies \(N'=-0.30N\), where \(t\) is measured in minutes. Is the model exponential growth or exponential decay? What are the units of the constant \(-0.30\)?

Hints

- Use the sign of the proportionality constant to decide whether the quantity grows or decays. - Compare the units of \(N'\) with the units of \(N\).

Solution

1. The rate is proportional to \(N\) with a negative proportionality constant, so the quantity decreases exponentially. 2. Because \(N'/N\) has units of inverse time, the constant has units \(\text{min}^{-1}\).

Answer

Exponential decay; the constant \(-0.30\) has units \(\text{min}^{-1}\).
55604912
An account balance \(B(t)\) has a continuous relative growth rate of \(4\%\) per year. Write the differential equation for \(B\).

Hints

- A relative growth rate compares the derivative with the current amount. - Convert \(4\%\) to a decimal before writing the proportional-rate equation.

Solution

1. A continuous relative growth rate means \(\frac{B'}{B}=0.04\). 2. Therefore, \(B'=0.04B\).

Answer

\(B'=0.04B\)
55605012
An exponentially decaying quantity has a half-life of \(6\) hours. What fraction of the initial amount remains after \(12\) hours?

Hints

- Determine how many half-life intervals fit into \(12\) hours. - Apply the same halving factor once for each interval.

Solution

1. Twelve hours is two half-lives. 2. Each half-life multiplies the amount by \(\frac{1}{2}\), so the remaining fraction is \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\).

Answer

\(\frac{1}{4}\) of the initial amount
52992912
A radioactive substance decays according to \(m(t)=m_0e^{-0.0231t}\), where \(t\) is measured in years and \(m(t)\) is measured in milligrams. a) What percent of the initial mass remains after \(25\) years? b) Find the half-life \(T_H\), rounded to two decimal places. c) If \(m_0=500\,\text{mg}\), find the instantaneous rate of change at \(t=50\). Interpret the sign.

Hints

- 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.

Solution

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.

Answer

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
53999612
A colony of virtual agents in a computer simulation is modeled by the differential equation \(N'(t)=0.18N(t)\), with \(N(0)=240\), where \(t\) is measured in hours. a) Write the particular exponential model. b) Find \(N(6)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- A constant relative rate gives a model of the form \(N(t)=N(0)e^{kt}\). - Use \(k=0.18\) and evaluate the model only after inserting the initial count \(240\). - The positive exponent coefficient determines growth and gives the continuous hourly relative rate.

Solution

1. The solution of \(N'=kN\) with initial value \(240\) is \(N(t)=240e^{0.18t}\). 2. \(N(6)=240e^{0.18\cdot 6}\approx 706.72\). 3. Since \(k=0.18\), the model represents growth with a continuous relative rate of \(18.0\%\) per hour.

Answer

a) \(N(t)=240e^{0.18t}\) b) \(N(6)\approx 706.72\,\text{agents}\) c) Growth with a continuous relative rate of \(18.0\%\) per hour.
53999712
The mass \(M(t)\) of a light-sensitive coating, measured in milligrams, is modeled by the differential equation \(M'(t)=-0.07M(t)\), with \(M(0)=85\), where \(t\) is measured in days. a) Write the particular exponential model. b) Find \(M(12)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- Use \(M(t)=M(0)e^{kt}\) with the given negative rate constant. - Insert \(M(0)=85\) before evaluating at \(t=12\). - A negative exponent coefficient signals decay; keep full precision until the requested rounding.

Solution

1. The solution of \(M'=kM\) with initial value \(85\) is \(M(t)=85e^{-0.07t}\). 2. \(M(12)=85e^{-0.07\cdot12}\approx36.70\). 3. Since \(k=-0.07\), the model represents decay with a continuous relative rate of \(7.0\%\) per day.

Answer

a) \(M(t)=85e^{-0.07t}\) b) \(M(12)\approx36.70\,\text{mg}\) c) Decay with a continuous relative rate of \(7.0\%\) per day.
53999812
The number of activated accounts in a controlled rollout is modeled by the differential equation \(A'(t)=0.09A(t)\), with \(A(0)=1200\), where \(t\) is measured in weeks. a) Write the particular exponential model. b) Find \(A(8)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- Build the particular model from \(A(0)=1200\) and the continuous rate \(0.09\). - Evaluate the exponential at \(t=8\) without rounding the exponent early. - Use the positive sign of the rate constant to classify the model and state the weekly relative rate.

Solution

1. The solution of \(A'=kA\) with initial value \(1200\) is \(A(t)=1200e^{0.09t}\). 2. \(A(8)=1200e^{0.09\cdot 8}\approx 2465.32\). 3. Since \(k=0.09\), the model represents growth with a continuous relative rate of \(9.0\%\) per week.

Answer

a) \(A(t)=1200e^{0.09t}\) b) \(A(8)\approx 2465.32\,\text{accounts}\) c) Growth with a continuous relative rate of \(9.0\%\) per week.
53999912
The amount \(B(t)\) of a biodegradable packing material, measured in kilograms, is modeled by the differential equation \(B'(t)=-0.12B(t)\), with \(B(0)=640\), where \(t\) is measured in months. a) Write the particular exponential model. b) Find \(B(5)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- Write the decay model with initial amount \(640\) and exponent coefficient \(-0.12\). - Substitute \(t=5\) into the particular model before rounding. - Interpret the negative coefficient as a continuous monthly relative decrease.

Solution

1. The solution of \(B'=kB\) with initial value \(640\) is \(B(t)=640e^{-0.12t}\). 2. \(B(5)=640e^{-0.12\cdot5}\approx351.24\). 3. Since \(k=-0.12\), the model represents decay with a continuous relative rate of \(12.0\%\) per month.

Answer

a) \(B(t)=640e^{-0.12t}\) b) \(B(5)\approx351.24\,\text{kg}\) c) Decay with a continuous relative rate of \(12.0\%\) per month.
54000012
The intensity \(I(x)\) of a laser pulse in an absorbing medium, measured in arbitrary intensity units, is modeled by the differential equation \(I'(x)=-0.035I(x)\), with \(I(0)=50\), where \(x\) is measured in meters. a) Write the particular exponential model. b) Find \(I(18)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- Use distance \(x\), not time \(t\), in the exponential solution for \(I'(x)=-0.035I(x)\). - The initial intensity \(50\) is the leading coefficient in the particular model. - The negative per-meter coefficient gives decay with distance; round only the final value at \(x=18\).

Solution

1. The solution of \(I'=kI\) with initial value \(50\) is \(I(x)=50e^{-0.035x}\). 2. \(I(18)=50e^{-0.035\cdot 18}\approx 26.63\). 3. Since \(k=-0.035\), the model represents decay with a continuous relative rate of \(3.5\%\) per meter.

Answer

a) \(I(x)=50e^{-0.035x}\) b) \(I(18)\approx 26.63\,\text{intensity units}\) c) Decay with a continuous relative rate of \(3.5\%\) per meter.
54000112
The number of cells in a plant-tissue model is modeled by the differential equation \(P'(t)=0.22P(t)\), with \(P(0)=1500\), where \(t\) is measured in days. a) Write the particular exponential model. b) Find \(P(4)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- A proportional-growth equation gives \(P(t)=P(0)e^{0.22t}\). - Use the initial cell count \(1500\) before evaluating at four days. - State the positive continuous daily relative rate separately from the computed population.

Solution

1. The solution of \(P'=kP\) with initial value \(1500\) is \(P(t)=1500e^{0.22t}\). 2. \(P(4)=1500e^{0.22\cdot 4}\approx 3616.35\). 3. Since \(k=0.22\), the model represents growth with a continuous relative rate of \(22.0\%\) per day.

Answer

a) \(P(t)=1500e^{0.22t}\) b) \(P(4)\approx 3616.35\,\text{cells}\) c) Growth with a continuous relative rate of \(22.0\%\) per day.
54000212
The amount \(Q(t)\) of a temporary marker in a stream simulation, measured in grams, is modeled by the differential equation \(Q'(t)=-0.16Q(t)\), with \(Q(0)=300\), where \(t\) is measured in hours. a) Write the particular exponential model. b) Find \(Q(9)\), rounded to two decimal places. c) State whether the model represents growth or decay and give the continuous relative rate.

Hints

- Use the exponential form determined by the IVP. - Evaluate at nine hours with the full exponential value. - The negative rate constant means decay at a continuous relative rate of magnitude \(0.16\) per hour.

Solution

1. The solution of \(Q'=kQ\) with initial value \(300\) is \(Q(t)=300e^{-0.16t}\). 2. \(Q(9)=300e^{-0.16\cdot9}\approx71.08\). 3. Since \(k=-0.16\), the model represents decay with a continuous relative rate of \(16.0\%\) per hour.

Answer

a) \(Q(t)=300e^{-0.16t}\) b) \(Q(9)\approx71.08\,\text{g}\) c) Decay with a continuous relative rate of \(16.0\%\) per hour.
54002712
An exponential model has \(P(2)=90\) and \(P(7)=90\). Can it have a nonzero growth or decay constant? Justify.

Hints

- Write the exponential model in the form \(P(t)=Ce^{kt}\). - Use both given values in the model. - Determine when a real exponential factor can equal \(1\).

Solution

1. Write the model as \(P(t)=Ce^{kt}\). Since \(P(2)=90\), \(C\ne0\). 2. The equal values give \(Ce^{2k}=Ce^{7k}\), so \(e^{5k}=1\). 3. For real \(k\), this requires \(k=0\). Therefore, the model is constant and cannot have a nonzero growth or decay constant.

Answer

No. Equality at two distinct times gives \(e^{5k}=1\), so \(k=0\) and the exponential model is constant.
54002812
A quantity satisfies \(Y'=kY\), \(Y(0)>0\), and \(Y'(4)<0\). What can be concluded about \(k\) and the long-term behavior?

Hints

- Because \(Y(0)>0\), the exponential solution remains positive for every finite time. - At \(t=4\), the sign of \(Y'=kY\) is therefore the sign of \(k\). - Use the sign of \(k\) to determine whether the quantity approaches zero or grows without bound as time increases.

Solution

1. Since \(Y(0)>0\), the exponential solution \(Y(t)=Y(0)e^{kt}\) remains positive, so \(Y(4)>0\). 2. From \(Y'(4)=kY(4)<0\) and \(Y(4)>0\), it follows that \(k<0\). 3. Therefore, \(Y(t)\) decays toward \(0\) as \(t\to\infty\).

Answer

Because \(Y(4)>0\), the sign of \(Y'(4)=kY(4)\) is the sign of \(k\). Thus \(k<0\), and \(Y(t)\) decays toward \(0\) as \(t\to\infty\).
54003012
A lab report claims a positive quantity follows \(R'=kR\), reaches \(0\) after \(10\) hours, and was positive initially. Explain why the claim is incompatible with the model.

Hints

- Write the exponential solution determined by the positive initial value. - Recall the sign of \(e^{kt}\) for finite real \(t\). - Distinguish approaching zero from reaching zero.

Solution

1. The solution with positive initial value is \(R(t)=R(0)e^{kt}\). 2. For every finite \(t\), both \(R(0)\) and \(e^{kt}\) are positive, so \(R(t)>0\). 3. Under decay, the quantity may approach \(0\) as \(t\to\infty\), but it cannot equal \(0\) after a finite time such as \(10\) hours.

Answer

A solution with \(R(0)>0\) has the form \(R(t)=R(0)e^{kt}>0\) for every finite \(t\). It may approach \(0\) under decay, but it cannot reach \(0\) after \(10\) hours.
54351112
A radioactive sample has mass \(24\,\text{g}\) when observation begins and is losing mass at \(6\,\text{g/day}\). Its mass follows \(M'=kM\), where \(t\) is measured in days. a) Determine \(k\) and write the mass model. b) Find the half-life exactly. c) Find the mass after \(3\) days.

Hints

- A loss rate must be represented by a negative derivative. - Use the initial mass and initial rate in \(M'=kM\). - For half-life, set the exponential factor equal to \(\frac{1}{2}\). - Substitute \(t=3\) only after writing the complete model.

Solution

1. The sample is losing mass, so \(M'(0)=-6\,\text{g/day}\). Using \(M'=kM\), \(-6=k(24)\), so \(k=-0.25\,\text{day}^{-1}\). 2. Therefore, \(M(t)=24e^{-0.25t}\) grams. 3. The half-life \(T\) satisfies \(e^{-0.25T}=\frac{1}{2}\), so \(T=4\ln 2\) days. 4. After \(3\) days, \(M(3)=24e^{-0.75}\,\text{g}\).

Answer

a) \(k=-0.25\,\text{day}^{-1}\) and \(M(t)=24e^{-0.25t}\) grams b) \(4\ln 2\) days c) \(24e^{-0.75}\,\text{g}\)
54351712
A savings account begins with a balance of \(\$5000\) and earns interest continuously at \(5\%\) per year. Rina writes the model \(B'=0.05t\). a) Explain Rina’s error and write the correct differential equation and initial condition. b) Find the initial rate of change of the balance. c) Using the exponential model, find when the balance reaches \(\$6000\). Give an exact answer.

Hints

- A continuous percentage rate is applied to the amount currently in the account. - Evaluate the correct differential equation at \(t=0\) for the initial rate. - Divide the target balance by the initial balance before taking a logarithm.

Solution

1. Continuous percentage growth is proportional to the current balance, not to time. The correct initial-value problem is \(B'=0.05B\), \(B(0)=5000\). 2. The initial rate is \(B'(0)=0.05\cdot5000=250\), so the balance initially increases at \(\$250\) per year. 3. The model is \(B(t)=5000e^{0.05t}\). Setting \(B(t)=6000\) gives \(e^{0.05t}=1.2\), so \(t=20\ln(1.2)\) years.

Answer

a) \(B'=0.05B\), \(B(0)=5000\); the rate must depend on the current balance. b) \(\$250\) per year c) \(20\ln(1.2)\) years
54354512
A lab report claims that a growing bacterial mass \(Q(t)\), measured in grams, follows \(Q'=kQ\), where \(t\) is measured in hours. The report lists \(Q(0)=9\,\text{g}\), \(Q'(0)=12\,\text{g/h}\), and \(Q''(0)=20\,\text{g/h}^2\). Determine whether the three measurements are consistent with one exponential model. If not, identify the required value of \(Q''(0)\).

Hints

- Use the initial mass and first derivative to determine the exponential rate constant. - Differentiate the rate equation once to relate \(Q''\) to \(Q\). - Compare the required second derivative, including its units, with the reported value.

Solution

1. From \(Q'(0)=kQ(0)\), \(12=9k\), so \(k=\frac{4}{3}\,\text{h}^{-1}\). 2. Differentiating \(Q'=kQ\) gives \(Q''=k^2Q\). Therefore, the first two measurements require \(Q''(0)=\left(\frac{4}{3}\right)^2(9)=16\,\text{g/h}^2\). 3. The reported value \(20\,\text{g/h}^2\) does not equal \(16\,\text{g/h}^2\), so the measurements are inconsistent.

Answer

No. The first two measurements require \(Q''(0)=16\,\text{g/h}^2\), not \(20\,\text{g/h}^2\).
54356012
A medication concentration \(C(t)\), in milligrams per liter, follows \(C'=kC\). The initial concentration is \(24\,\text{mg/L}\), and the half-life is \(6\) hours. a) Determine \(k\) and state its units. b) Complete the concentrations at \(t=0,6,12,18\) hours. c) Find the instantaneous rate of change at \(t=12\).

Hints

- Translate the half-life into an equation for the exponential factor over \(6\) hours. - Use repeated halving rather than evaluating four separate exponentials. - Apply \(C'=kC\) to the concentration at \(t=12\).

Solution

1. A half-life of \(6\) hours gives \(e^{6k}=\frac{1}{2}\), so \(k=-\frac{\ln 2}{6}\,\text{h}^{-1}\). 2. The concentration halves every \(6\) hours, so the values are \(24\), \(12\), \(6\), and \(3\) milligrams per liter at \(t=0,6,12,18\), respectively. 3. At \(t=12\), \(C'(12)=kC(12)=-\frac{\ln 2}{6}\cdot6=-\ln 2\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\).

Answer

a) \(k=-\frac{\ln 2}{6}\,\text{h}^{-1}\) b) \(C(0)=24\), \(C(6)=12\), \(C(12)=6\), and \(C(18)=3\) milligrams per liter c) \(C'(12)=-\ln 2\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\)
54358512
A radioactive tracer sample is decreasing exponentially according to \(Q'=kQ\), where \(Q(t)\) is measured in grams and \(t\) in years. At \(t=2\), \(Q(2)=5\) grams and \(Q''(2)=20\) grams per year squared. Determine \(Q(t)\) and find \(Q'(0)\).

Hints

- Express the second derivative of an exponential model in terms of the model itself. - Use the fact that the sample is decreasing to select the sign of the rate constant. - Center the exponential formula at the time where the amount is known.

Solution

1. For an exponential model, \(Q''=k^2Q\). Thus \(20=5k^2\), so \(k^2=4\). 2. Because the sample is decreasing, \(k=-2\) per year. 3. Using the value at \(t=2\), \(Q(t)=5e^{-2(t-2)}\) grams. 4. Then \(Q'(0)=-2Q(0)=-10e^4\) grams per year.

Answer

\(Q(t)=5e^{-2(t-2)}\) grams, and \(Q'(0)=-10e^4\) grams per year.
54359812
The number of views on a new instructional video is modeled exponentially. The table shows two values. <table> <tr><th>Time \(t\) (hours)</th><th>Views \(V(t)\)</th></tr> <tr><td>0</td><td>400</td></tr> <tr><td>1</td><td></td></tr> <tr><td>2</td><td></td></tr> <tr><td>3</td><td>3200</td></tr> </table> Complete the table, write the model in the form \(V'=kV\), and state the doubling time.

Hints

- Find the total multiplication factor from \(t=0\) to \(t=3\). - Take the cube root to obtain the factor for one hour. - Convert a base-2 exponential to base \(e\) to identify \(k\).

Solution

1. Over \(3\) hours, the view count is multiplied by \(3200/400=8\). 2. The hourly growth factor is \(\sqrt[3]{8}=2\), so the missing values are \(V(1)=800\) and \(V(2)=1600\). 3. Since \(V(t)=400\cdot2^t=400e^{(\ln2)t}\), the differential equation is \(V'=(\ln2)V\). 4. A factor of \(2\) occurs every \(1\) hour, so the doubling time is \(1\) hour.

Answer

The missing table values are \(800\) and \(1600\). The model is \(V'=(\ln2)V\), with doubling time \(1\) hour.
54360612
A bacteria culture grows exponentially according to \(Q'=kQ\), where \(Q(t)\) is measured in thousands of cells and \(t\) in hours. The counts at \(t=0\) and \(t=2\) satisfy \(Q(0)+Q(2)=10\) and \(Q(2)-Q(0)=6\). Determine \(Q(t)\) and state the doubling time.

Hints

- Solve the sum-and-difference equations for the two counts first. - Use their ratio to find the growth factor across two hours. - Convert the two-hour factor into a one-hour doubling statement.

Solution

1. Solving the two equations gives \(Q(0)=2\) and \(Q(2)=8\). 2. The two-hour growth factor is \(Q(2)/Q(0)=4\), so \(e^{2k}=4\) and \(k=\ln2\) per hour. 3. Therefore, \(Q(t)=2e^{(\ln2)t}=2\cdot2^t\) thousand cells. 4. The model doubles whenever \(t\) increases by \(1\) hour, so the doubling time is \(1\) hour.

Answer

\(Q(t)=2\cdot2^t\) thousand cells, and the doubling time is \(1\) hour.
54361312
A savings account balance \(Q(t)\), measured in dollars, follows \(Q'=kQ\), where \(t\) is measured in years. At \(t=1\), the balance is \(\$2000\) and is increasing at \(\$400\) per year. a) Determine \(k\) and \(Q(t)\). b) Noah suggests forming the equation \(Q''(1)+Q'(1)=480\). Explain why this sum is not dimensionally meaningful unless coefficients with units are supplied.

Hints

- Use the known balance and first derivative in \(Q'=kQ\). - Center the exponential formula at the time of the known balance. - Compare the units of a first derivative with those of a second derivative.

Solution

1. From \(Q'=kQ\), \(400=k\cdot2000\), so \(k=0.20\) per year. 2. Centering the model at \(t=1\) gives \(Q(t)=2000e^{0.20(t-1)}\) dollars. 3. The quantity \(Q'(1)\) has units of dollars per year, while \(Q''(1)\) has units of dollars per year squared. Quantities with different units cannot be added directly. 4. For reference, \(Q''(1)=kQ'(1)=80\) dollars per year squared, but \(80+400\) does not represent a single physical rate.

Answer

a) \(k=0.20\) per year and \(Q(t)=2000e^{0.20(t-1)}\) dollars. b) \(Q'\) and \(Q''\) have different units, so their numerical values cannot be added directly.
54363412
A radioactive sample decays exponentially. After \(1000\) years, \(70\%\) of the original amount remains. Determine the continuous decay rate \(k\), find the half-life, and determine the percentage remaining after \(3000\) years.

Hints

- Express the remaining fraction as an exponential multiplier. - Set the same model equal to \(0.5\) to find the half-life. - Use repeated \(1000\)-year factors for the final percentage.

Solution

1. With \(Q(t)=Q_0e^{kt}\), the data give \(e^{1000k}=0.70\). Thus \(k=\frac{\ln(0.70)}{1000}\) per year. 2. The half-life \(T\) satisfies \(e^{kT}=0.5\), so \(T=\frac{1000\ln(0.5)}{\ln(0.70)}\approx1943.36\) years. 3. Three successive \(1000\)-year intervals multiply the amount by \(0.70^3=0.343\). Therefore, \(34.3\%\) remains after \(3000\) years.

Answer

\(k=\frac{\ln(0.70)}{1000}\approx-0.0003567\) per year; half-life \(\frac{1000\ln(0.5)}{\ln(0.70)}\approx1943.36\) years; \(34.3\%\) remains after \(3000\) years.
54365612
An app's active-user count \(Q(t)\), measured in thousands of users, grows exponentially according to \(Q'=kQ\), where \(t\) is measured in months. Its accumulated relative growth over the first \(6\) months is \(\int_0^6\frac{Q'(t)}{Q(t)}\,dt=3\), and at month \(6\) the user count is increasing at \(20\) thousand users per month. Determine \(Q(t)\).

Hints

- In an exponential model, relative growth rate is constant. - Use the rate measurement to recover the user count at month \(6\). - Center the exponential formula at the known endpoint.

Solution

1. Since \(Q'/Q=k\), the accumulated relative growth is \(6k=3\), so \(k=\frac12\) per month. 2. From \(Q'(6)=kQ(6)\), \(20=\frac12Q(6)\), so \(Q(6)=40\) thousand users. 3. Centering the exponential model at month \(6\) gives \(Q(t)=40e^{(t-6)/2}\) thousand users.

Answer

\(Q(t)=40e^{(t-6)/2}\) thousand users.
54366412
A bacteria culture grows exponentially according to \(Q'=kQ\), where \(Q(t)\) is measured in thousands of cells and \(t\) in hours. Over every \(2\)-hour interval, the increase in the culture is three times the amount present at the start of that interval. At \(t=1\), the culture is growing at \(12\ln2\) thousand cells per hour. Determine \(Q(t)\).

Hints

- Rewrite the stated increase as a total multiplicative factor. - Convert the two-hour multiplier into a continuous rate. - Use the growth-rate measurement to find the culture size at \(t=1\).

Solution

1. The time-shift condition is \(Q(t+2)-Q(t)=3Q(t)\), so \(Q(t+2)=4Q(t)\). 2. Therefore, \(e^{2k}=4\), giving \(k=\ln2\) per hour. 3. From \(Q'(1)=kQ(1)\), \(12\ln2=(\ln2)Q(1)\), so \(Q(1)=12\) thousand cells. 4. Thus \(Q(t)=12e^{(t-1)\ln2}=12\cdot2^{t-1}\) thousand cells.

Answer

\(Q(t)=12\cdot2^{t-1}\) thousand cells.
54368412
An investment account grows continuously with no deposits or withdrawals. Its balance rises from \(\$5000\) to \(\$6500\) in \(4\) years. Determine the continuous growth rate \(k\), write the balance model \(B(t)\), and find how long it takes the original balance to double.

Hints

- Use the ratio of the two balances to find the four-year growth factor. - Convert that factor into a continuous rate. - Set the model's multiplier equal to \(2\) for the doubling time.

Solution

1. Write \(B(t)=5000e^{kt}\). The four-year value gives \(6500=5000e^{4k}\). 2. Thus \(e^{4k}=1.3\), so \(k=\frac14\ln(1.3)\) per year. 3. The model is \(B(t)=5000e^{(\ln1.3)t/4}\) dollars. 4. The doubling time \(T\) satisfies \(e^{kT}=2\), so \(T=\frac{4\ln2}{\ln1.3}\approx10.57\) years.

Answer

\(k=\frac14\ln(1.3)\approx0.06559\) per year, \(B(t)=5000e^{(\ln1.3)t/4}\) dollars, and the doubling time is \(\frac{4\ln2}{\ln1.3}\approx10.57\) years.
54369112
A bacteria culture is claimed to follow one exact exponential model \(Q'=kQ\), where \(Q\) is measured in thousands of cells and \(t\) in hours. A lab reports \(Q(0)=4\), \(Q'(0)=2\), \(Q(3)=10\), and \(Q'(3)=4\). Determine whether all four measurements can come from one exponential model.

Hints

- Divide each reported growth rate by the corresponding culture size. - An exponential model has one constant relative growth rate. - Compare the two values of \(Q'/Q\) before attempting to build a model.

Solution

1. In an exponential model, the relative growth rate \(Q'/Q\) must be the same at every time. 2. At \(t=0\), the data give \(k=Q'(0)/Q(0)=2/4=0.5\) per hour. 3. At \(t=3\), the data give \(k=Q'(3)/Q(3)=4/10=0.4\) per hour. 4. Since the two exact relative growth rates are unequal, no single exponential model fits all four measurements.

Answer

No. The measurements imply \(k=0.5\) per hour at \(t=0\) but \(k=0.4\) per hour at \(t=3\).
54370312
A bacteria culture grows exponentially. At \(t=1\) hour, it contains \(6\) thousand cells. From \(t=1\) to \(t=3\), the culture gains a total of \(18\) thousand cells: \(\int_1^3 Q'(t)\,dt=18\). Determine \(Q(t)\).

Hints

- Convert the integral of the growth rate into a change in culture size. - Use the resulting endpoint ratio to find the rate constant. - Center the model at the time where the culture size is given.

Solution

1. By the Fundamental Theorem of Calculus, \(Q(3)-Q(1)=18\), so \(Q(3)=24\) thousand cells. 2. The two-hour multiplier is \(24/6=4\), so \(e^{2k}=4\) and \(k=\ln2\) per hour. 3. Therefore, \(Q(t)=6e^{(t-1)\ln2}=6\cdot2^{t-1}\) thousand cells.

Answer

\(Q(t)=6\cdot2^{t-1}\) thousand cells.
54371012
A bacteria culture grows exponentially according to \(Q'=kQ\), where \(Q(t)\) is measured in thousands of cells and \(t\) in hours. Initially, \(Q(0)=2\). At a positive time \(T\), the culture contains \(10\) thousand cells and is growing at \(10\) thousand cells per hour. Determine \(k\), \(T\), and \(Q(t)\).

Hints

- Use the growth rate and culture size at the same time in \(Q'=kQ\). - Apply the initial value after identifying the rate constant. - Solve the endpoint equation for time.

Solution

1. At time \(T\), \(10=k(10)\), so \(k=1\) per hour. 2. The model is therefore \(Q(t)=2e^t\) thousand cells. 3. The condition \(Q(T)=10\) gives \(2e^T=10\), so \(T=\ln5\) hours.

Answer

\(k=1\) per hour, \(T=\ln5\) hours, and \(Q(t)=2e^t\) thousand cells.
55605112
A quantity follows an exponential model \(Q'=kQ\). Measurements are shown below. <table> <tr><th>\(t\) (years)</th><th>\(Q(t)\)</th></tr> <tr><td>0</td><td>80</td></tr> <tr><td>2</td><td>40</td></tr> <tr><td>4</td><td>20</td></tr> </table> Determine \(k\), including units, and write the exponential model for \(Q(t)\).

Hints

- Compare consecutive measurements to identify the repeated multiplicative factor. - Relate that factor over a two-year interval to \(e^{2k}\). - Use the value at \(t=0\) as the initial amount in the model.

Solution

1. The quantity is multiplied by \(\frac{1}{2}\) every \(2\) years, so \(e^{2k}=\frac{1}{2}\). 2. Therefore, \(k=-\frac{\ln 2}{2}\,\text{year}^{-1}\). 3. Since \(Q(0)=80\), the model is \(Q(t)=80e^{-(\ln 2)t/2}\).

Answer

\(k=-\frac{\ln 2}{2}\,\text{year}^{-1}\) \(Q(t)=80e^{-(\ln 2)t/2}\)
52641512
A laboratory tracks the area of a bacterial culture. At \(t=0\), the culture covers \(1500\,\text{mm}^2\), and its area grows by \(3.5\%\) per hour. 1. Write an exponential model \(A(t)=ba^t\), where \(t\) is measured in hours. 2. Rewrite the model as \(A(t)=be^{kt}\). Find \(k\) to four decimal places. 3. Use \(A'(t)\) to find the instantaneous growth rate after \(10\) hours.

Hints

- 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.

Solution

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, \(A'(10)=1500\ln(1.035)(1.035)^{10}\approx72.79\,\text{mm}^2/\text{h}\).

Answer

1. \(A(t)=1500(1.035)^t\) 2. \(k=\ln(1.035)\approx0.0344\) 3. \(A'(10)\approx72.79\,\text{mm}^2/\text{h}\)
52643412
The mass of a radioactive substance decreases exponentially according to \(m(t)=m_0a^t\), where \(t\) is measured in days and \(m(t)\) is measured in milligrams. After \(2\) days, the mass is \(20\,\text{mg}\). After \(5\) days, the mass is \(2.5\,\text{mg}\). a) Find the daily decay factor \(a\) and the initial mass \(m_0\). b) Find the half-life of the substance. c) Rewrite the model in the form \(m(t)=m_0e^{kt}\). Then find the instantaneous rate of change of the mass after exactly \(1\) day.

Hints

- 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.

Solution

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, \(m'(1)=40\ln(0.5)\approx -27.73\,\text{mg/day}\).

Answer

a) \(a=0.5\); \(m_0=80\,\text{mg}\) b) \(1\,\text{day}\) c) \(m(t)=80e^{\ln(0.5)t}\); \(m'(1)=40\ln(0.5)\approx -27.73\,\text{mg/day}\)
52643912
Researchers measured the area of a shallow lake covered by algae over several days. <table> <tr> <td>Time \(t\) (days)</td> <td>0</td> <td>2</td> <td>4</td> <td>6</td> <td>8</td> </tr> <tr> <td>Area \(A(t)\) (\(\text{m}^2\))</td> <td>\(5.0\)</td> <td>\(6.1\)</td> <td>\(7.4\)</td> <td>\(9.0\)</td> <td>\(11.0\)</td> </tr> </table> a) Explain why an exponential function is a reasonable model for the data. b) Find the growth factor for each \(2\)-day interval and the corresponding daily growth factor. Give the approximate daily percent increase. c) Use the growth factor from part b) to write a model of the form \(A(t)=A_0e^{kt}\). d) Find the growth-rate function \(A'(t)\), and find the instantaneous growth rate at \(t=5\). e) Find the doubling time for the algae-covered area.

Hints

- 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.

Solution

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.

Answer

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 \(6.97\,\text{days}\)
52644012
After a patient takes a medication, the concentration of the active ingredient in the patient's blood is measured at several times. <table> <tr> <td>Time \(t\) (hours)</td> <td>0</td> <td>3</td> <td>6</td> <td>9</td> <td>12</td> </tr> <tr> <td>Concentration \(C(t)\) (\(\text{mg/L}\))</td> <td>\(120\)</td> <td>\(84.0\)</td> <td>\(58.8\)</td> <td>\(41.2\)</td> <td>\(28.8\)</td> </tr> </table> a) Use calculations to show that exponential decay is a reasonable model for the data. b) Give the decay factor for each \(3\)-hour interval and the corresponding hourly decay factor. By approximately what percent does the concentration decrease each hour? c) Write a model of the form \(C(t)=C_0e^{kt}\). d) Find the instantaneous rate of change of the concentration at \(t=6\). e) Find the half-life of the medication in the blood.

Hints

- 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.

Solution

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.

Answer

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 \(5.83\,\text{hours}\)
52648112
A radioactive isotope used in medical imaging has a half-life of \(6\) hours. Immediately after it is administered, its activity in a patient's body is \(450\,\text{MBq}\). a) Write a function \(A(t)=be^{kt}\) that models the activity after \(t\) hours. b) Interpret the parameter \(b\), and find the percent decrease in activity per hour. c) Find when the activity falls below \(20\,\text{MBq}\). d) Find \(A'(t)\). Calculate and interpret the instantaneous rates of change at \(t=0\) and \(t=12\).

Hints

- 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.

Solution

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.

Answer

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 \(k=\frac{\ln(0.5)}{6}\); \(A'(0)\approx -51.99\,\text{MBq/h}\), \(A'(12)\approx -13.00\,\text{MBq/h}\)
52648212
An industrial filtration system reduces the concentration of a pollutant in a treatment tank. The concentration decreases exponentially. At \(t=0\), the concentration is \(320\,\text{mg/L}\). After \(10\) minutes, it is \(200\,\text{mg/L}\). a) Find a model of the form \(c(t)=c_0e^{kt}\), where \(t\) is measured in minutes. b) Find the half-life of the pollutant concentration. c) Find when the concentration falls below \(10\,\text{mg/L}\). d) Find the rate-of-change function \(c'(t)\). Calculate and interpret the instantaneous rate of change after \(20\) minutes.

Hints

- 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.

Solution

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.

Answer

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 \(k=\frac{\ln(0.625)}{10}\); \(c'(20)\approx -5.88\,\text{mg}/(\text{L}\cdot\text{min})\)
52649712
Two bacterial cultures are observed in a laboratory. Culture A starts with \(1200\) cells and grows by \(4.2\%\) per hour. Culture B starts with \(5000\) cells and decreases by \(3.5\%\) per hour because of an antibiotic. a) Model each culture with a function of the form \(f(t)=be^{kt}\), where \(t\) is measured in hours. b) Find when culture A reaches \(2000\) cells. c) Find when the two cultures have the same number of cells. d) Find the rate-of-change function for each culture. Calculate and interpret the rate of change of culture B at \(t=0\) and \(t=10\).

Hints

- 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.

Solution

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.

Answer

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
52650012
A bottle of soda is removed from a refrigerator and placed in a warm room. Its temperature is modeled by \(g(t)=A-Be^{-kt}\), where \(t\) is measured in minutes and \(g(t)\) is measured in degrees Celsius. a) Initially, the soda is \(7\,{}^\circ\text{C}\), and the room is \(25\,{}^\circ\text{C}\). Find \(A\) and \(B\). b) After \(20\) minutes, the soda is \(16\,{}^\circ\text{C}\). Find \(k\). c) For a different drink, let \(A=25\), \(B=18\), and \(k=0.04\,\text{min}^{-1}\). Find when its temperature is increasing at an instantaneous rate of \(0.25\,{}^\circ\text{C}/\text{min}\).

Hints

- 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.

Solution

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.

Answer

a) \(A=25\), \(B=18\) b) \(k\approx0.0347\,\text{min}^{-1}\) c) About \(26.4\) minutes
52660312
A bacterial culture covers an area of \(250\,\text{mm}^2\) at the start of an experiment. After \(5\) hours, it covers \(600\,\text{mm}^2\). Assume that the area grows exponentially. a) Find a model of the form \(f(t)=ae^{kt}\), where \(t\) is measured in hours. b) Find the doubling time of the area. c) Find when the culture reaches an area of \(2500\,\text{mm}^2\). d) Find the instantaneous growth rate at \(t=8\).

Hints

- 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.

Solution

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, \(f'(8)\approx 177.64\,\text{mm}^2/\text{h}\).

Answer

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) \(f'(8)\approx 177.64\,\text{mm}^2/\text{h}\)
52660412
A radioactive isotope decays exponentially. At the beginning of an observation, \(120\,\text{mg}\) is present. After \(10\) days, \(45\,\text{mg}\) remains. a) Find a model of the form \(m(t)=m_0e^{-\lambda t}\), where \(t\) is measured in days. b) Find the half-life of the isotope. c) Find when the mass decreases to \(10\,\text{mg}\). d) Find the instantaneous decay rate at \(t=0\) and at \(t=20\).

Hints

- 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.

Solution

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

Answer

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) \(m'(0)\approx -11.77\,\text{mg/day}\); \(m'(20)\approx -1.66\,\text{mg/day}\)
52770512
A metal part is removed from a furnace at \(95\,{}^\circ\text{C}\) and placed in a room maintained at \(20\,{}^\circ\text{C}\). Its temperature satisfies \(T'(t)=k(20-T(t))\), where \(t\) is the number of minutes since removal. Immediately after removal, the cooling rate is \(T'(0)=-4.5\,{}^\circ\text{C}/\text{min}\). Find \(k\) and a function \(T(t)\) for the temperature.

Hints

- Substitute the initial temperature and initial cooling rate into the differential equation. - The temperature difference from the room temperature decays exponentially. - Use the initial condition to determine the coefficient of the exponential term.

Solution

1. The initial temperature is \(T(0)=95\). Substitute \(t=0\) into the differential equation: \(-4.5=k(20-95)=-75k\). 2. Therefore, \(k=0.06\,\text{min}^{-1}\). 3. The solution that approaches the room temperature and satisfies \(T(0)=95\) is \(T(t)=20+(95-20)e^{-0.06t}=20+75e^{-0.06t}\).

Answer

\(k=0.06\,\text{min}^{-1}\); \(T(t)=20+75e^{-0.06t}\)
52770612
A frozen pizza at \(-18\,{}^\circ\text{C}\) is placed in an oven preheated to \(220\,{}^\circ\text{C}\). Its temperature satisfies Newton's law of heating, \(T'(t)=k(220-T(t))\), where \(t\) is measured in minutes. At \(t=0\), the instantaneous heating rate is \(11.9\,{}^\circ\text{C}/\text{min}\). a) Find \(k\) and the corresponding temperature function \(T(t)\). b) Find when the pizza reaches an internal temperature of \(75\,{}^\circ\text{C}\). Round the time to two decimal places.

Hints

- Substitute the initial temperature and heating rate into the differential equation. - The temperature difference from the oven temperature decays exponentially. - Isolate the exponential expression before taking a logarithm.

Solution

1. Since \(T(0)=-18\) and \(T'(0)=11.9\), the differential equation gives \(11.9=k(220-(-18))=238k\). Thus, \(k=0.05\,\text{min}^{-1}\). 2. The solution approaching the oven temperature is \(T(t)=220-(220-(-18))e^{-0.05t}=220-238e^{-0.05t}\). 3. Set \(T(t)=75\): \(75=220-238e^{-0.05t}\). Then \(e^{-0.05t}=\frac{145}{238}\), so \(t=\frac{\ln\left(\frac{145}{238}\right)}{-0.05}\approx 9.91\) minutes.

Answer

a) \(k=0.05\,\text{min}^{-1}\); \(T(t)=220-238e^{-0.05t}\) b) \(t\approx9.91\,\text{min}\)
52992312
A medication is eliminated from the body so that the amount decreases by about \(12\%\) each hour. At \(t=0\), a patient has \(400\,\text{mg}\) of the medication in the bloodstream. Model the amount by \(N(t)=N_0e^{-kt}\), where \(t\) is measured in hours. 1. Find the decay constant \(k\) and write the model. 2. Find when one-fourth of the initial amount remains. 3. Find the instantaneous rate of change exactly \(5\) hours after the observation begins. Explain the meaning of its sign.

Hints

- 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.

Solution

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.

Answer

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
52992412
Radiocarbon dating uses the decay of carbon-14, \({}^{14}\text{C}\), to estimate the age of organic material. The half-life of carbon-14 is approximately \(5730\) years. Its decay is modeled by \(N(t)=N_0e^{-kt}\), where \(t\) is measured in years. 1. Find the decay constant \(k\). 2. A wood fragment contains \(65\%\) of its original carbon-14. Estimate the age of the fragment. 3. Show that the instantaneous rate of change \(N'(t)\) is proportional to the amount \(N(t)\). Give the constant of proportionality.

Hints

- 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 \(N(t)\).

Solution

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 \(-k\).

Answer

1. \(k\approx 0.00012097\,\text{yr}^{-1}\) 2. About \(3561\,\text{years}\) 3. \(N'(t)=-kN(t)\); constant of proportionality: \(-k\approx -0.00012097\,\text{yr}^{-1}\)
53277712
A cup of hot coffee is placed in a room at a constant temperature of \(70\,{}^\circ\text{F}\). Its temperature is modeled by \(f(t)=ae^{-kt}+c\), where \(t\ge0\) is measured in minutes and \(f(t)\) is measured in degrees Fahrenheit. Immediately after the coffee is poured, its temperature is \(180\,{}^\circ\text{F}\), and it is cooling at an instantaneous rate of \(11\,{}^\circ\text{F}\) per minute. a) Find \(a\), \(c\), and \(k\), including the units of \(k\). b) Find the coffee's temperature after \(10\) minutes. c) When will the coffee cool to \(125\,{}^\circ\text{F}\)? Round to two decimal places.

Hints

- Translate the initial temperature, long-term temperature, and initial cooling rate into equations. - Determine which parameter gives the long-term temperature. - Use the derivative to represent the initial cooling rate and track the time unit of its coefficient. - After finding the complete model, substitute the given time or target temperature as needed.

Solution

1. The coffee approaches room temperature as \(t\to\infty\). Since \(e^{-kt}\to0\), the model approaches \(c\), so \(c=70\). 2. Use the initial temperature: \(f(0)=a+70=180\), so \(a=110\). 3. Differentiate: \(f'(t)=-110ke^{-kt}\). The initial cooling rate is \(f'(0)=-11\), so \(-110k=-11\), giving \(k=0.1\,\text{min}^{-1}\). Thus \(f(t)=110e^{-0.1t}+70\). 4. After \(10\) minutes, \(f(10)=110e^{-1}+70\approx110.47\,{}^\circ\text{F}\). 5. Set the temperature equal to \(125\): \(110e^{-0.1t}+70=125\). Then \(e^{-0.1t}=\frac{1}{2}\), so \(t=10\ln2\approx6.93\) minutes.

Answer

a) \(a=110\,{}^\circ\text{F}\), \(c=70\,{}^\circ\text{F}\), and \(k=0.1\,\text{min}^{-1}\), so \(f(t)=110e^{-0.1t}+70\). b) \(f(10)\approx110.47\,{}^\circ\text{F}\) c) \(t\approx6.93\,\text{min}\)
53450312
A cup of hot tea cools in a room. Its temperature is modeled by \(T(t)=ae^{-kt}+T_R\), where \(t\) is measured in minutes, \(T(t)\) is measured in degrees Celsius, and \(T_R\) is the room temperature. The graph shows the cooling process. a) Use the graph to find \(T_R\) and the tea's initial temperature. Then determine \(a\). b) After \(10\) minutes, the difference between the tea's temperature and the room temperature is half its initial value. Find the exact value of \(k\). c) Find the instantaneous rate of change of the tea's temperature at \(t=0\). Round to two decimal places and include units.
Figure for problem 534503

Hints

- 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.

Solution

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, \(T'(0)=-70\left(\frac{\ln 2}{10}\right)=-7\ln 2\approx -4.85\,{}^\circ\text{C}/\text{min}\).

Answer

a) \(T_R=20\,{}^\circ\text{C}\), \(T(0)=90\,{}^\circ\text{C}\), and \(a=70\) b) \(k=\frac{\ln 2}{10}\) c) \(T'(0)\approx -4.85\,{}^\circ\text{C}/\text{min}\)
54000312
A digital archive's indexed-file count follows \(F'=kF\) and begins at \(F(0)=4000\), where \(t\) is measured in days. The quantity doubles every \(15\) days. a) Find \(k\) exactly. b) Write the particular model. c) Find the number of indexed files after \(45\) days, rounded to two decimal places.

Hints

- Translate the doubling statement into a multiplier over one period. - Compare that multiplier with the exponential factor over the same period. - Use the initial number of files as the model's coefficient.

Solution

1. The doubling condition gives \(e^{15k}=2\), so \(k=\frac{\ln 2}{15}\,\text{day}^{-1}\). 2. Substitute this value into \(F(t)=4000e^{kt}\) to obtain \(F(t)=4000e^{\left(\frac{\ln 2}{15}\right)t}\). 3. \(F(45)=4000e^{\left(\frac{\ln 2}{15}\right)\cdot45}=4000e^{3\ln2}=4000\cdot 2^3=32{,}000.00\).

Answer

a) \(k=\frac{\ln 2}{15}\,\text{day}^{-1}\) b) \(F(t)=4000e^{\left(\frac{\ln 2}{15}\right)t}\) c) \(F(45)=32{,}000.00\,\text{indexed files}\)
54000412
The amount \(L(t)\) of a fluorescent label, measured in milligrams, follows \(L'=kL\) and begins at \(L(0)=72\), where \(t\) is measured in hours. The quantity has a half-life of \(8\) hours. a) Find \(k\) exactly. b) Write the particular model. c) Find the amount after \(20\) hours, rounded to two decimal places.

Hints

- A half-life of \(8\) hours means \(L(8)=36\), so compare \(72e^{8k}\) with \(36\). - Solve the resulting exponential equation for \(k\) using natural logarithms. - Evaluate the particular model at \(t=20\) and round only the final amount.

Solution

1. The half-life condition gives \(e^{8k}=\frac{1}{2}\), so \(k=-\frac{\ln2}{8}\,\text{h}^{-1}\). 2. Substitute this value into \(L(t)=72e^{kt}\) to obtain \(L(t)=72e^{-\left(\frac{\ln2}{8}\right)t}\). 3. \(L(20)=72e^{-\left(\frac{\ln2}{8}\right)\cdot20}\approx12.73\).

Answer

a) \(k=-\frac{\ln2}{8}\,\text{h}^{-1}\) b) \(L(t)=72e^{-\left(\frac{\ln2}{8}\right)t}\) c) \(L(20)\approx12.73\,\text{mg}\)
54000512
A seedling count in a greenhouse model follows \(S'=kS\) and begins at \(S(0)=125\), where \(t\) is measured in weeks. The quantity doubles every \(6\) weeks. a) Find \(k\) exactly. b) Write the particular model. c) Find the seedling count after \(15\) weeks, rounded to two decimal places.

Hints

- Express the six-week doubling condition as \(S(6)=2S(0)\). - Substitute the exponential form \(S(t)=125e^{kt}\) and solve the resulting equation for \(k\). - Use \(15/6\) doubling periods as a reasonableness check on the final count.

Solution

1. The doubling condition gives \(e^{6k}=2\), so \(k=\frac{\ln 2}{6}\,\text{week}^{-1}\). 2. Substitute this value into \(S(t)=125e^{kt}\) to obtain \(S(t)=125e^{\left(\frac{\ln 2}{6}\right)t}\). 3. \(S(15)=125e^{\left(\frac{\ln 2}{6}\right)\cdot15}\approx 707.11\).

Answer

a) \(k=\frac{\ln 2}{6}\,\text{week}^{-1}\) b) \(S(t)=125e^{\left(\frac{\ln 2}{6}\right)t}\) c) \(S(15)\approx 707.11\,\text{seedlings}\)
54001012
The quality index \(Q(t)\) for a paper-preservation treatment, measured in arbitrary index units, is modeled by \(Q'=kQ\). Measurements give \(Q(0)=100\) and \(Q(6)=82\), where \(t\) is measured in days. a) Determine \(k\). b) Write the particular model. c) Predict \(Q(14)\), rounding \(k\) and the final prediction to four and two decimal places, respectively.

Hints

- Use \(Q(6)/Q(0)\) to isolate the six-day exponential multiplier. - Take a logarithm and divide by the elapsed time to determine \(k\). - Because the later index is smaller, check that \(k\) is negative and use its unrounded value for \(Q(14)\).

Solution

1. From \(82=100e^{6k}\), \(k=\frac{1}{6}\ln\left(\frac{82}{100}\right)\approx -0.0331\,\text{day}^{-1}\). 2. Substitute the exact value of \(k\) to obtain \(Q(t)=100e^{\left[\frac{1}{6}\ln\left(\frac{82}{100}\right)\right]t}\). 3. Using the unrounded rate, \(Q(14)=100e^{\left[\frac{1}{6}\ln\left(\frac{82}{100}\right)\right]\cdot14}\approx 62.94\).

Answer

a) \(k=\frac{1}{6}\ln\left(\frac{82}{100}\right)\approx -0.0331\,\text{day}^{-1}\) b) \(Q(t)=100e^{\left[\frac{1}{6}\ln\left(\frac{82}{100}\right)\right]t}\) c) \(Q(14)\approx 62.94\,\text{index units}\)
54001312
A simulated population \(N(t)\) of autonomous vehicles is modeled by \(N'=kN\). Measurements give \(N(0)=900\) and \(N(4)=1350\), where \(t\) is measured in months. a) Determine \(k\). b) Write the particular model. c) Predict \(N(10)\), rounding \(k\) and the final prediction to four and two decimal places, respectively.

Hints

- Cancel the initial vehicle count by comparing \(N(4)\) with \(N(0)\). - Divide the logarithm of the observed multiplier by four months. - The larger later count implies positive growth; use the full rate in the ten-month model evaluation.

Solution

1. From \(1350=900e^{4k}\), \(k=\frac{1}{4}\ln\left(\frac{1350}{900}\right)\approx 0.1014\,\text{month}^{-1}\). 2. Substitute the exact value of \(k\) to obtain \(N(t)=900e^{\left[\frac{1}{4}\ln\left(\frac{1350}{900}\right)\right]t}\). 3. Using the unrounded rate, \(N(10)=900e^{\left[\frac{1}{4}\ln\left(\frac{1350}{900}\right)\right]\cdot10}\approx 2480.11\).

Answer

a) \(k=\frac{1}{4}\ln\left(\frac{1350}{900}\right)\approx 0.1014\,\text{month}^{-1}\) b) \(N(t)=900e^{\left[\frac{1}{4}\ln\left(\frac{1350}{900}\right)\right]t}\) c) \(N(10)\approx 2480.11\,\text{vehicles}\)
54001712
A museum audio guide's active-device count is modeled by \(A(t)=800e^{0.06t}\), where \(t\) is measured in days. Find the time when the count first reaches \(1500\). Round to two decimal places, and show the equation used.

Hints

- Set \(800e^{0.06t}\) equal to the target count \(1500\). - Divide by the initial count before taking a logarithm. - Since the model grows and the target exceeds the initial value, the resulting time should be positive.

Solution

1. Set the model equal to the target: \(800e^{0.06t}=1500\). 2. Divide by \(800\) and take natural logarithms: \(0.06t=\ln\left(\frac{1500}{800}\right)\). 3. \(t=\frac{\ln\left(\frac{1500}{800}\right)}{0.06}\approx 10.48\,\text{days}\).

Answer

\(800e^{0.06t}=1500\), so \(t=\frac{\ln\left(\frac{1500}{800}\right)}{0.06}\approx 10.48\,\text{days}\).
54001812
A protective coating's remaining thickness index is modeled by \(H(t)=120e^{-0.045t}\), where \(t\) is measured in months. Find the time when the index first falls to \(70\). Round to two decimal places, and show the equation used.

Hints

- Set the coating model equal to the target index \(70\). - After dividing by \(120\), use a logarithm to bring down the exponent \(-0.045t\). - The target is below the initial value, so a positive time is consistent with decay.

Solution

1. Set the model equal to the target: \(120e^{-0.045t}=70\). 2. Divide by \(120\) and take natural logarithms: \(-0.045t=\ln\left(\frac{70}{120}\right)\). 3. \(t=\frac{\ln\left(\frac{70}{120}\right)}{-0.045}\approx 11.98\,\text{months}\).

Answer

\(120e^{-0.045t}=70\), so \(t=\frac{\ln\left(\frac{70}{120}\right)}{-0.045}\approx 11.98\,\text{months}\).
54002412
A model satisfies \(Q'=0.08Q\) and \(Q(0)=500\). Kai writes \(Q(t)=500(1.08)^t\). Explain the mismatch and give the model that satisfies the differential equation.

Hints

- Compare the relative rate \(Q'/Q\) implied by each exponential form. - Rewrite \((1.08)^t\) using base \(e\) before comparing rate constants. - Keep the initial value \(500\) unchanged while correcting the exponential factor.

Solution

1. Solving \(Q'=0.08Q\) gives \(Q(t)=Ce^{0.08t}\), and \(Q(0)=500\) gives \(C=500\). 2. Therefore, the required model is \(Q(t)=500e^{0.08t}\). 3. The expression \(500(1.08)^t=500e^{(\ln1.08)t}\) is also an exponential function of real \(t\), but its continuous relative rate is \(\ln(1.08)\), not \(0.08\). At integer times it matches a model with an \(8\%\) multiplicative increase per time unit, which is a different rate specification.

Answer

The differential equation requires the continuous relative rate \(0.08\), so \(Q(t)=500e^{0.08t}\). The model \(500(1.08)^t\) instead has continuous relative rate \(\ln(1.08)\), so it satisfies \(Q'=\ln(1.08)Q\), not \(Q'=0.08Q\).
54002512
Two quantities satisfy \(A'=0.04A\) and \(B'=0.07B\), with \(A(0)=1000\) and \(B(0)=600\), where \(t\) is measured in time units. Determine which quantity is larger initially, whether \(B\) eventually exceeds \(A\), and the crossover time.

Hints

- Write both exponential models from their initial values and continuous rates. - Set the models equal to find the crossover time. - Compare the growth constants to determine which model is larger afterward.

Solution

1. The models are \(A(t)=1000e^{0.04t}\) and \(B(t)=600e^{0.07t}\), so initially \(A(0)=1000>600=B(0)\). 2. At the crossover, \(1000e^{0.04t}=600e^{0.07t}\). Thus \(e^{0.03t}=\frac{5}{3}\), so \(t=\frac{\ln\left(\frac{5}{3}\right)}{0.03}\approx17.03\). 3. Because \(B\) has the larger growth constant, \(B(t)>A(t)\) after the crossover.

Answer

Initially, \(A(0)=1000>600=B(0)\). The crossover occurs at \(t=\frac{\ln\left(\frac{5}{3}\right)}{0.03}\approx17.03\,\text{time units}\), and \(B\) is larger after that time.
54002612
A decay model is \(M(t)=M_0e^{-0.12t}\), where \(t\) is measured in time units. Interpret \(\frac{M'(t)}{M(t)}\) and state whether the percentage lost over one full time unit is exactly \(12\%\).

Hints

- Find the relative rate by dividing the derivative by the current amount. - A continuous rate determines an exponential multiplier over one time unit. - Convert the remaining multiplier into the fraction lost.

Solution

1. Differentiate the model: \(M'(t)=-0.12M_0e^{-0.12t}=-0.12M(t)\). 2. Therefore, \(\frac{M'(t)}{M(t)}=-0.12\,\text{time unit}^{-1}\). This is the continuous relative rate, and its negative sign indicates decay. 3. Over one time unit, the multiplier is \(e^{-0.12}\), so the fraction lost is \(1-e^{-0.12}\approx0.1131\), or about \(11.31\%\), not exactly \(12\%\).

Answer

\(\frac{M'(t)}{M(t)}=-0.12\,\text{time unit}^{-1}\) is the continuous relative rate and indicates decay. The percentage lost in one time unit is \(100\left(1-e^{-0.12}\right)\%\approx11.31\%\), not exactly \(12\%\).
54002912
Model A is \(Q_A(t)=200e^{0.05t}\). Model B begins at \(200\) and doubles every \(12\) time units. Which model has the larger continuous growth constant?

Hints

- Read Model A's continuous growth constant from its exponent. - Translate Model B's doubling time into an exponential equation. - Compare the two constants in the same units.

Solution

1. Model A has continuous growth constant \(k_A=0.05\,\text{time unit}^{-1}\). 2. For Model B, the doubling condition gives \(e^{12k_B}=2\), so \(k_B=\frac{\ln2}{12}\approx0.0578\,\text{time unit}^{-1}\). 3. Since \(0.0578>0.05\), Model B has the larger continuous growth constant.

Answer

Model B. Its continuous growth constant is \(\frac{\ln2}{12}\approx0.0578\,\text{time unit}^{-1}\), which is greater than Model A's \(0.05\,\text{time unit}^{-1}\).
54346812
An invasive plant patch covers \(40\,\text{m}^2\) when monitoring begins. Its area is modeled by \(Q'=kQ\). During the first \(4\) weeks, the patch’s average rate of increase is \(30\,\text{m}^2\) per week. a) Determine \(k\) exactly and state its units. b) Find the doubling time. c) Find the instantaneous growth rate at the end of week \(4\).

Hints

- Convert the average growth rate into the patch’s area after \(4\) weeks. - Compare the final area with the initial area in the exponential model. - A doubling time is the time required for the exponential factor to equal \(2\). - Use \(Q'=kQ\) after finding the area at week \(4\).

Solution

1. The average-rate condition gives \(\frac{Q(4)-40}{4}=30\), so \(Q(4)=160\). 2. Since \(Q(t)=40e^{kt}\), \(160=40e^{4k}\). Thus \(e^{4k}=4\), so \(k=\frac{\ln 2}{2}\,\text{week}^{-1}\). 3. The doubling time \(T\) satisfies \(e^{kT}=2\). Therefore, \(T=\frac{\ln 2}{k}=2\) weeks. 4. The instantaneous growth rate at week \(4\) is \(Q'(4)=kQ(4)=\frac{\ln 2}{2}(160)=80\ln 2\,\text{m}^2/\text{week}\).

Answer

a) \(k=\frac{\ln 2}{2}\,\text{week}^{-1}\) b) \(2\) weeks c) \(Q'(4)=80\ln 2\,\text{m}^2/\text{week}\)
54347612
A loaf of bread is removed from an oven at \(190\,^{\circ}\text{F}\) and placed in a room kept at \(70\,^{\circ}\text{F}\). Its temperature follows Newton’s law of cooling, \(T'=k(T-70)\), where \(t\) is measured in minutes. After \(10\) minutes, the bread is at \(130\,^{\circ}\text{F}\). a) Find \(k\) and the temperature model \(T(t)\). b) Find the temperature after \(20\) minutes. c) When will the bread reach \(90\,^{\circ}\text{F}\)? Round to two decimal places.

Hints

- Model the difference between the bread’s temperature and room temperature. - Use the temperature after \(10\) minutes to determine the exponential factor over one ten-minute interval. - For part b, notice that \(20\) minutes represents two equal cooling intervals. - For part c, isolate the exponential factor before using logarithms.

Solution

1. The temperature difference from the room follows \(T(t)-70=Ce^{kt}\). Since \(T(0)=190\), \(C=120\). 2. Use \(T(10)=130\): \(60=120e^{10k}\), so \(e^{10k}=\frac{1}{2}\) and \(k=-\frac{\ln 2}{10}\,\text{min}^{-1}\). 3. Therefore, \(T(t)=70+120e^{-(\ln 2)t/10}=70+120\cdot2^{-t/10}\). 4. At \(t=20\), \(T(20)=70+120\cdot2^{-2}=100\,^{\circ}\text{F}\). 5. For \(T=90\), \(20=120\cdot2^{-t/10}\), so \(2^{-t/10}=\frac{1}{6}\). Thus \(t=\frac{10\ln 6}{\ln 2}\approx25.85\) minutes.

Answer

a) \(k=-\frac{\ln 2}{10}\,\text{min}^{-1}\) and \(T(t)=70+120\cdot2^{-t/10}\) b) \(100\,^{\circ}\text{F}\) c) Approximately \(25.85\) minutes
54348412
The rate at which an online video is receiving views is modeled by \(Q'=kQ\), where \(Q(t)\) is measured in views per hour and \(t\) is measured in hours. Measurements give \(Q(0)=80\), \(Q(5)=320\), and \(\int_0^5 Q(t)\,dt=\frac{600}{\ln 2}\) total views. Determine \(k\), verify that all three measurements are consistent, and write the model for \(Q(t)\). State the units of \(k\).

Hints

- The integral of a view rate over time represents a total number of views. - Integrate the differential equation over the measured five-hour interval. - Use the Fundamental Theorem of Calculus on the derivative side. - Check the endpoint ratio against the exponential factor produced by the rate constant.

Solution

1. Integrating \(Q'=kQ\) from \(0\) to \(5\) gives \(Q(5)-Q(0)=k\int_0^5Q(t)\,dt\). 2. Thus \(240=k\frac{600}{\ln 2}\), so \(k=\frac{2\ln 2}{5}\,\text{h}^{-1}\). 3. The endpoint ratio predicted by this rate is \(e^{5k}=e^{2\ln 2}=4\), which agrees with \(\frac{320}{80}=4\). 4. Therefore, \(Q(t)=80e^{(2\ln 2/5)t}\) views per hour. The integral has units of views, so it is consistent with the stated total number of views.

Answer

\(k=\frac{2\ln 2}{5}\,\text{h}^{-1}\), and \(Q(t)=80e^{(2\ln 2/5)t}\) views per hour. All three measurements are consistent.
54349212
A medication amount \(Q(t)\), in milligrams, follows the exponential model \(Q'=kQ\), where \(t\) is measured in hours. At \(t=3\), the amount is \(20\,\text{mg}\), and the tangent line to the graph at that time crosses the t-axis at \(t=5\). a) Determine \(k\) and state its units. b) Write the model \(Q(t)\). c) Find the medication’s half-life.

Hints

- Express the tangent-line slope at \(t=3\) using \(Q'=kQ\). - Use the t-axis intercept in the tangent-line equation. - Anchor the exponential model at the measured point \((3,20)\). - For half-life, set the exponential factor over a time interval equal to \(\frac{1}{2}\).

Solution

1. The tangent line at \(t=3\) is \(L(t)=20+Q'(3)(t-3)=20+20k(t-3)\). 2. Since \(L(5)=0\), \(0=20+40k\), so \(k=-\frac{1}{2}\,\text{h}^{-1}\). 3. Using \(Q(3)=20\), the model is \(Q(t)=20e^{-\frac{1}{2}(t-3)}\) milligrams. 4. The half-life \(T\) satisfies \(e^{-T/2}=\frac{1}{2}\), so \(T=2\ln 2\) hours.

Answer

a) \(k=-\frac{1}{2}\,\text{h}^{-1}\) b) \(Q(t)=20e^{-\frac{1}{2}(t-3)}\) milligrams c) \(2\ln 2\) hours
54350412
A rectangular display panel has length \(A(t)\) centimeters and width \(B(t)\) centimeters. The length grows according to \(A'=0.12A\), while the width shrinks according to \(B'=-0.03B\), where \(t\) is measured in years. At \(t=4\), both dimensions are \(200\,\text{cm}\). Let \(R(t)=A(t)B(t)\) be the panel’s area. a) Find a differential equation and initial value for \(R\). b) Determine \(R(10)\) and state its units. c) Explain why the area grows even though one dimension shrinks.

Hints

- Differentiate the product of the two dimensions. - Replace each derivative using its own exponential rate equation. - Anchor the area model at the time when both dimensions are known. - Compare the two signed continuous relative rates to explain the direction of area change.

Solution

1. By the product rule, \(R'=A'B+AB'=(0.12-0.03)AB=0.09R\). 2. Since \(A(4)=B(4)=200\,\text{cm}\), \(R(4)=40{,}000\,\text{cm}^2\). 3. Therefore, \(R(t)=40{,}000e^{0.09(t-4)}\), so \(R(10)=40{,}000e^{0.54}\,\text{cm}^2\). 4. The length has continuous relative rate \(0.12\) per year and the width has continuous relative rate \(-0.03\) per year. Their sum is \(0.09\) per year, so the area grows.

Answer

a) \(R'=0.09R\), \(R(4)=40{,}000\,\text{cm}^2\) b) \(R(10)=40{,}000e^{0.54}\,\text{cm}^2\) c) The signed continuous relative rates add to \(0.12-0.03=0.09\) per year, so the area has a positive continuous relative growth rate.
54352412
The rate \(Q(t)\) at which a news story is receiving new views is measured in views per hour and follows \(Q'=kQ\). The viewing rate is decreasing. Two hours after publication, \(Q(2)=5\) views per hour, and the model predicts \(\int_2^\infty Q(t)\,dt=20\) additional views after that time. Determine \(k\), state its units, and write the model for \(Q(t)\).

Hints

- Write the exponential model using the known viewing rate at \(t=2\). - The improper integral converges only for a negative rate constant. - Interpret the integral of a rate as an accumulated number of views. - Evaluate the future accumulation in terms of \(k\) before using the stated total.

Solution

1. Anchored at \(t=2\), the model is \(Q(t)=5e^{k(t-2)}\), with \(k<0\). 2. The integral of the viewing rate represents the total number of additional views. Thus \(\int_2^\infty5e^{k(t-2)}\,dt=-\frac{5}{k}\). 3. Setting \(-\frac{5}{k}=20\) gives \(k=-\frac{1}{4}\,\text{h}^{-1}\). 4. Therefore, \(Q(t)=5e^{-(t-2)/4}\) views per hour.

Answer

\(k=-\frac{1}{4}\,\text{h}^{-1}\), and \(Q(t)=5e^{-(t-2)/4}\) views per hour.
54353912
A pollutant concentration \(Q(t)\), in milligrams per liter, follows \(Q'=kQ\), where \(t\) is measured in hours. The concentration is decreasing. Measurements give \(Q'(2)=-12\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\) and \(Q'(8)=-3\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\). Determine \(k\), \(Q(2)\), and a formula for \(Q(t)\). State the units of \(k\).

Hints

- Take a ratio of the two rate measurements to eliminate the unknown concentration scale. - Use the six-hour time gap in the exponential factor. - Recover the concentration from its rate at the same time using \(Q'=kQ\).

Solution

1. Derivatives in an exponential model have the same exponential factor as the quantity, so \(\frac{Q'(8)}{Q'(2)}=e^{6k}\). 2. Thus \(\frac{1}{4}=e^{6k}\), giving \(k=-\frac{\ln 2}{3}\,\text{h}^{-1}\). 3. From \(Q'(2)=kQ(2)\), \(-12=-\frac{\ln 2}{3}Q(2)\), so \(Q(2)=\frac{36}{\ln 2}\,\text{mg/L}\). 4. Therefore, \(Q(t)=\frac{36}{\ln 2}e^{-\frac{(\ln 2)(t-2)}{3}}\) milligrams per liter.

Answer

\(k=-\frac{\ln 2}{3}\,\text{h}^{-1}\), \(Q(2)=\frac{36}{\ln 2}\,\text{mg/L}\), and \(Q(t)=\frac{36}{\ln 2}e^{-\frac{(\ln 2)(t-2)}{3}}\) milligrams per liter.
54355212
A bacterial culture’s mass \(Q(t)\), in grams, follows \(Q'=kQ\), where \(t\) is measured in hours. The culture has mass \(2\,\text{g}\) at \(t=0\) and \(18\,\text{g}\) at \(t=4\). a) Determine the mass at \(t=2\) without first solving for \(k\). b) Determine \(k\) and state its units. c) Write the exponential model.

Hints

- For exponential growth, the value at the midpoint time is the geometric mean of the endpoint values. - Form the ratio of the two measured masses to isolate the rate constant. - Use the measured mass at \(t=0\) in the final model.

Solution

1. Exponential values at equally spaced times satisfy \(Q(2)^2=Q(0)Q(4)=36\). Since mass is positive, \(Q(2)=6\,\text{g}\). 2. The ratio \(\frac{Q(4)}{Q(0)}=e^{4k}=9\), so \(k=\frac{\ln9}{4}=\frac{\ln3}{2}\,\text{h}^{-1}\). 3. Using the initial mass, \(Q(t)=2e^{(\ln3)t/2}\) grams.

Answer

a) \(Q(2)=6\,\text{g}\) b) \(k=\frac{\ln3}{2}\,\text{h}^{-1}\) c) \(Q(t)=2e^{(\ln3)t/2}\) grams
54357412
A rectangular digital image has width \(A(t)\) and height \(B(t)\), measured in centimeters. The dimensions change exponentially according to \(A'=aA\) and \(B'=bB\), where \(t\) is measured in hours. Over every \(4\)-hour interval, the aspect ratio \(A/B\) is multiplied by \(16\), while the area \(AB\) is multiplied by \(\frac14\). Determine \(a\) and \(b\), and state their units.

Hints

- Express the aspect ratio and area as exponential quantities. - Translate each four-hour multiplier into an equation involving \(a\) and \(b\). - Solve the resulting pair of linear equations and attach reciprocal-hour units.

Solution

1. The aspect ratio changes by \(\frac{A(t+4)/B(t+4)}{A(t)/B(t)}=e^{4(a-b)}=16\), so \(a-b=\ln 2\). 2. The area changes by \(\frac{A(t+4)B(t+4)}{A(t)B(t)}=e^{4(a+b)}=\frac14\), so \(a+b=-\frac12\ln 2\). 3. Solving the two equations gives \(a=\frac14\ln 2\,\text{h}^{-1}\) and \(b=-\frac34\ln 2\,\text{h}^{-1}\).

Answer

\(a=\frac{\ln 2}{4}\,\text{h}^{-1}\) and \(b=-\frac{3\ln 2}{4}\,\text{h}^{-1}\).
54358112
A bacteria culture contains \(7\) thousand cells at \(t=0\) hours and is modeled by \(Q'=kQ\). The tangent lines to the population graph at \(t=0\) and \(t=2\) have the same slope. Elena divides the equation \(7k=7ke^{2k}\) by \(7k\), obtains \(e^{2k}=1\), and concludes that the model has no possible value of \(k\). Identify her error, then determine \(k\) and \(Q(t)\).

Hints

- Write the exponential solution determined by the initial value. - When an equation contains a factor of \(k\), check \(k=0\) before dividing by it. - For real \(k\), \(e^{2k}=1\) only when \(k=0\).

Solution

1. The model has the form \(Q(t)=7e^{kt}\), so the two tangent slopes are \(Q'(0)=7k\) and \(Q'(2)=7ke^{2k}\). 2. Equal slopes give \(7k=7ke^{2k}\). Elena divided by \(k\) without first considering \(k=0\), so her division could discard a valid case. 3. If \(k\ne0\), division gives \(e^{2k}=1\), which implies \(k=0\), contradicting \(k\ne0\). Thus no nonzero value works. 4. For \(k=0\), both tangent slopes are \(0\), and \(Q(t)=7e^0=7\).

Answer

Elena's error is dividing by \(k\) before checking whether \(k=0\). The only possible value is \(k=0\), and \(Q(t)=7\) thousand cells.
54362112
A medication amount \(Q(t)\), measured in milligrams, decreases exponentially according to \(Q'=kQ\), where \(t\) is measured in hours. The geometric mean of the amounts measured at \(t=0\) and \(t=4\) is \(9\) milligrams. At \(t=2\), the amount is decreasing at \(6\) milligrams per hour. Determine \(Q(t)\) and the medication's half-life.

Hints

- Use the exponential midpoint property to find the amount at \(t=2\). - Apply \(Q'=kQ\) where both the amount and its rate are known. - Set the exponential decay factor equal to \(1/2\) to find the half-life.

Solution

1. For an exponential model, the midpoint value is the geometric mean of values at equally spaced times. Therefore, \(Q(2)=9\) milligrams. 2. From \(Q'(2)=kQ(2)\), \(-6=9k\), so \(k=-\frac23\) per hour. 3. Centering at \(t=2\) gives \(Q(t)=9e^{-\frac23(t-2)}\) milligrams. 4. The half-life \(T\) satisfies \(e^{-2T/3}=1/2\), so \(T=\frac32\ln2\) hours.

Answer

\(Q(t)=9e^{-\frac23(t-2)}\) milligrams, with half-life \(\frac32\ln2\approx1.04\) hours.
54362912
A bacteria culture grows exponentially. Its tripling time is \(5\) hours longer than its doubling time. Determine its continuous growth rate \(k\), including units, and express its tenfold-growth time in exact form.

Hints

- Write the time needed to reach a general multiplicative factor. - Translate the difference between tripling and doubling times into an equation for \(k\). - Use the same rate constant for the tenfold factor.

Solution

1. For rate constant \(k>0\), the doubling and tripling times are \(\frac{\ln2}{k}\) hours and \(\frac{\ln3}{k}\) hours. 2. Their difference is \(\frac{\ln(3/2)}{k}=5\), so \(k=\frac{\ln(3/2)}{5}\) per hour. 3. The tenfold-growth time is \(\frac{\ln10}{k}=\frac{5\ln10}{\ln(3/2)}\) hours.

Answer

\(k=\frac{\ln(3/2)}{5}\) per hour, and the tenfold-growth time is \(\frac{5\ln10}{\ln(3/2)}\) hours.
54364012
A medication amount \(Q(t)\), measured in milligrams, decreases exponentially according to \(Q'=kQ\), where \(t\) is measured in hours. Its elimination rate \(Q'(t)\) is multiplied by \(\frac14\) over every \(6\)-hour interval, and \(Q'(2)=-10\) milligrams per hour. Determine \(k\), \(Q(0)\), and \(Q(8)\), including units.

Hints

- The derivative of an exponential model has the same time multiplier as the model. - Use the differential equation to recover the amount at \(t=2\). - Apply the exponential multiplier forward and backward over the stated time differences.

Solution

1. The derivative has the same exponential time factor as the amount, so \(e^{6k}=\frac14\). Therefore, \(k=-\frac{\ln2}{3}\) per hour. 2. From \(Q'(2)=kQ(2)\), \(Q(2)=\frac{-10}{-\ln2/3}=\frac{30}{\ln2}\) milligrams. 3. Moving back \(2\) hours multiplies the amount by \(e^{-2k}=2^{2/3}\), so \(Q(0)=\frac{30\cdot2^{2/3}}{\ln2}\) milligrams. 4. Moving forward \(6\) hours multiplies the amount by \(\frac14\), so \(Q(8)=\frac{15}{2\ln2}\) milligrams.

Answer

\(k=-\frac{\ln2}{3}\) per hour, \(Q(0)=\frac{30\cdot2^{2/3}}{\ln2}\) milligrams, and \(Q(8)=\frac{15}{2\ln2}\) milligrams.
54367012
A medication amount is claimed to follow the exact exponential model \(Q'=kQ\). A lab reports \(Q(0)=8\) milligrams, \(Q'(0)=-4\) milligrams per hour, and \(Q(2)=4\) milligrams. Determine whether all three measurements can belong to one exact exponential model. If not, find the model implied by the first two measurements and the value it predicts at \(t=2\).

Hints

- Use the initial value and initial derivative in \(Q'=kQ\). - Build the model before checking the later measurement. - Compare the rate constant implied by the endpoint values with the one implied by the initial derivative.

Solution

1. The first two measurements give \(-4=k(8)\), so \(k=-1/2\) per hour. 2. The model implied by those measurements is \(Q(t)=8e^{-t/2}\) milligrams. 3. It predicts \(Q(2)=8e^{-1}=8/e\approx2.94\) milligrams, not \(4\) milligrams. 4. Equivalently, the values \(Q(0)=8\) and \(Q(2)=4\) would imply \(k=-\ln2/2\), which differs from \(-1/2\). Thus the three exact measurements are inconsistent.

Answer

No single exact exponential model fits all three measurements. The first two imply \(Q(t)=8e^{-t/2}\) milligrams, which predicts \(Q(2)=8/e\approx2.94\) milligrams.
54367712
A solar array's power output \(Q(t)\), measured in kilowatts, increases exponentially according to \(Q'=kQ\), where \(t\) is measured in hours. The output rises from \(3\) kilowatts at \(t=0\) to \(12\) kilowatts at time \(T\). During that interval, the array produces \(18\) kilowatt-hours of energy: \(\int_0^T Q(t)\,dt=18\). Determine \(k\) and \(T\), including units.

Hints

- Interpret the integral of power over time as energy. - Express the definite integral using the two endpoint power values and \(k\). - Use the endpoint ratio after finding \(k\).

Solution

1. For \(Q'=kQ\), an antiderivative of \(Q\) is \(Q/k\). Therefore, \(\int_0^TQ(t)\,dt=\frac{Q(T)-Q(0)}{k}\). 2. Thus \(18=\frac{12-3}{k}\), so \(k=1/2\) per hour. 3. The endpoint ratio gives \(e^{kT}=12/3=4\). 4. Therefore, \(T=\frac{\ln4}{1/2}=4\ln2\) hours.

Answer

\(k=\frac12\) per hour, and \(T=4\ln2\approx2.77\) hours.
54369712
The area \(Q(t)\) of a mold colony, measured in square centimeters, grows exponentially according to \(Q'=kQ\), where \(t\) is measured in days. The colony's area is increasing at \(5\) square centimeters per day at \(t=1\) and at \(40\) square centimeters per day at \(t=4\). Determine \(k\), \(Q(0)\), and \(Q(t)\), including units.

Hints

- Compare the two area-growth rates over their time difference. - Use \(Q'=kQ\) to recover the colony area where a growth rate is known. - Move backward one day to find the initial area.

Solution

1. The derivative has the same exponential rate as the area, so \(Q'(4)/Q'(1)=e^{3k}=8\). 2. Thus \(k=\ln2\) per day. 3. From \(Q'(1)=kQ(1)\), \(Q(1)=5/\ln2\) square centimeters. 4. Moving back one day gives \(Q(0)=Q(1)e^{-k}=5/(2\ln2)\) square centimeters. 5. Therefore, \(Q(t)=\frac{5}{2\ln2}2^t\) square centimeters.

Answer

\(k=\ln2\) per day, \(Q(0)=\frac{5}{2\ln2}\) square centimeters, and \(Q(t)=\frac{5}{2\ln2}2^t\) square centimeters.
54359212
An investment account has no deposits or withdrawals and follows the exponential model \(Q'=kQ\), with \(Q(0)>0\). Its average dollar growth rate from \(t=0\) to \(t=2\) years equals its instantaneous dollar growth rate at \(t=1\). Prove that the balance must be constant. Use exponential functions only; do not use hyperbolic functions.

Hints

- Write the balance as \(Ae^{kt}\) and form the two rates directly. - After canceling the positive centered value \(Ae^k\), move all terms to one side. - Differentiate the resulting function and use \(e^k+e^{-k}\ge2\).

Solution

1. Write \(Q(t)=Ae^{kt}\), where \(A>0\). 2. The average rate on \([0,2]\) is \(\frac{Ae^{2k}-A}{2}\), and the instantaneous rate at \(t=1\) is \(kAe^k\). 3. Equating the rates and dividing by the positive number \(Ae^k\) gives \(e^k-e^{-k}=2k\). 4. Let \(g(k)=e^k-e^{-k}-2k\). Then \(g(0)=0\) and \(g'(k)=e^k+e^{-k}-2\). 5. Since \(e^k e^{-k}=1\), the arithmetic-geometric mean inequality gives \(e^k+e^{-k}\ge2\), with equality only when \(k=0\). Thus \(g\) is increasing and can equal \(0\) only at \(k=0\). 6. Therefore, \(Q(t)=A\) is constant.

Answer

The rate constant must be \(k=0\), so the account balance is \(Q(t)=A\) for some positive constant \(A\).

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