Exponential Models with Differential Equations - AP Calculus BC

A quantity satisfies $\frac{dy}{dt}=\frac{2}{3}y$ with $y(0)=27$ . Which is $y(t)$ ?
Answer: $27e^{2t/3}$ This problem demonstrates exponential modeling with a fractional growth rate. The differential equation dy/dt = (2/3)y with y(0) = 27 represents moderate exponential growth with fractional rate constant 2/3. The solution y(t) = $27e^{(2/3)t}$ shows quantity increasing exponentially from initial value 27, with the fractional coefficient creating steady but not rapid growth. This models systems with moderate positive feedback or growth under constrained resources. Choice C incorrectly reciprocates the fractional rate, which would create much faster growth inconsistent with the original differential equation. For fractional growth rates, maintain the fraction as the coefficient in the exponent of the exponential function.
An investment satisfies $\frac{dV}{dt}=0.12V$ with $V(0)=500$ . Find $V(1)$ .
Answer: $500e^{0.12}$ This problem requires evaluating an exponential investment model at a specific time. From dV/dt = 0.12V with V(0) = 500, we establish V(t) = $500e^{0.12t}$ as the continuous compound interest formula. To find V(1), we substitute t = 1: V(1) = $500e^{0.12×1}$ = $500e^{0.12}$ . This represents the account value after one year of continuous compounding at 12% annual rate. Choice C incorrectly reciprocates the rate in the exponent, which has no basis in the differential equation solution. For investment problems with continuous compounding, always use the form Principal× $e^{rate×time}$ for evaluation.
A culture satisfies $\frac{dP}{dt}=\frac{3}{2}P$ with $P(0)=4$ . Find $P(2)$ .
Answer: $4e^{3}$ This problem requires evaluating an exponential growth model at a specific time. From $\frac{dP}{dt} = \frac{3}{2}P$ with $P(0) = 4$ , we get $P(t) = 4e^{\frac{3}{2}t}$ . To find P(2): $P(2) = 4e^{\frac{3}{2} \times 2} = 4e^3$ . The fractional growth rate $\frac{3}{2}$ multiplied by time 2 gives the exponent 3, representing rapid culture growth. This calculation shows how bacterial or cellular cultures can grow exponentially under optimal conditions. Choice C gives $4e^{4/3}$ , which would result from incorrectly computing $(\frac{3}{2}) \times 2$ as $\frac{4}{3}$ instead of 3. When evaluating exponential models, carefully multiply the rate constant by the time value to determine the correct exponent.
A quantity satisfies $\frac{dy}{dt}=-\frac{3}{4}y$ with $y(0)=16$ . Which is $y(t)$ ?
Answer: $y(t)=16e^{-3t/4}$ This problem involves exponential decay with a fractional coefficient. The differential equation $dy/dt = -(3/4)y$ with $y(0) = 16$ represents quantity decrease at a fractional rate. The solution $y(t) = 16e^{-(3/4)t}$ shows exponential decay where the fractional coefficient creates the exponent $-3t/4$ . The negative sign ensures decreasing quantity over time, while fraction $3/4$ determines the decay speed. Choice C incorrectly uses a positive exponent, which would model quantity increase contrary to the negative rate in the differential equation. For decay problems with fractional rates, ensure the negative sign appears in the exponential function to represent the decreasing behavior correctly.
A colony count $C$ obeys $\frac{dC}{dt}=\frac{1}{2}C$ with $C(0)=8$ . Which is $C(t)$ ?
Answer: $C(t)=8e^{t/2}$ This problem involves solving an exponential growth model with fractional growth rate. The differential equation dC/dt = (1/2)C indicates exponential growth with rate k = 1/2. The general solution is C(t) = $Ce^{t/2}$ , where C is determined by the initial condition. Using C(0) = 8, we get C = 8, so C(t) = $8e^{t/2}$ . Choice C incorrectly uses 2 as the growth rate instead of 1/2, giving $8e^{2t}$ , which would represent much faster growth. When the differential equation has a fractional coefficient like dy/dt = (1/n)y, the solution is $Ce^{t/n}$ , not $Ce^{nt}$ .
A quantity satisfies $\frac{dy}{dt}=\frac{1}{5}y$ with $y(0)=25$ . Find $y(5)$ .
Answer: $25e^{1}$ This problem requires evaluating an exponential growth model with a fractional rate. From dy/dt = (1/5)y with y(0) = 25, we get y(t) = $25e^{(1/5)t}$ . To find y(5): y(5) = $25e^{(1/5)×5}$ = $25e^1$ . The fractional growth rate 1/5 multiplied by time 5 gives the simple exponent 1, making the evaluation straightforward. This represents moderate growth over time period 5, reaching 25e times the original amount. Choice C incorrectly uses the time value as part of a larger exponent, which has no basis in the solution method for this differential equation. When working with fractional rates, carefully multiply by the time value to determine the final exponent.
A material decays by $\frac{dM}{dt}=-\frac{2}{5}M$ with $M(0)=15$ . Which is $M(t)$ ?
Answer: $M(t)=15e^{-2t/5}$ This problem models exponential decay with a fractional rate constant. The differential equation dM/dt = -(2/5)M with M(0) = 15 represents material degradation at a fractional rate. The solution M(t) = $15e^{-(2/5)t}$ shows decay where the fractional coefficient creates the exponent -2t/5. The negative sign ensures decreasing material over time, while the fraction 2/5 determines the decay speed. Choice C incorrectly uses a positive exponent, which would model material accumulation contrary to the negative rate in the differential equation. For decay problems with fractional rates, write the exponent as -rt where r is the positive fractional rate constant.
A quantity satisfies $\frac{dy}{dt}=\ln 2\, y$ with $y(0)=5$ . Find $y(1)$ .
Answer: $10$ This problem involves exponential growth with a logarithmic rate constant. From $dy/dt = (\ln 2)y$ with $y(0) = 5$ , we get $y(t) = 5e^{(\ln 2)t}$ . To find $y(1)$ : $y(1) = 5e^{(\ln 2) \times 1} = 5e^{\ln 2}$ . Using the fundamental property $e^{\ln a} = a$ , we have $e^{\ln 2} = 2$ , so $y(1) = 5 \times 2 = 10$ . The logarithmic rate constant $\ln 2$ creates a specific growth pattern where the quantity doubles over unit time. Choice B gives $5 \ln 2$ , which incorrectly treats the exponential function as a linear function. When rate constants involve logarithms, use the property $e^{\ln a} = a$ to evaluate the exponential expression.
A drug amount $A$ obeys $\frac{dA}{dt}=-\frac{1}{5}A$ and $A(0)=30$ . What is $A(t)$ ?
Answer: $A(t)=30e^{-t/5}$ This problem involves solving an exponential decay model from a differential equation. The equation dA/dt = -1/5 · A shows that the rate of change is proportional to A with negative constant k = -1/5, indicating exponential decay. The general solution to dy/dt = ky is y(t) = $Ce^{kt}$ , so here A(t) = $Ce^{-t/5}$ . Using the initial condition A(0) = 30, we find C = 30, giving A(t) = $30e^{-t/5}$ . Choice A incorrectly uses -5 as the exponent instead of -1/5, which would represent much faster decay. When the differential equation has dy/dt = ky with negative k, the solution is always $Ce^{kt}$ with exponential decay.
A sample satisfies $\frac{dM}{dt}=-kM$ , $M(0)=100$ , and $M(5)=100e^{-10}$ . Find $k$ .
Answer: $k=2$ This problem involves finding the decay constant from given conditions. From dM/dt = -kM with M(0) = 100 and M(5) = $100e^{-10}$ , we know M(t) = $100e^{-kt}$ . Substituting the second condition: $100e^{-k×5}$ = $100e^{-10}$ , which simplifies to $e^{-5k}$ = $e^{-10}$ . Therefore, -5k = -10, giving k = 2. The negative exponent in the given condition M(5) = $100e^{-10}$ indicates decay, confirming our use of -k in the general solution. Choice C gives k = 1/2, which would result in M(5) = $100e^{-2.5}$ , not the required $100e^{-10}$ . When finding decay constants, match the coefficient of time in the exponents to solve for k.
A bank account satisfies $\frac{dB}{dt}=\frac{1}{4}B$ with $B(0)=16$ . What is $B(8)$ ?
Answer: $16e^{2}$ This problem requires evaluating an exponential growth model at a specific time. From dB/dt = (1/4)B with B(0) = 16, we have B(t) = $16e^{(1/4)t}$ . To find B(8), we substitute: B(8) = $16e^{(1/4)×8}$ = $16e^2$ . The fractional growth rate 1/4 multiplied by time 8 gives the exponent 2. This represents moderate growth over a longer time period, reaching $16e^2$ times the original amount. Choice C incorrectly uses the initial value as part of the exponent, which has no basis in exponential growth theory. When evaluating exponential models, always multiply the rate constant by the time value to determine the exponent.
A population satisfies $\frac{dN}{dt}=kN$ , $N(0)=3$ , and $N(4)=3e^{2}$ . Find $k$ .
Answer: $k=\frac{1}{2}$ This problem requires finding the growth constant using population data over time. From dN/dt = kN with N(0) = 3 and N(4) = $3e^2$ , we know N(t) = $3e^{kt}$ . Substituting the condition at t = 4: $3e^{k×4}$ = $3e^2$ , which simplifies to $e^{4k}$ = $e^2$ . Therefore, 4k = 2, giving k = 1/2. The time interval of 4 units with exponential factor $e^2$ determines the growth rate per unit time. Choice A gives k = 1/2 correctly, while choice B would produce N(4) = $3e^8$ instead of the required $3e^2$ . When finding growth constants, divide the logarithmic change by the time interval to get the rate per unit time.
A substance amount $Q$ satisfies $\frac{dQ}{dt}=-kQ$ , $Q(0)=12$ , and $Q(2)=3$ . What is $Q(t)$ ?
Answer: $Q(t)=12e^{-(\ln 4/2)t}$ This problem requires recognizing and solving an exponential decay model using differential equations. The equation dQ/dt = -kQ (k>0) indicates decay, solved by Q= $Ce^{-kt}$ . Q(0)=12 gives C=12; Q(2)=3 implies $12e^{-2k}$ =3, $e^{-2k}$ =1/4, -2k=ln(1/4), k=(ln4)/2, so Q= $12e^{-(ln4/2)t}$ . This fits the quartering. A tempting distractor like C, $3e^{-(ln4/2)t}$ , fails by using the wrong initial value 3 instead of 12. To recognize exponential models, look for rates of change proportional to the current value, leading to solutions of the form $Ce^{kt}$ .
A value satisfies $\frac{dy}{dt}=ky$ , $y(0)=9$ , and $y(1)=9e^{-1}$ . What is $k$ ?
Answer: $k=-1$ This problem involves finding the decay constant from given boundary conditions. From dy/dt = ky with y(0) = 9 and y(1) = $9e^{-1}$ , we know y(t) = $9e^{kt}$ . Substituting the second condition: $9e^{k×1}$ = $9e^{-1}$ , which simplifies to $e^k$ = $e^{-1}$ . Therefore, k = -1. The negative value confirms this is a decay process, where the quantity decreases over time. Choice C gives k = 1, which would produce y(1) = $9e^1$ instead of the required $9e^{-1}$ , representing growth rather than decay. When determining rate constants from exponential data, match the exponents in the general solution kt with the given exponential expression.
A quantity satisfies $\frac{dy}{dt}=-\frac{1}{6}y$ with $y(0)=12$ . Find $y(6)$ .
Answer: $\frac{12}{e}$ This problem requires evaluating an exponential decay model with a fractional rate. From dy/dt = -(1/6)y with y(0) = 12, we get y(t) = $12e^{-(1/6)t}$ . To find y(6): y(6) = $12e^{-(1/6)×6}$ = $12e^{-1}$ = 12/e. The fractional decay rate -1/6 multiplied by time 6 gives the exponent -1, and $e^{-1}$ = 1/e. This demonstrates how fractional rates over longer time periods can produce simple exponential values. Choice C uses a positive exponent, which would incorrectly model growth rather than decay as specified by the negative rate constant. For fractional decay rates, ensure proper multiplication with time to determine the final negative exponent.
A population satisfies $\frac{dN}{dt}=kN$ and $N(0)=30$ , with $N(2)=60$ . What is $k$ ?
Answer: $k=\frac{\ln(2)}{2}$ This problem requires finding the growth constant k from given population data. The general solution to dN/dt = kN is N(t) = $Ce^{kt}$ , and with N(0) = 30, we get N(t) = $30e^{kt}$ . Using N(2) = 60, we have 60 = $30e^{2k}$ , which simplifies to $e^{2k}$ = 2. Taking natural logarithm of both sides: 2k = ln(2), so k = ln(2)/2. Choice A (k = ln(2)) forgets to divide by 2, missing that the exponent contains 2k, not just k. When finding growth constants from doubling conditions, remember to account for the time variable in the exponent.
A population satisfies $\frac{dN}{dt}=kN$ , $N(0)=5$ , and $N(2)=20$ . What is $N(t)$ ?
Answer: $N(t)=5e^{(\ln 4/2)t}$ This problem requires finding the explicit exponential model using two data points. From dN/dt = kN with N(0) = 5 and N(2) = 20, we first find k. Since N(t) = $5e^{kt}$ and N(2) = 20, we have $5e^{2k}$ = 20, so $e^{2k}$ = 4. Taking logarithms: 2k = ln(4), giving k = (ln 4)/2. Therefore, N(t) = $5e^{(ln 4/2)t}$ . We can verify: N(2) = $5e^{(ln 4/2)×2}$ = $5e^{ln 4}$ = 5×4 = 20 ✓. Choice C incorrectly doubles ln 4 instead of halving it, which would make the growth too rapid. When finding exponential models from data points, use the ratio of final to initial values to determine $e^{kt}$ , then solve for k.
A population $N$ satisfies $\frac{dN}{dt}=kN$ and $N(0)=120$ , $N(2)=480$ . What is $N(t)$ ?
Answer: $N(t)=120e^{(\ln 4/2)t}$ This problem requires recognizing and solving an exponential growth model using differential equations. The equation dN/dt = kN indicates proportional growth, solved by separation: dN/N = k dt, integrating to ln|N| = kt + C, so N = $Ce^{kt}$ . Using N(0)=120 gives C=120; then N(2)=480 implies $120e^{2k}$ =480, so $e^{2k}$ =4, k=(ln4)/2, yielding N(t)= $120e^{(ln4/2)t}$ . A tempting distractor like A, $120e^{2t}$ , fails by incorrectly assuming k=2 without solving for it from the given data. To recognize exponential models, look for rates of change proportional to the current value, leading to solutions of the form $Ce^{kt}$ .
A cooling model satisfies $\frac{dT}{dt}=-0.1T$ with $T(0)=300$ . Which is $T(t)$ ?
Answer: $T(t)=300e^{-0.1t}$ This problem demonstrates exponential decay modeling in cooling applications. The differential equation dT/dt = -0.1T with T(0) = 300 represents Newton's law of cooling, where temperature decreases at a rate proportional to the current temperature. Following the standard solution method, we get T(t) = $300e^{-0.1t}$ , with the negative exponent indicating temperature decrease over time. The coefficient -0.1 becomes the exponent's coefficient, maintaining the negative sign for decay. Choice C incorrectly uses a positive exponent, which would model heating rather than cooling. For cooling problems, always ensure the solution reflects decreasing temperature through a negative exponent in the exponential function.
A drug amount satisfies $\frac{dA}{dt}=-0.5A$ with $A(0)=8$ . Find $A(4)$ .
Answer: $8e^{-2}$ This problem involves evaluating an exponential decay model for pharmacokinetics. From dA/dt = -0.5A with A(0) = 8, we get A(t) = $8e^{-0.5t}$ . To find A(4): A(4) = $8e^{-0.5×4}$ = $8e^{-2}$ . This represents drug elimination following first-order kinetics, where half of the remaining drug is eliminated per unit time proportional to the rate constant. The calculation shows the drug amount after 4 time units of elimination. Choice C uses a positive exponent, which would incorrectly model drug accumulation rather than elimination from the body. For pharmacokinetic decay problems, ensure the negative rate constant produces decreasing drug concentrations over time.

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