Exponential Growth and Decay Models - Calculus 2

1. A tank initially contains 1000 gallons of pure water. Salt water containing 2 pounds of salt per gallon flows in at 5 gallons per minute, and the well-mixed solution flows out at the same rate. The amount of salt $S(t)$ satisfies $\frac{dS}{dt} = 10 - \frac{S}{200}$. How much salt will be in the tank when the concentration reaches 1.5 pounds per gallon?
   Answer: $S = 1500$ pounds (concentration × volume) This is a classic mixing problem that tests your understanding of differential equations and equilibrium concepts. When you encounter these problems, focus on what the question is actually asking rather than getting caught up in solving the differential equation. The key insight is recognizing what "concentration reaches 1.5 pounds per gallon" means. Since the tank always contains 1000 gallons (inflow rate equals outflow rate), you can find the amount of salt directly: $S = \text{concentration} \times \text{volume} = 1.5 \times 1000 = 1500$ pounds. This straightforward calculation gives you the answer without needing to solve the differential equation. Let's examine why the other answers are wrong. Answer A (1200 pounds) represents 60% of the equilibrium value, which you'd get by incorrectly assuming the question asks for a percentage of maximum capacity. Answer C (1800 pounds) represents 90% of equilibrium—another percentage trap that doesn't relate to the actual concentration given. Answer D (2000 pounds) is the equilibrium value itself, which occurs when $\frac{dS}{dt} = 0$, meaning $10 = \frac{S}{200}$, so $S = 2000$. This would correspond to the final concentration of 2 pounds per gallon, not the 1.5 pounds per gallon specified. Strategy tip: In mixing problems, always identify what the question is asking before diving into complex calculations. Often, the relationship between concentration, volume, and total amount ($S = C \times V$) provides a direct path to the answer, especially when the volume remains constant.
2. A culture of bacteria grows at a rate proportional to the number of bacteria present. The number of bacteria triples in 6 hours. By what factor will the number of bacteria increase in 18 hours?
   Answer: 27 Let $P(t)$ be the number of bacteria. The model is $P(t) = P_0 e^{kt}$. We are given that $P(6) = 3P_0$. So, $3P_0 = P_0 e^{6k}$, which implies $e^{6k} = 3$. We want to find the factor of increase in 18 hours, which is $\frac{P(18)}{P_0}$. $P(18) = P_0 e^{18k} = P_0 e^{6k \cdot 3} = P_0 (e^{6k})^3$. Since we know $e^{6k} = 3$, we have $P(18) = P_0 (3)^3 = 27P_0$. Therefore, the number of bacteria will increase by a factor of 27.
3. Two populations, $P_1(t)$ and $P_2(t)$, grow exponentially with respective positive growth constants $k_1$ and $k_2$. Initially, at $t=0$, the size of the first population is twice the size of the second, i.e., $P_1(0) = 2P_2(0)$. After 5 years, the populations are equal. Which equation correctly relates the growth constants?
   Answer: $k_2 - k_1 = \frac{\ln(2)}{5}$ The population models are $P_1(t) = P_1(0)e^{k_1 t}$ and $P_2(t) = P_2(0)e^{k_2 t}$. Let $P_2(0) = A$, so $P_1(0) = 2A$. At $t=5$, we have $P_1(5) = P_2(5)$. Substituting the models gives $(2A)e^{5k_1} = (A)e^{5k_2}$. Dividing by $A$ (since $A>0$) gives $2e^{5k_1} = e^{5k_2}$. Taking the natural logarithm of both sides gives $\ln(2e^{5k_1}) = \ln(e^{5k_2})$, which simplifies to $\ln(2) + 5k_1 = 5k_2$. Rearranging the terms to relate $k_1$ and $k_2$ gives $5k_2 - 5k_1 = \ln(2)$, or $k_2 - k_1 = \frac{\ln(2)}{5}$.
4. A quantity $y(t)$ follows the law of exponential decay, $\frac{dy}{dt} = -ky$ for a constant $k>0$. Which of the following linear, homogeneous, second-order differential equations with constant coefficients is also satisfied by $y(t)$?
   Answer: $\frac{d^2y}{dt^2} + k \frac{dy}{dt} = 0$ Given $\frac{dy}{dt} = -ky$. We can differentiate this equation with respect to $t$ to find a second-order equation: $\frac{d}{dt}\left(\frac{dy}{dt}\right) = \frac{d}{dt}(-ky)$, which gives $\frac{d^2y}{dt^2} = -k \frac{dy}{dt}$. Rearranging gives $\frac{d^2y}{dt^2} + k \frac{dy}{dt} = 0$. This is a linear, homogeneous, second-order differential equation with constant coefficients. Choices A and C have incorrect forms, and choice D has the wrong sign.
5. The rate of change of a quantity $y$ is directly proportional to $y$. It is known that at time $t_1$, the value of the quantity is $a$, and at time $t_2$, its value is $b$. What is the value of the quantity at time $2t_2 - t_1$?
   Answer: $\frac{b^2}{a}$ The governing equation is $\frac{dy}{dt} = ky$, so $y(t) = y_0 e^{kt}$. We are given $y(t_1) = a = y_0 e^{kt_1}$ and $y(t_2) = b = y_0 e^{kt_2}$. Dividing these gives $\frac{b}{a} = \frac{y_0 e^{kt_2}}{y_0 e^{kt_1}} = e^{k(t_2 - t_1)}$. We want to find $y(2t_2 - t_1) = y_0 e^{k(2t_2 - t_1)}$. We can rewrite the exponent as $k(t_2 + (t_2 - t_1))$. So, $y(2t_2-t_1) = y_0 e^{kt_2} e^{k(t_2 - t_1)}$. We know $y_0 e^{kt_2} = y(t_2) = b$ and $e^{k(t_2 - t_1)} = \frac{b}{a}$. Therefore, $y(2t_2 - t_1) = b \cdot \frac{b}{a} = \frac{b^2}{a}$.
6. Newton's Law of Cooling states that $\frac{dT}{dt} = -k(T - T_{amb})$, where $T$ is temperature, $T_{amb}$ is ambient temperature, and $k > 0$. A cup of coffee at 90°C is placed in a 20°C room. After 10 minutes, the temperature is 70°C. What will the temperature be after 20 minutes?
   Answer: $T(20) = 20 + 50 \cdot (\frac{5}{7})^2 = 20 + \frac{1250}{49} \approx 45.5°C$ Newton's Law of Cooling problems require solving a separable differential equation. When you see $\frac{dT}{dt} = -k(T - T_{amb})$, recognize this as exponential decay toward the ambient temperature. Start by separating variables and integrating: $\frac{dT}{T - T_{amb}} = -k \, dt$. This gives us $T(t) = T_{amb} + (T_0 - T_{amb})e^{-kt}$, where $T_0$ is the initial temperature. With $T_{amb} = 20°C$ and $T_0 = 90°C$, we have $T(t) = 20 + 70e^{-kt}$. Using the condition that $T(10) = 70°C$: $70 = 20 + 70e^{-10k}$ $50 = 70e^{-10k}$ $e^{-10k} = \frac{5}{7}$ For $t = 20$ minutes: $T(20) = 20 + 70e^{-20k} = 20 + 70(e^{-10k})^2 = 20 + 70(\frac{5}{7})^2 = 20 + 70 \cdot \frac{25}{49} = 20 + \frac{1250}{49} \approx 45.5°C$. This matches answer C. Answer A incorrectly uses $70 \cdot e^{-2k \cdot 10}$ instead of the proper exponential form with the calculated decay constant. Answer B assumes linear cooling, but Newton's Law describes exponential decay, not linear change. Answer D uses the correct exponential structure but incorrectly substitutes $\frac{50}{70}$ directly for $e^{-10k}$, missing that we need $e^{-10k} = \frac{5}{7}$. Study tip: Always find the decay constant using given conditions before plugging in your target time. Don't assume linear relationships in exponential decay problems.
7. A cup of coffee at 90°C is placed in a room. After 10 minutes, its temperature is 70°C. After another 10 minutes, its temperature is 55°C. Assuming the coffee cools according to Newton's Law of Cooling, what is the ambient temperature of the room?
   Answer: 10°C Let $T(t)$ be the temperature and $T_s$ be the room temperature. Newton's Law of Cooling is $\frac{dT}{dt} = k(T - T_s)$, with solution $T(t) = T_s + (T_0 - T_s)e^{kt}$. We have $T(0)=90$, $T(10)=70$, and $T(20)=55$. From these, we get the equations: $70 = T_s + (90-T_s)e^{10k}$ and $55 = T_s + (90-T_s)e^{20k}$. Let $x = e^{10k}$. The equations become $70 - T_s = (90-T_s)x$ and $55 - T_s = (90-T_s)x^2$. From the first equation, $x = \frac{70-T_s}{90-T_s}$. Substituting this into the second equation gives $55 - T_s = (90 - T_s) \left(\frac{70-T_s}{90-T_s}\right)^2$, which simplifies to $(55-T_s)(90-T_s) = (70-T_s)^2$. Expanding both sides yields $4950 - 145T_s + T_s^2 = 4900 - 140T_s + T_s^2$. Solving for $T_s$ gives $50 = 5T_s$, so $T_s = 10$.
8. An object is cooling in an environment with an ambient temperature of 20°C. At 2:00 PM, its temperature is 60°C. At 3:00 PM, its temperature is 40°C. What was the object's temperature at 1:00 PM?
   Answer: 100°C When you encounter a cooling problem like this, you're dealing with Newton's Law of Cooling, which states that temperature follows an exponential decay model: $T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}})e^{-kt}$ Let's set up the problem systematically. The ambient temperature is 20°C, so our equation becomes $T(t) = 20 + (T_0 - 20)e^{-kt}$, where $T_0$ is the initial temperature and $k$ is the cooling constant. Using the given data points: at $t = 1$ hour (3:00 PM), $T = 40°C$, and at $t = 0$ (2:00 PM), $T = 60°C$. From the 2:00 PM data: $60 = 20 + (T_0 - 20)e^0$, which gives us $T_0 - 20 = 40$. From the 3:00 PM data: $40 = 20 + 40e^{-k}$, so $e^{-k} = 0.5$. To find the temperature at 1:00 PM ($t = -1$), we need $e^{k} = 2$. Therefore: $T(-1) = 20 + 40e^{k} = 20 + 40(2) = 100°C$. Choice A (80°C) assumes linear cooling rather than exponential. Choice B (90°C) might come from averaging incorrectly. Choice C (120°C) results from doubling the temperature difference without accounting for the exponential nature. The correct answer is D (100°C). Remember: cooling problems always involve exponential decay, not linear relationships. When temperatures are given at regular intervals, use the exponential pattern to work backward or forward in time.
9. A 100-liter tank initially contains pure water. A salt solution with a concentration of 0.5 kg/L is added at a rate of 4 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate of 4 L/min. How many minutes does it take for the concentration of salt in the tank to reach 0.25 kg/L?
   Answer: $25 \ln(2)$ Let $A(t)$ be the amount of salt in kg. The volume is constant at $V=100$ L. The rate of change is $\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})$. Rate in = $(0.5 \text{ kg/L}) \times (4 \text{ L/min}) = 2$ kg/min. Rate out = $(\frac{A(t)}{100} \text{ kg/L}) \times (4 \text{ L/min}) = \frac{A}{25}$. The DE is $\frac{dA}{dt} = 2 - \frac{A}{25}$. The solution to this linear DE with $A(0)=0$ is $A(t) = 50(1 - e^{-t/25})$. We want the time $t$ when the concentration is 0.25 kg/L, which means $A(t) = 0.25 \times 100 = 25$ kg. So, $25 = 50(1 - e^{-t/25}) \Rightarrow 0.5 = 1 - e^{-t/25} \Rightarrow e^{-t/25} = 0.5$. Taking the natural log, $-t/25 = \ln(0.5) = -\ln(2)$. Thus, $t = 25 \ln(2)$.
10. The value of a machine, $V(t)$ in dollars, depreciates according to the differential equation $\frac{dV}{dt} = -0.15V$, where $t$ is in years. The initial value is $V(0) = \$20,000$. What is the correct interpretation of the quantity $\frac{1}{5}\int_0^5 V(t) dt$?
    Answer: The average value of the machine over the first 5 years. The expression $\frac{1}{b-a}\int_a^b f(x) dx$ is the definition of the average value of a function $f(x)$ on the interval $[a, b]$. In this context, with $V(t)$ representing the value of the machine at time $t$, the expression $\frac{1}{5-0}\int_0^5 V(t) dt$ represents the average value of the machine over the time interval from $t=0$ to $t=5$ years. The other choices represent different quantities: total depreciation is $V(0)-V(5)$, instantaneous rate is $V'(5)$, and value at 5 years is $V(5)$.
11. A bank account with continuous compounding at an annual rate $r$ is modeled by $\frac{dB}{dt} = rB$. If an initial deposit of $P$ quadruples in 18 years, in how many years will the initial deposit triple?
    Answer: $18 \frac{\ln(3)}{\ln(4)}$ years When you encounter differential equations modeling exponential growth like $\frac{dB}{dt} = rB$, you're dealing with continuous compounding problems that follow the pattern $B(t) = Pe^{rt}$. The key insight is using given information to find relationships between different time scenarios. Since the account quadruples in 18 years, we have $4P = Pe^{18r}$, which simplifies to $4 = e^{18r}$. Taking the natural logarithm: $\ln(4) = 18r$, so $r = \frac{\ln(4)}{18}$. Now we need to find when the account triples: $3P = Pe^{rt}$, giving us $3 = e^{rt}$. Taking the natural logarithm: $\ln(3) = rt = t \cdot \frac{\ln(4)}{18}$. Solving for t: $t = \frac{18\ln(3)}{\ln(4)}$. Looking at the wrong answers: Choice A (9 years) incorrectly assumes tripling takes half the time of quadrupling, which only works for linear growth, not exponential. Choice B (12 years) might come from incorrectly thinking tripling takes $\frac{3}{4}$ the time of quadrupling ($18 \times \frac{3}{4} = 13.5$, rounded to 12). Choice C has the logarithms inverted—this would give the wrong relationship between the growth rates. The correct answer is D: $18 \frac{\ln(3)}{\ln(4)}$ years. Study tip: In exponential growth problems, always use the given condition to find the growth rate first, then apply it to the desired scenario. Remember that time ratios in exponential growth involve logarithms, not simple proportions.
12. An account balance $B(t)$ earns interest at a rate proportional to the current balance, with an annual proportionality constant $r$. Simultaneously, continuous withdrawals are made at a rate proportional to the square root of the balance, with a proportionality constant $w$. Which differential equation correctly models the balance $B(t)$ over time?
    Answer: $\frac{dB}{dt} = rB - w\sqrt{B}$ The differential equation is built by considering the rates of increase and decrease of the balance. The phrase 'earns interest at a rate proportional to the current balance' translates to an interest rate term of $+rB$. The phrase 'continuous withdrawals are made at a rate proportional to the square root of the balance' translates to a withdrawal term of $-w\sqrt{B}$. Combining these two terms gives the net rate of change of the balance: $\frac{dB}{dt} = rB - w\sqrt{B}$.
13. The rate of formation of a chemical product is proportional to the amount of unreacted substrate remaining. Initially, there are 100 grams of substrate. After 1 hour, 50 grams remain. What is the amount of substrate remaining after a total of 3 hours?
    Answer: 12.5 grams Let $S(t)$ be the amount of substrate remaining. The problem states $\frac{dS}{dt} = -kS$ for some $k>0$. We are given $S(0) = 100$ and $S(1) = 50$. Since the amount was reduced by half in 1 hour, the half-life is 1 hour. The amount of substrate remaining after $t$ hours is given by $S(t) = 100 (1/2)^{t/1} = 100 (1/2)^t$. We want to find the amount remaining after 3 hours, which is $S(3) = 100 (1/2)^3 = 100 (1/8) = 12.5$ grams.
14. A 1000-liter tank initially contains 500 liters of brine with 50 kg of dissolved salt. Brine containing 0.2 kg of salt per liter is pumped into the tank at a rate of 10 L/min. The well-mixed solution is pumped out at a rate of 5 L/min. Which differential equation models the amount of salt $A(t)$ in kilograms in the tank at time $t$ in minutes?
    Answer: $\frac{dA}{dt} = 2 - \frac{A}{100+t}$ The rate of change of salt is $\frac{dA}{dt} = (\text{rate in}) - (\text{rate out})$. The rate of salt flowing in is $(0.2 \text{ kg/L}) \times (10 \text{ L/min}) = 2 \text{ kg/min}$. The volume of the solution in the tank changes over time. Initially $V(0)=500$ L. Since liquid flows in at 10 L/min and out at 5 L/min, the net change is +5 L/min. So, $V(t) = 500 + 5t$. The concentration of salt at time $t$ is $C(t) = \frac{A(t)}{V(t)} = \frac{A(t)}{500+5t}$ kg/L. The rate of salt flowing out is $C(t) \times (\text{outflow rate}) = \frac{A(t)}{500+5t} \times 5 = \frac{5A(t)}{500+5t} = \frac{A(t)}{100+t}$. Combining these gives $\frac{dA}{dt} = 2 - \frac{A}{100+t}$.
15. Carbon-14 dating is used to determine the age of an ancient artifact. The decay model is $\frac{dC}{dt} = -\lambda C$ with half-life 5730 years. If the artifact currently contains 25% of the original carbon-14, approximately how old is the artifact?
    Answer: 11,460 years (exactly 2 half-lives) If 25% remains, then 75% has decayed. Since $C(t) = C_0 e^{-\lambda t}$, we have $0.25C_0 = C_0 e^{-\lambda t}$, so $0.25 = e^{-\lambda t}$. Taking ln: $\ln(0.25) = -\lambda t$, giving $t = \frac{\ln(4)}{\lambda}$. Since $\lambda = \frac{\ln(2)}{5730}$, we get $t = \frac{\ln(4) \cdot 5730}{\ln(2)} = \frac{2\ln(2) \cdot 5730}{\ln(2)} = 11,460$ years. This is exactly 2 half-lives since $(1/2)^2 = 1/4 = 0.25$. Choice B uses the correct formula but miscalculates. Choice C assumes 2.5 half-lives. Choice D assumes 4 half-lives.
16. A hot object is placed in a room with a constant ambient temperature. According to Newton's Law of Cooling, the temperature difference between the object and the room decreases by 50% in the first 10 minutes. By what percentage does this temperature difference decrease over the first 20 minutes?
    Answer: 75% Let $D(t) = T(t) - T_s$ be the temperature difference. It follows the model $\frac{dD}{dt} = kD$, with solution $D(t) = D_0 e^{kt}$. A 50% decrease in 10 minutes means $D(10) = 0.5 D_0$. This is the definition of half-life, so the half-life of the temperature difference is 10 minutes. After another 10 minutes (for a total of 20 minutes), another half-life has passed. The difference will be halved again: $D(20) = 0.5 D(10) = 0.5(0.5 D_0) = 0.25 D_0$. If 25% of the difference remains, the total decrease is $100$.
17. A chemical reaction follows first-order kinetics: $\frac{d[A]}{dt} = -k[A]$, where $[A]$ is the concentration of reactant A. If the reaction is 80% complete after 45 minutes, what is the half-life of the reaction?
    Answer: $t_{1/2} = \frac{45\ln(2)}{\ln(5)} \approx 19.4$ minutes When you encounter first-order kinetics problems, you're dealing with exponential decay where the rate of change is proportional to the current amount. The key is solving the differential equation and connecting different time points. Starting with $\frac{d[A]}{dt} = -k[A]$, separation of variables gives us $\frac{d[A]}{[A]} = -k \, dt$. Integrating both sides yields $\ln[A] = -kt + C$, or $[A] = [A]_0 e^{-kt}$ where $[A]_0$ is the initial concentration. If the reaction is 80% complete after 45 minutes, then 20% of the original reactant remains: $[A] = 0.2[A]_0$. Substituting into our equation: $0.2[A]_0 = [A]_0 e^{-45k}$, which simplifies to $0.2 = e^{-45k}$. Taking the natural logarithm: $\ln(0.2) = -45k$, so $k = -\frac{\ln(0.2)}{45} = \frac{\ln(5)}{45}$. For half-life, we need the time when $[A] = 0.5[A]_0$: $0.5 = e^{-kt_{1/2}}$. This gives us $\ln(0.5) = -kt_{1/2}$, so $t_{1/2} = \frac{\ln(2)}{k} = \frac{45\ln(2)}{\ln(5)}$, which is answer D. A incorrectly multiplies by $\ln(2)$ instead of dividing by the rate constant. B naively assumes linear decay rather than exponential. C uses $\ln(0.2)$ in the denominator but forgets that $\ln(0.2) = -\ln(5)$, missing the sign change. Remember: for first-order kinetics, always solve for the rate constant first using given information, then apply it to find other time-related quantities.
18. The population $P(t)$ of a certain species grows in such a way that the rate of change of its relative growth rate is proportional to the current population. The relative growth rate is defined as $\frac{1}{P}\frac{dP}{dt}$. Which differential equation correctly models this scenario, where $c$ is a constant of proportionality?
    Answer: $\frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right) = cP$ Let's translate the statement piece by piece. The 'relative growth rate' is given as $R = \frac{1}{P}\frac{dP}{dt}$. The 'rate of change of its relative growth rate' is the derivative of $R$ with respect to time, $\frac{dR}{dt} = \frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right)$. The statement says this quantity is 'proportional to the current population', which means it equals $cP$ for some constant $c$. Therefore, the correct differential equation is $\frac{d}{dt}\left( \frac{1}{P}\frac{dP}{dt} \right) = cP$.
19. A quantity $y$ grows exponentially according to $y(t) = y_0 e^{kt}$. It is observed that the quantity doubles from its initial value at time $t=T$, and it triples from its initial value at time $t=T+5$. What is the value of the doubling time $T$?
    Answer: $5 \frac{\ln(2)}{\ln(3) - \ln(2)}$ We are given two conditions: $y(T) = 2y_0$ and $y(T+5) = 3y_0$. Using the solution form, these become $y_0 e^{kT} = 2y_0$ and $y_0 e^{k(T+5)} = 3y_0$. The first equation simplifies to $e^{kT} = 2$. The second becomes $e^{kT}e^{5k} = 3$. Substituting $e^{kT}=2$ into the second simplified equation gives $2e^{5k} = 3$, so $e^{5k} = 1.5$. From $e^{kT}=2$, we have $kT = \ln(2)$, so $k = \frac{\ln(2)}{T}$. From $e^{5k} = 1.5$, we have $5k = \ln(1.5)$, so $k = \frac{\ln(1.5)}{5}$. Equating the two expressions for $k$ gives $\frac{\ln(2)}{T} = \frac{\ln(1.5)}{5}$. Solving for $T$ yields $T = 5 \frac{\ln(2)}{\ln(1.5)} = 5 \frac{\ln(2)}{\ln(3/2)} = 5 \frac{\ln(2)}{\ln(3) - \ln(2)}$.
20. The concentration of a drug in the bloodstream, $C(t)$, decays exponentially after an injection. The initial concentration is 1.2 mg/L. After 4 hours, the concentration is 0.3 mg/L. To be effective, the concentration must be at least 0.1 mg/L. For how many total hours after the injection is the drug effective?
    Answer: $4 + 2\log_2(3)$ The model is $C(t) = C_0 e^{kt}$ with $C_0 = 1.2$. We are given $C(4) = 0.3$. So, $0.3 = 1.2 e^{4k}$, which gives $e^{4k} = 0.3/1.2 = 1/4$. Taking the natural log, $4k = \ln(1/4) = -2\ln(2)$, so $k = -\frac{\ln(2)}{2}$. This implies a half-life of 2 hours. We need to find the time $t$ for which $C(t) = 0.1$. So, $0.1 = 1.2 e^{kt}$, which gives $e^{kt} = \frac{0.1}{1.2} = \frac{1}{12}$. Taking the log, $kt = \ln(1/12) = -\ln(12)$. Substituting for $k$, $(-\frac{\ln(2)}{2})t = -\ln(12)$. Solving for $t$ gives $t = \frac{2\ln(12)}{\ln(2)} = 2\log_2(12)$. We can simplify this: $t = 2\log_2(4 \cdot 3) = 2(\log_2(4) + \log_2(3)) = 2(2 + \log_2(3)) = 4 + 2\log_2(3)$.

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