Savings with Continuous Compounding

You deposit 1000 dollars in an account that pays 5% interest compounded continuously. How much is in the account after 10 years? Round to the nearest cent.

Identify the values: $P = 1000$, $r = 0.05$, $t = 10$ = Values identified

Find the exponent $rt$: $rt = 0.05 \times 10 = 0.5$ = $rt = 0.5$

Find $e^{0.5}$ with the calculator: $2.71828^{0.5} \approx 1.648721$ = $e^{0.5} \approx 1.648721$

Multiply by $P$: $A = 1000 \times 1.648721 \approx 1648.72$ = $A \approx 1648.72$

Answer: After 10 years the account holds about **1648.72 dollars**.

Continuous vs Annual Compounding

You invest 2000 dollars at 6% for 5 years. Compare compounding once a year with compounding continuously. Round to the nearest cent.

Annual compounding: use $A = P(1 + r)^t$: $A = 2000 \times 1.06^{5} \approx 2000 \times 1.338226$ = $A \approx 2676.45$

Continuous compounding: find the exponent: $rt = 0.06 \times 5 = 0.3$ = $rt = 0.3$

Use $A = Pe^{rt}$: $A = 2000 \times e^{0.3} \approx 2000 \times 1.349859$ = $A \approx 2699.72$

Compare the two results: $2699.72 - 2676.45 = 23.27$ = Continuous gives 23.27 more

Answer: With the same rate and time, continuous compounding gives **2699.72 dollars**, annual compounding **2676.45 dollars**: continuous earns **23.27 dollars** more.

Bacteria Growing Continuously

A culture starts with 500 bacteria and grows continuously at a rate of 20% per hour. How many bacteria are there after 10 hours? Round to the nearest whole number.

Write the model: $N(t) = 500 e^{0.2t}$ = $P = 500$, $r = 0.2$

Find the exponent for $t = 10$: $0.2 \times 10 = 2$ = Exponent $= 2$

Find $e^{2}$: $2.71828^{2} \approx 7.38905$ = $e^{2} \approx 7.38905$

Multiply and round: $N(10) = 500 \times 7.38905 \approx 3694.5$ = $N(10) \approx 3695$

Answer: After 10 hours there are about **3695 bacteria**, more than 7 times the starting amount.

Radioactive Decay

A sample contains 80 grams of a radioactive substance that decays continuously at a rate of 3% per year. How much is left after 20 years? Round to two decimal places.

Write the decay model: $A(t) = 80 e^{-0.03t}$ = Negative exponent: decay

Find the exponent for $t = 20$: $-0.03 \times 20 = -0.6$ = Exponent $= -0.6$

Find $e^{-0.6}$: $2.71828^{-0.6} \approx 0.548812$ = $e^{-0.6} \approx 0.548812$

Multiply by the initial amount: $A(20) = 80 \times 0.548812 \approx 43.90$ = $A(20) \approx 43.90$ g

Answer: After 20 years about **43.90 grams** remain, so about $80 - 43.90 = 36.10$ grams have decayed.

Finding the Initial Amount

How much must you deposit today, at 4% compounded continuously, to have 5000 dollars after 8 years? Round to the nearest cent.

Write the equation: $5000 = P e^{0.04 \times 8} = P e^{0.32}$ = $P$ is unknown

Find $e^{0.32}$: $2.71828^{0.32} \approx 1.37713$ = $e^{0.32} \approx 1.37713$

Divide both sides by $e^{0.32}$: $P = \frac{5000}{e^{0.32}}$ = $P = 5000 \div e^{0.32}$

Calculate: $P = 5000 \div 2.71828^{0.32} \approx 3630.75$ = $P \approx 3630.75$

Answer: You need to deposit about **3630.75 dollars** today. Check: $3630.75 \times 1.37713 \approx 5000$.

Mistake: Writing the rate as a percent in the exponent: $1000 e^{5 \times 10}$

Why: The formula needs the rate as a decimal. $e^{50}$ is a gigantic number (about $5 \times 10^{21}$), which makes no sense for a savings account.

Correct: Convert first: $5\% = 0.05$, so $A = 1000 e^{0.05 \times 10} = 1000 e^{0.5}$.

Mistake: Forgetting the minus sign for decay: $80 e^{0.03 \times 20}$

Why: Without the minus sign the exponent is positive and the amount grows to about 145.77 grams, more than the 80 grams you started with.

Correct: Decay needs a negative exponent: $A = 80 e^{-0.03 \times 20} = 80 e^{-0.6} \approx 43.90$.

Mistake: Multiplying before raising to the power: $(1000 \times 2.71828)^{0.5}$

Why: The exponent belongs only to $e$, not to $P$. Order of operations says powers come before multiplication.

Correct: First compute $e^{rt}$, then multiply by $P$. On the calculator: $1000 \times e^{(0.05 \times 10)}$, with brackets around the exponent.

Mistake: Using $(1 + r)^t$ when the problem says compounded continuously

Why: $(1 + r)^t$ is compounding once per period. Continuous compounding always gives a little more at the same rate.

Correct: Look for the words continuously or continuous rate: they mean $Pe^{rt}$. Once a year means $P(1 + r)^t$.

Banking: Continuously Compounded Savings

Banks compare offers using the most generous possible compounding, which is continuous. It shows the ceiling of what a given rate can earn.

At 5% compounded continuously, 1 dollar grows to $e^{0.05} \approx 1.0513$ dollars in a year, so the effective yearly rate is about 5.13%, a bit more than 5%.

Biology: Bacteria in the Lab

Microbiologists model a bacteria culture that has plenty of food with $N(t) = N_0 e^{rt}$, where $r$ is the continuous growth rate per hour.

A culture of 1000 bacteria growing at a continuous rate of 30% per hour has $1000 e^{0.6} \approx 1822$ bacteria after 2 hours.

Medicine: Radioactive Tracers

Hospitals use radioactive substances such as iodine-131 for diagnosis and therapy. The amount left in a sample follows $A = Pe^{-kt}$, so doctors know how much remains on any day.

Iodine-131 decays continuously at about 8.6% per day. After 1 day, $e^{-0.086} \approx 0.918$, so about 91.8% of the sample is left.

Q: Why do we use $e$ for continuous growth?

A: Compounding $n$ times a year at rate $r$ gives $\left(1 + \frac{r}{n}\right)^{nt}$. As $n$ gets larger and larger, this approaches $e^{rt}$. So $e$ is what you get when growth happens at every instant.

Q: How much more does continuous compounding really give?

A: Not much more, but always more. At 5% for one year, annual compounding multiplies by $1.05$ and continuous compounding by $e^{0.05} \approx 1.0513$. Over long periods and large amounts the difference becomes noticeable.

Q: Is $A = Pe^{-kt}$ the same as $A = Pe^{rt}$ with a negative $r$?

A: Yes. Some books write decay with a positive decay rate $k$ and a minus sign, others use one formula $Pe^{rt}$ and let $r$ be negative, for example $r = -0.03$. Both give the same result.

Q: How do I find how long it takes to reach a certain amount?

A: That means solving $A = Pe^{rt}$ for $t$, which needs the natural logarithm $\ln$. You will learn it in the logarithm lessons. For now you can estimate $t$ by trying values in a table.