Question

Use \(y=y_{0} e^{k t}\). If \(y=1000\) at \(t=3\) and \(y=3000\) at \(t=4\), what was \(y_{0}\) at \(t=0 ?\)

Short answer

The initial value \( y_{0} \) at \( t=0 \) is approximately 37.04.

Step-by-step solution

Understand the equation

The equation given is an exponential growth formula: \[ y = y_{0} e^{kt} \]where \( y \) is the value at time \( t \), \( y_0 \) is the initial value (at \( t = 0 \)), \( e \) is Euler's number (approximately 2.71828), and \( k \) is the rate constant.

Identify the given values

From the problem, we have two conditions: \( y = 1000 \) when \( t = 3 \) and \( y = 3000 \) when \( t = 4 \). We will use these values to form two equations and solve for \( y_0 \) and \( k \).

Set up equations with given values

Using \( y = y_{0} e^{kt} \), plug in the values for \( t = 3 \) and \( t = 4 \):1. \( 1000 = y_{0} e^{3k} \)2. \( 3000 = y_{0} e^{4k} \).

Solve for \( e^{k} \) as a ratio

Divide the second equation by the first to eliminate \( y_0 \):\[ \frac{3000}{1000} = \frac{y_{0} e^{4k}}{y_{0} e^{3k}} \]\[ 3 = e^{k} \]This means \( e^{k} = 3 \).

Find \( y_0 \) using \( e^{3k} \)

Substitute \( e^{k} = 3 \) back into the first equation (\( 1000 = y_{0} e^{3k} \)):\[ 1000 = y_{0} \cdot (3)^3 \]\[ 1000 = y_{0} \cdot 27 \]Solve for \( y_{0} \):\[ y_{0} = \frac{1000}{27} \]

Calculate \( y_0 \)

Finally, calculate \( y_{0} \):\[ y_{0} \approx 37.04 \]

Key concepts

Initial Value

In many real-world problems dealing with exponential growth, determining the initial value is crucial. The initial value, denoted as \( y_0 \), is essentially the starting point of your data set. It represents the quantity present at the beginning, at \( t = 0 \), before any growth has occurred.
In the given exercise, the initial value \( y_0 \) is what we are solving for. With the exponential growth formula \( y = y_0 e^{kt} \), it can be found by substituting known data points into the equation, and then solving for \( y_0 \). This is typically done after isolating \( y_0 \) on one side of the equation and plugging in other known variables like \( e^k \) and any \( y \) values at specific times \( t \).
Understanding the initial value is essential in contexts ranging from population studies to financial investments, as it gives a baseline from which growth begins.
When working on such problems, remember:

• Identify and gather known variables first.
• Set up equations accordingly using the exponential model.
• Solve step-by-step for the initial value \( y_0 \).

Rate Constant

The rate constant, symbolized by \( k \) in exponential growth equations, indicates the rate at which a quantity grows or decays over a certain period. It determines how quickly growth occurs and is pivotal for understanding the dynamics of exponential processes.
In our scenario, to figure out \( k \), we utilized the known values at specific times. By forming equations using these values and seeing their ratio, we deduced \( e^k = 3 \). This step is essential as the rate constant multiplies time \( t \) in the exponent, directly manipulating the growth curve.
Finding the rate constant involves:

• Setting up equations with data points at different times.
• Using known values to form a ratio, simplifying expressions where necessary.
• Solving thereof for \( e^k \), which can then be used to further solve for \( k \).

Comprehending \( k \) is crucial because

• A higher \( k \) implies a faster growth rate.
• A lower \( k \) indicates slower growth.

This understanding can help predict outcomes and decision-making in various fields.

Euler's Number

Euler's number, denoted as \( e \), is an indispensable constant in mathematics, especially when dealing with exponential growth and decay. It is approximately 2.71828 and serves as the base for natural logarithms, making it crucial in mathematical models of growth.
In the provided problem, \( e \) acts as the base for the exponent in \( y = y_0 e^{kt} \), which is a standard expression for exponential growth. The uniqueness of \( e \) lies in its mathematical properties that make processes involving continuous growth smooth and predictable.
Euler's number simplifies calculations of natural growth and decay phenomena. Here's how:

• Acts as a base in exponential functions, which model every conceivable growth and decline scenario from population biology to finance.
• Makes derivations and integration straightforward, pivotal for calculus.

Understanding \( e \) helps in restoring the accuracy and realism in mathematical models dealing with real-world applications, as it naturally models systems where growth compounds continuously.