Homework 5.5: Exponential and Compounding Models

Vocabulary terms, formulas, and parameters extracted from Homework 5.5 covering compound interest, exponential decay, logistic population models, and continuous bacterial growth.

Quarterly Compounding

An interest compounding frequency applied $4$ times per year, utilized in Problem 1 to calculate the growth of Mark's fixed annual $2.5\%$ investment of \fresh13{,}000 over $5$ years.

Finn's Annuity

An investment account that accumulates to a balance of \fresh41{,}966.78 after $8$ years of weekly compounding at a fixed annual interest rate of $6.99\%$.

Continuously Compounded Interest

Interest calculated continuously according to the exponential formula $A = P e^{rt}$, used in Problem 3 to determine that Grace's \fresh20{,}000 deposit at a $5\%$ fixed rate yields approximately \fresh31{,}366.20 after $9$ years.

Plutonium 241

A fissile isotope partly used in commercial power reactors that has a half-life of about $14.3$ years.

Half-Life (Plutonium 241)

The period of time, specified as $14.3$ years, required for half of a quantity of Plutonium 241 in open storage to decay.

Bee Population Function $B(t)$

The model $B(t) = \frac{3400}{5 + 15e^{-0.01t}}$ representing the population of a certain species of bees in a large public park after $t$ days.

Initial Bee Population

The value of $B(0)$ in the logistic model, calculated as $\frac{3400}{5 + 15(1)} = \frac{3400}{20} = 170\text{ bees}$.

Limiting Population (Bee Model)

The horizontal asymptote that the bee population approaches as time continues indefinitely ($t \rightarrow \infty$), evaluated as $\frac{3400}{5} = 680$.

Continuous Bacteria Growth Rate

The continuous rate of $4.6\%$ per day at which an initial population of $820$ bacteria expands, expressed by the continuous growth model $A = 820e^{0.046t}$.