Problem 19
Question
Represent exponential growth or decay. What is the initial quantity? What is the growth rate? State if the growth rate is continuous. $$P=15 e^{-0.06 t}$$
Step-by-Step Solution
Verified Answer
Initial quantity is 15, decay rate is -0.06, and it is continuous.
1Step 1: Identify the Mathematical Model
The equation given is in the form \( P = P_0 e^{rt} \), where \( P_0 \) is the initial quantity, \( e \) is the base of the natural logarithm, \( r \) is the rate, and \( t \) is time. This is a model for exponential decay due to the negative sign in the exponent.
2Step 2: Determine the Initial Quantity
The initial quantity \( P_0 \) is the constant value in front of the exponential term. In the equation \( P = 15 e^{-0.06 t} \), \( P_0 = 15 \). Thus, the initial quantity is 15 units.
3Step 3: Find the Growth or Decay Rate
The rate \( r \) is the coefficient of \( t \) in the exponent. Here, \( r = -0.06 \). The negative sign indicates that this is a decay rate rather than a growth rate.
4Step 4: State the Continuity of the Rate
Since the given function uses the base \( e \), the decay rate \( r = -0.06 \) can be considered a continuous rate. This is because exponential functions with base \( e \) naturally model continuous processes.
Key Concepts
Initial QuantityContinuous RateDecay Rate
Initial Quantity
In any exponential decay scenario, identifying the initial quantity is a crucial first step. The initial quantity is the starting point or the original amount before any decay begins. In mathematical terms, when the time (\( t \)) is zero, the resulting value from the equation represents this initial quantity.
The given formula, \( P = 15 e^{-0.06 t} \), provides us with this information directly. Here, the initial quantity \( P_0 \) is represented by the number 15, as it is the coefficient in front of the exponential expression. This value tells us that at time zero, the quantity starts at 15 units.
The given formula, \( P = 15 e^{-0.06 t} \), provides us with this information directly. Here, the initial quantity \( P_0 \) is represented by the number 15, as it is the coefficient in front of the exponential expression. This value tells us that at time zero, the quantity starts at 15 units.
- The initial quantity is an essential component in predicting how a quantity will decay over time.
- It serves as a baseline for observing the reduction that will occur as time progresses.
Continuous Rate
The concept of continuous rate in exponential functions is key to understanding how changes occur over time. Exponential functions that utilize the mathematical constant \( e \) are particularly suited for modeling processes that happen continuously, such as radioactive decay, population growth, or interest compounding.
In the equation \( P = 15 e^{-0.06 t} \), the presence of \( e \) indicates that the changes to the quantity occur in a continuous manner. This means the quantity doesn’t change abruptly at set intervals but rather diminishes smoothly over time.
In the equation \( P = 15 e^{-0.06 t} \), the presence of \( e \) indicates that the changes to the quantity occur in a continuous manner. This means the quantity doesn’t change abruptly at set intervals but rather diminishes smoothly over time.
- A continuous rate implies an ongoing, uninterrupted process.
- This contrasts with discrete changes which happen at specific, separate times.
Decay Rate
The decay rate in an exponential decay equation signifies how quickly a quantity is reducing over time. It is typically represented as the coefficient in the exponent of the decay expression.
In \( P = 15 e^{-0.06 t} \), the rate of decay is \( -0.06 \). The negative sign is a clear indicator that we are dealing with decay rather than growth.
In \( P = 15 e^{-0.06 t} \), the rate of decay is \( -0.06 \). The negative sign is a clear indicator that we are dealing with decay rather than growth.
- The absolute value, \( 0.06 \), represents the speed of decay - a larger absolute value implies faster decay.
- The negative sign denotes that the quantity is decreasing over time, as opposed to increasing, which would affect the equation differently.
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