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Amy Ferguson Moncure

Calculus BC · Unit 7

Exponential Models

Use half-life, doubling time, or two measurements. For cooling, follow the gap from the surrounding temperature.

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Key Ideas Easy Medium Hard

\(y=y_0e^{kt}\)

What to Remember

Proportional growth or decay

\(\dfrac{dy}{dt}=ky\) has solution

\(y=y_0 e^{kt}\)

\(k>0\) grows. \(k<0\) decays.

Half-life and doubling are special values of k

Half-life \(h\):

\(y=y_0\left(\dfrac{1}{2}\right)^{t/h}\)

Doubling time \(d\):

\(y=y_0\, 2^{t/d}\)

Two measurements find k

Put the later amount over the earlier amount. The elapsed time is the exponent’s input.

\(\dfrac{y(t_2)}{y(t_1)}=e^{k(t_2-t_1)}\)

Newton’s law uses the gap from the room

\(\dfrac{dT}{dt}=k(A-T)\) makes \(T-A\) an exponential. The object approaches the surrounding temperature \(A\).

Level 1

Easy

The half-life or the doubling time is given.

Example
A population doubles every 3 years. It starts at 50. Find the population after 6 years.

\(P=50\cdot 2^{t/3}\)

\(P(6)=50\cdot 2^{6/3}\)

\(50\cdot 2^{2}\)

\(200\)

Answer: \(200\)

1
A sample has a half-life of 4 hours and starts at 80 mg. How much is left after 12 hours?
Show Answer

\(A=80\left(\dfrac{1}{2}\right)^{t/4}\)

\(A(12)=80\left(\dfrac{1}{2}\right)^{3}\)

\(80\cdot \dfrac{1}{8}\)

\(10\)

Answer: 10 mg

2
\(dy/dt=0.3y\) and \(y(0)=20\). Write \(y(t)\).
Show Answer

\(y=20e^{0.3t}\)

Answer: \(y=20e^{0.3t}\)

3
\(P(0)=100\) and \(P(2)=400\), with \(dP/dt=kP\). Find \(P(3)\).
Show Answer

\(400=100e^{2k}\)

\(e^{2k}=4\)

\(k=\ln 2\)

\(P(t)=100\cdot 2^{t}\)

\(P(3)=800\)

Answer: \(800\)

4
Which equation describes growth proportional to the amount present?
A. \(dP/dt=0.04P\)
B. \(dP/dt=0.04t\)
C. \(dP/dt=0.04\)
D. \(dP/dt=0.04(P-t)\)
Show Answer

Proportional to \(P\) means \(P\) is a factor on the right.

A positive coefficient means growth.

Answer: A. \(dP/dt=0.04P\)

5
Write \(k\) for a half-life of 6 years in \(y=y_0e^{kt}\).
Show Answer

\(\dfrac{1}{2}=e^{6k}\)

\(k=\dfrac{\ln(1/2)}{6}\)

\(k=-\dfrac{\ln 2}{6}\)

Answer: \(k=-\dfrac{\ln 2}{6}\)

Level 2

Medium

Square a growth factor, or follow the gap from room temperature.

Example
\(P(0)=100\), \(P(2)=300\), and \(dP/dt=kP\). Find \(P(4)\).

\(300=100e^{2k}\)

\(e^{2k}=3\)

\(P(4)=100e^{4k}=100(e^{2k})^2\)

\(100\cdot 9\)

\(900\)

Answer: \(900\)

1
A room is \(70^\circ\). An object starts at \(150^\circ\) and is \(110^\circ\) after 10 minutes. Assume \(dT/dt=k(70-T)\). When is the object \(90^\circ\)?
Show Answer

Let \(U=T-70\).

\(U(0)=80\) and \(U(10)=40\).

The gap halves every 10 minutes.

\(T=70+80\left(\dfrac{1}{2}\right)^{t/10}\)

Set \(90=70+80\left(\dfrac{1}{2}\right)^{t/10}\).

\(\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^{t/10}\)

\(t=20\)

Answer: 20 minutes

2
A culture triples every 5 hours and starts at 200. Find the amount after 10 hours.
Show Answer

\(B=200\cdot 3^{t/5}\)

\(B(10)=200\cdot 3^{2}\)

\(1800\)

Answer: \(1800\)

3
A quantity is 40 at \(t=0\) and 10 at \(t=4\). Find the half-life.
Show Answer

\(10=40\left(\dfrac{1}{2}\right)^{4/h}\)

\(\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^{4/h}\)

\(2=\dfrac{4}{h}\)

\(h=2\)

Answer: \(2\)

4
A population satisfies \(dP/dt=0.05P\). After one year, the factor is
A. \(e^{0.05}\)
B. \(1.05\)
C. \(0.05\)
D. \(5\)
Show Answer

\(P=P_0e^{0.05t}\)

At \(t=1\), the factor is \(e^{0.05}\).

That is not the same as adding 5 percent once.

Answer: A. \(e^{0.05}\)

5
Write the temperature model if the surrounding temperature is \(20^\circ\) and \(dT/dt\) is proportional to \(20-T\).
Show Answer

\(\dfrac{dT}{dt}=k(20-T)\)

Answer: \(\dfrac{dT}{dt}=k(20-T)\)

Level 3

Hard

Use a measured ratio more than once, or follow the temperature gap.

Example
A sample starts at 200 mg and is 150 mg after 3 hours. It decays proportionally to the amount present. Find the amount after 9 hours.

\(150=200e^{3k}\)

\(e^{3k}=\dfrac{3}{4}\)

\(A(9)=200e^{9k}=200(e^{3k})^3\)

\(200\left(\dfrac{3}{4}\right)^3\)

\(200\cdot \dfrac{27}{64}\)

\(\dfrac{675}{8}\)

Answer: \(\dfrac{675}{8}\) mg

1
Coffee starts at \(80^\circ\) in a \(20^\circ\) room and is \(50^\circ\) after 5 minutes. When is it \(35^\circ\)?
Show Answer

\(U=T-20\)

\(U(0)=60\) and \(U(5)=30\).

The gap halves every 5 minutes.

\(T=20+60\left(\dfrac{1}{2}\right)^{t/5}\)

\(35=20+60\left(\dfrac{1}{2}\right)^{t/5}\)

\(\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^{t/5}\)

\(t=10\)

Answer: 10 minutes

2
Population \(P\) doubles every 4 years. Population \(Q\) triples every 4 years. They start equal. Find \(Q/P\) after 8 years.
Show Answer

After 8 years, \(P\) has doubled twice, so it is 4 times its start.

\(Q\) has tripled twice, so it is 9 times its start.

\(\dfrac{Q}{P}=\dfrac{9}{4}\)

Answer: \(\dfrac{9}{4}\)

3
\(dT/dt=-k(T-70)\), \(T(0)=130\), and \(T(5)=100\). Find \(T(10)\).
Show Answer

\(U=T-70\)

\(U(0)=60\) and \(U(5)=30\).

\(U(10)=15\)

\(T(10)=85\)

Answer: \(85\)

4
For the soup model \(T=70+80\left(\dfrac{1}{2}\right)^{t/10}\), find \(\lim_{t\to\infty} T(t)\) and say what it means.
Show Answer

\(\left(\dfrac{1}{2}\right)^{t/10}\) goes to 0.

The limit is 70.

The soup approaches the surrounding temperature.

Answer: \(70\)

5
Two samples of the same substance start at 80 mg and 48 mg. The half-life is 5 days. How much of each sample is left after 5 days?
Show Answer

Five days is one half-life.

The larger sample: \(80\cdot \dfrac{1}{2}=40\).

The smaller sample: \(48\cdot \dfrac{1}{2}=24\).

Answer: 40 mg and 24 mg

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