An equation that relates a function with its derivatives.

The rate of change of a quantity $y$ with respect to time $t$.

The rate constant, indicating the rate of growth (if $k > 0$) or decay (if $k < 0$).

The initial value of the quantity $y$ at time $t = 0$.

A mathematical model that describes the growth or decay of a quantity over time, where the rate of change is proportional to the current amount.

Euler's number, the base of the natural logarithm, approximately equal to 2.71828; important in continuous growth/decay situations.

The measure of how a quantity is changing with respect to another quantity, typically time; represented by a derivative.

The rate of growth is a constant multiple of the current value.

The value of the function at a specific point, often at $t=0$, used to find a particular solution.

A technique where terms involving one variable are isolated on one side of the equation, and terms involving the other variable are on the other side, allowing for integration.

$y = y_0 e^{kt}$

$k = \frac{1}{t} \ln(\frac{y(t)}{y_0})$

$\ln|y| + C$

$y = y_0 cdot e^{kt}$

Using the formula $k = \frac{\ln(\frac{y}{y_0})}{t}$, where $y$ is the population at time $t$, and $y_0$ is the initial population.

$A(t) = A_0 e^{kt}$, where $A_0$ is the initial amount and $k$ is the decay constant.

$\frac{dy}{dt}$

$\frac{dy}{dt} = ky$

$y = e^{kt+C} = e^C e^{kt}$

$e^C = |y_0|$

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Divide both sides by $y_0$. 3. Take the natural logarithm of both sides. 4. Solve for $k$.

1. Write the equation $y = y_0 e^{kt}$. 2. Determine $y_0$ (initial population). 3. Find $k$ (growth rate). 4. Substitute $y_0$ and $k$ into the equation.

1. Determine the exponential model $y = y_0 e^{kt}$. 2. Substitute the known values of $y_0$, $k$, and $t$ (future time). 3. Calculate $y$ to find the future population.

1. Identify the initial amount $y_0$. 2. Use the given information to find the decay constant $k$. 3. Substitute $y_0$ and $k$ into the equation $y = y_0 e^{kt}$. 4. Solve for the desired variable.

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Divide both sides by $y_0$. 3. Take the natural logarithm of both sides. 4. Solve for $t$ using $t = \frac{\ln(\frac{y}{y_0})}{k}$.

1. Use the half-life formula $k = \frac{\ln(2)}{T}$, where $T$ is the half-life. 2. Substitute the given half-life value into the formula. 3. Calculate $k$.

1. Substitute the given values into $y = y_0 e^{kt}$. 2. Solve for $y_0$ using $y_0 = \frac{y}{e^{kt}}$.

1. Set $y = 2y_0$ in the equation $y = y_0 e^{kt}$. 2. Simplify to $2 = e^{kt}$. 3. Take the natural logarithm of both sides. 4. Solve for $t$ using $t = \frac{\ln(2)}{k}$.

1. Identify the initial amount $A_0$. 2. Use the given information to find the constant $k$. 3. Write the equation $A(t) = A_0 e^{kt}$.

1. Assign an initial value to the bacteria, $y_0$. 2. Note that after 5 hours, $y = 2y_0$. 3. Use the exponential growth equation to find $k$. 4. Use the exponential growth equation to find $t$ when $y = 4y_0$.