Exponential Models with Differential Equations - AP Calculus AB

A savings balance $B(t)$ satisfies $\frac{dB}{dt}=0.07B$ . Which function solves for $B(t)$ ?
Answer: $B(t)=Ce^{0.07t}$ This problem involves solving an exponential differential equation representing compound interest growth. When we have $\frac{dB}{dt}=0.07B$ , the savings balance grows at a rate proportional to its current amount with rate 0.07. This represents continuous compound interest where larger balances earn proportionally more interest. The differential equation $\frac{dy}{dt} = ky$ has the general solution $y = Ce^{kt}$ , so with $k = 0.07$ , we get $B(t) = Ce^{0.07t}$ . Choice A would represent simple interest (linear growth), but continuous compounding requires exponential growth. To identify compound interest: when growth rate is proportional to current balance ( $\frac{dB}{dt} = rB$ ), the solution is always exponential ( $Ce^{rt}$ ).
A population grows at a rate proportional to its size: $\frac{dN}{dt}=kN$ with $k>0$ . Which must be true?
Answer: $N(t)=Ce^{kt}$ This problem presents the fundamental exponential growth differential equation dN/dt = kN with k > 0. When the growth rate is proportional to the current population size, we get exponential behavior described by N(t) = $Ce^{kt}$ . The positive constant k ensures the population grows over time, as $e^{kt}$ increases when k > 0 and t increases. This proportional relationship is what distinguishes exponential growth from other types—the larger the population, the faster it grows. Choice E (N(t) = $Ck^t$ ) might seem plausible but represents a different exponential base (k instead of e), which doesn't solve our differential equation. Remember: differential equations of the form dy/dt = ky always have solutions y = $Ce^{kt}$ , where e is the natural exponential base.
A culture's growth rate is always $7\%$ of its current size, so $\frac{dP}{dt}=0.07P$ . Which is the general solution?
Answer: $P(t)=Ce^{0.07t}$ This problem explicitly states that growth rate is a percentage of current size, the defining characteristic of exponential growth. The equation dP/dt = 0.07P means the population grows at 7% of its current value per unit time. The general solution to this exponential differential equation is P(t) = $Ce^{0.07t}$ , where C represents the initial population size. This captures how larger populations grow faster in absolute terms while maintaining the same relative growth rate. Choice A (0.07t + C) would mean the population grows by a constant 0.07 units per time period, not 7% of its current size. The key recognition pattern: when a rate is described as a percentage or proportion of the current value, you have an exponential differential equation.
A quantity $y$ satisfies $\frac{dy}{dx}=y$ . Which is the general solution for $y(x)$ ?
Answer: $y=Ce^{x}$ This problem involves solving the fundamental exponential differential equation. When we have $\frac{dy}{dx}=y$ , the quantity grows at a rate equal to its current value, which is the most basic form of exponential growth. This means the derivative equals the function itself, a defining property of the exponential function. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$ , and with $k = 1$ , we get $y = Ce^{x}$ . Choice A represents addition of an exponential term rather than multiplication by an exponential factor. To recognize the natural exponential: when $\frac{dy}{dx} = y$ , the solution is always $Ce^{x}$ , the natural exponential function family.
A quantity $y$ satisfies $\frac{dy}{dx}=-\ln(2)\,y$ . Which is the general solution?
Answer: $y=Ce^{-(\ln 2)x}$ This problem involves solving an exponential differential equation representing exponential decay with rate ln(2). When we have $\frac{dy}{dx}=-\ln(2)y$ , the quantity decreases at a rate proportional to its current value with constant $-\ln(2)$ . The negative constant indicates exponential decay, and the ln(2) factor relates to half-life calculations in exponential decay. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$ , so with $k = -\ln(2)$ , we get $y = Ce^{-(\ln 2)x}$ . Choice D represents linear decay, but the proportional relationship requires exponential behavior. To handle negative logarithmic constants: $-\ln(2)$ produces exponential decay solutions with the corresponding negative exponent.
A bacteria culture satisfies $\frac{dP}{dt}=0.4P$ with $P(0)=200$ . Which function gives $P(t)$ ?
Answer: $P(t)=200e^{0.4t}$ This problem tests your understanding of exponential differential equations, where the rate of change is proportional to the current value. The equation dP/dt = 0.4P means the population grows at a rate equal to 0.4 times its current size, which is the hallmark of exponential growth. When we solve this type of equation, we get P(t) = $Ce^{0.4t}$ , and using the initial condition P(0) = 200, we find C = 200, giving us P(t) = $200e^{0.4t}$ . Choice A (200 + 0.4t) represents linear growth, not exponential, because it would mean dP/dt = 0.4 (constant rate), not 0.4P (proportional rate). To recognize exponential models, look for differential equations where dy/dt = ky—these always yield solutions of the form y = $Ce^{kt}$ .
A pollutant amount $Q$ satisfies $\frac{dQ}{dt}=kQ$ with $k<0$ . Which form must $Q(t)$ have?
Answer: $Q(t)=Ce^{kt}$ This question tests understanding of exponential differential equations with a negative growth constant. The equation dQ/dt = kQ with k < 0 describes pollutant decay where the removal rate is proportional to the current amount. The general solution is Q(t) = $Ce^{kt}$ , which represents exponential decay since k is negative. Writing k < 0 as k = -|k| makes the decay explicit: Q(t) = $Ce^{-|k|t}$ . Choice D ( $Ce^{-kt}$ ) might seem correct but would actually represent growth since -k would be positive when k is negative. The crucial insight: in exponential differential equations, the solution always takes the form y = $Ce^{kt}$ regardless of k's sign—the sign of k determines whether it's growth or decay.
A sample's mass $m(t)$ decays with $\frac{dm}{dt}=-2m$ . Which is the general solution?
Answer: $m(t)=Ce^{-2t}$ This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dm}{dt}=-2m$ , the mass decreases at a rate proportional to its current amount with proportionality constant -2. The negative constant indicates decay, as larger masses lose material proportionally faster. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , so with $k = -2$ , we get $m(t) = Ce^{-2t}$ . Choice B would represent exponential growth rather than decay since it has a positive exponent. To distinguish exponential growth from decay: negative coefficients in $\frac{dy}{dt} = ky$ produce solutions with negative exponents ( $Ce^{-|k|t}$ ).
A population satisfies $\frac{dP}{dt}=0.18P$ . Which is the general solution for $P(t)$ ?
Answer: $P(t)=Ce^{0.18t}$ This problem involves solving an exponential differential equation where the rate of change is proportional to the current value. When we have $\frac{dP}{dt}=0.18P$ , this means the population grows at a rate proportional to its current size with proportionality constant 0.18. This type of differential equation has solutions of the form $P(t) = Ce^{kt}$ where $k$ is the proportionality constant. Since our constant is positive (0.18), we get exponential growth with $P(t) = Ce^{0.18t}$ . Choice A represents linear growth, not exponential growth as required by the proportional relationship. To recognize exponential models: positive constants in $\frac{dy}{dt} = ky$ give growth ( $Ce^{kt}$ ), negative constants give decay ( $Ce^{-kt}$ ).
A cooling object satisfies $\frac{dT}{dt}=-0.25T$ . Which is the general solution for $T(t)$ ?
Answer: $T(t)=Ce^{-0.25t}$ This problem involves solving an exponential differential equation where the cooling rate is proportional to the current temperature. When we have $\frac{dT}{dt}=-0.25T$ , the temperature decreases at a rate proportional to its current value with constant -0.25. This negative proportionality constant indicates exponential decay, as higher temperatures cool proportionally faster. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , so with $k = -0.25$ , we get $T(t) = Ce^{-0.25t}$ . Choice D represents linear cooling, but Newton's law of cooling requires the proportional relationship that produces exponential decay. To recognize cooling/heating models: negative constants in proportional rate equations always produce exponential decay solutions.
A radioactive sample satisfies $\frac{dM}{dt}=-\lambda M$ for constant $\lambda>0$ . Which describes $M(t)$ ?
Answer: $M(t)=Ce^{-\lambda t}$ This question tests exponential decay in the context of radioactive decay, a classic application of exponential differential equations. The equation dM/dt = -λM tells us the decay rate is proportional to the current amount, with λ > 0 ensuring decay (negative rate). Solving this differential equation gives M(t) = $Ce^{-λt}$ , where C represents the initial amount of radioactive material. The exponential function with negative exponent captures how the material decreases over time, never quite reaching zero but approaching it asymptotically. Choice A (M₀ - λt) would represent linear decay with constant rate λ, not the proportional decay that characterizes radioactive materials. For any differential equation dy/dt = -ky with k > 0, the solution is y = $Ce^{-kt}$ , representing exponential decay.
A culture grows at a rate proportional to its size: $\frac{dN}{dt}=kN$ . Which is the solution form?
Answer: $N(t)=Ce^{kt}$ This problem involves solving an exponential differential equation representing proportional growth. When we have $\frac{dN}{dt}=kN$ , the culture grows at a rate proportional to its current size, which is the defining characteristic of exponential growth. Each individual organism contributes proportionally to the reproduction rate, making larger populations grow proportionally faster. The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , giving us $N(t) = Ce^{kt}$ . Choice A represents linear growth, which would occur if the growth rate were constant rather than proportional to population size. To recognize biological exponential growth: when reproduction rate depends on current population size, the mathematical model is always $\frac{dN}{dt} = kN$ with exponential solutions.
A quantity $y$ satisfies $\frac{dy}{dx}=-y$ . Which is the general solution for $y(x)$ ?
Answer: $y=Ce^{-x}$ This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dy}{dx}=-y$ , the quantity decreases at a rate equal to its current value, representing the most basic form of exponential decay. The negative sign indicates that the derivative is opposite to the current value, causing exponential decrease. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$ , so with $k = -1$ , we get $y = Ce^{-x}$ . Choice C represents linear decay, but the proportional relationship (rate equals current value) requires exponential behavior. To recognize natural exponential decay: when $\frac{dy}{dx} = -y$ , the solution is always $Ce^{-x}$ , representing decay at the natural rate.
A quantity $y$ satisfies $\frac{dy}{dx}=\frac{1}{4}y$ . Which is the general solution?
Answer: $y=Ce^{x/4}$ This problem involves solving an exponential differential equation where the rate of change is proportional to the current quantity. When we have $\frac{dy}{dx}=\frac{1}{4}y$ , the quantity grows at a rate proportional to its current value with proportionality constant $\frac{1}{4}$ . This positive constant indicates exponential growth, though at a slower rate than unit growth. The standard solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$ , so with $k = \frac{1}{4}$ , we get $y = Ce^{x/4}$ . Choice A represents linear growth, which would occur if the rate of change were constant rather than proportional to the current value. To recognize fractional exponential growth: any positive proportionality constant, including fractions, produces exponential growth solutions.
A quantity decreases proportionally to its amount: $\frac{dy}{dt}=-ky$ , $k>0$ . Which is the general solution?
Answer: $y=Ce^{-kt}$ This problem involves solving an exponential differential equation representing proportional decay. When we have $\frac{dy}{dt}=-ky$ with $k>0$ , the quantity decreases at a rate proportional to its current amount. The negative sign with positive $k$ ensures decay behavior, as larger quantities experience proportionally larger decreases. The general solution to $\frac{dy}{dt} = -ky$ is $y = Ce^{-kt}$ , representing exponential decay. Choice A would represent exponential growth since it lacks the negative exponent required for decay. To identify exponential decay: negative coefficients in proportional rate equations ( $\frac{dy}{dt} = -ky$ ) always produce decay solutions with negative exponents ( $Ce^{-kt}$ ).
A quantity $y$ satisfies $\frac{dy}{dx}=-\pi y$ . Which is the general solution for $y(x)$ ?
Answer: $y=Ce^{-\pi x}$ This problem involves solving an exponential differential equation representing exponential decay with rate π. When we have $\frac{dy}{dx}=-\pi y$ , the quantity decreases at a rate proportional to its current value with constant $-π$ . The negative constant indicates exponential decay at π times the current value per unit change. The general solution to $\frac{dy}{dx} = ky$ is $y = Ce^{kx}$ , so with $k = -π$ , we get $y = Ce^{-πx}$ . Choice B represents linear decay, but the proportional relationship requires exponential rather than linear behavior. To handle negative transcendental constants: $-π$ in the differential equation produces exponential decay with the negative exponent $-πx$ .
A sample satisfies $\frac{dM}{dt}=-\frac{1}{3}M$ . Which is the general solution for $M(t)$ ?
Answer: $M(t)=Ce^{-t/3}$ This problem involves solving an exponential differential equation representing exponential decay. When we have $\frac{dM}{dt}=-\frac{1}{3}M$ , the sample mass decreases at a rate proportional to its current amount with constant $-\frac{1}{3}$ . The negative fractional constant indicates exponential decay at one-third the rate of the current mass. The general solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , so with $k = -\frac{1}{3}$ , we get $M(t) = Ce^{-t/3}$ . Choice A would represent exponential growth rather than decay since it lacks the negative exponent. To handle fractional decay constants: negative fractions in proportional rate equations produce exponential decay solutions with the corresponding fractional negative exponent.
A drug amount $A(t)$ obeys $\frac{dA}{dt}=-0.6A$ . Which form must $A(t)$ take?
Answer: $A(t)=Ce^{-0.6t}$ This problem involves solving an exponential differential equation where the rate of change is proportional to the current amount. When we have $\frac{dA}{dt}=-0.6A$ , the drug amount decreases at a rate proportional to the current amount with proportionality constant -0.6. This negative constant indicates exponential decay, as each unit of drug causes a proportional decrease in the total. The general solution to $\frac{dy}{dt} = ky$ is always $y = Ce^{kt}$ , so with $k = -0.6$ , we get $A(t) = Ce^{-0.6t}$ . Choice B would represent linear decay, but the proportional relationship requires exponential behavior. To identify the correct exponential form: negative constants in $\frac{dy}{dt} = ky$ always produce decay solutions $Ce^{-|k|t}$ .
A concentration $C(t)$ satisfies $\frac{dC}{dt}=-0.1C$ . Which is the general solution form?
Answer: $C(t)=Ce^{-0.1t}$ This problem involves solving an exponential differential equation where the concentration decreases proportionally. When we have $\frac{dC}{dt}=-0.1C$ , the concentration decreases at a rate proportional to its current value with constant -0.1. This negative proportionality constant indicates exponential decay, typical of dilution or consumption processes. The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , so with $k = -0.1$ , we get $C(t) = Ce^{-0.1t}$ . Choice C represents linear decay, but concentration changes typically follow exponential patterns due to proportional consumption rates. To recognize concentration models: negative constants in proportional equations always yield exponential decay solutions.
A medication amount $A$ obeys $\frac{dA}{dt}=-0.2A$ . Which is the general solution for $A(t)$ ?
Answer: $A(t)=Ce^{-0.2t}$ This question involves solving an exponential differential equation where the medication amount decreases proportionally to its current value. The equation dA/dt = -0.2A indicates exponential decay because the rate is negative and proportional to A itself. Solving this differential equation yields A(t) = $Ce^{-0.2t}$ , where C is determined by initial conditions. The negative coefficient (-0.2) in the differential equation becomes the exponent in the exponential function, maintaining the decay behavior. Choice B (C - 0.2t) would represent linear decay with constant rate -0.2, not the proportional decay described by our differential equation. When you see dy/dt = -ky (with k > 0), the solution is always y = $Ce^{-kt}$ , representing exponential decay.

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