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Magazine Chapter 10: Problem 19
Suppose \(Q=C e^{k t}\) satisfies the differential equation $$ \frac{d Q}{d t}=-0.03 Q $$ What (if anything) does this tell you about the values of \(C\) and \(k\) ?
Short Answer
Expert verified
\( k = -0.03 \); \( C \) is any non-zero constant.
Step by step solution
01
Write Down the Given Differential Equation
We are given that \( \frac{dQ}{dt} = -0.03 Q \). This equation describes how \( Q \) changes with respect to time \( t \).
02
Analyze the General Solution Structure
The problem states that \( Q = C e^{kt} \) is a solution. This form suggests that \( Q \) evolves exponentially with time, where \( C \) is a constant and \( k \) is the exponential growth rate.
03
Differentiate the General Solution with Respect to t
To find \( \frac{dQ}{dt} \) for the general solution \(Q = C e^{kt}\), differentiate with respect to \(t\): \( \frac{dQ}{dt} = C k e^{kt} \).
04
Equate Derivatives and Solve for k
Since \( Q = C e^{kt} \) is a solution to the differential equation, substitute the expressions:\( C k e^{kt} = -0.03 C e^{kt} \).Cancel \( C e^{kt} \) (provided \(C eq 0\)), we get \( k = -0.03 \).
05
Consider the Value of C
The equation \( k = -0.03 \) does not place any restrictions on the value of \( C \). Thus, \( C \) can be any non-zero constant since the general form \( Q = C e^{k t} \) requires it to be non-zero for the equation to hold meaningful form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Growth
Exponential growth is a fundamental concept in mathematics, often observed in natural and physical systems. When the quantity increases or decreases at a rate proportional to its current value, it is described by exponential growth or decay. The function \( Q = C e^{kt} \) represents this behavior:
- \( C \) is the initial amount (when \( t = 0 \)), and it represents the starting value of \( Q \).
- \( e \) is the base of natural logarithms, approximately equal to 2.718.
- \( k \) is the growth constant, which determines the rate of change; if \( k > 0 \), we have exponential growth, while \( k < 0 \) signifies exponential decay.
- \( t \) is the independent variable, typically representing time.
Constant Solutions
A constant solution is a special case in differential equations where the solution does not change over time. In other words, the derivative of the function is zero. For our equation \( \frac{dQ}{dt} = -0.03 Q \), a constant solution would only occur if \( Q = 0 \).
- Since \( Q \) would be zero from the start, it remains zero as time progresses.
- This is possible in situations where the initial conditions dictate that the quantity being measured doesn't change.
- In many real-world scenarios, however, constant solutions are not the solutions of interest, especially in modeling growth or decay, as they don't exhibit dynamic behavior.
Differentiation
Differentiation is a fundamental mathematical operation used to find the rate at which a function is changing. The derivative of a function reflects its rate of change concerning an independent variable, often time in growth models. The differentiation process involves:
Differentiation allows us to replace the original equation's left side with an expression involving \( C \), \( k \), and \( t \), facilitating the comparison needed to solve for constants like \( k \).
- Applying the power rule, product rule, or chain rule, depending on the function's form.
- For our exponential function \( Q = C e^{kt} \), the differentiation yields \( \frac{dQ}{dt} = Ck e^{kt} \).
- This expression evaluates the rate of change, showing how \( Q \) evolves over time.
Differentiation allows us to replace the original equation's left side with an expression involving \( C \), \( k \), and \( t \), facilitating the comparison needed to solve for constants like \( k \).
Initial Value Problems
Initial value problems (IVPs) in differential equations involve finding a solution satisfying both the differential equation and an initial condition. An initial condition is often given as \( Q(0) = Q_0 \), where \( Q_0 \) is known.
- IVPs specify the starting point or initial state of the system being modeled, vital in predicting future behavior.
- The given condition \( Q = C e^{kt} \) with no specific starting point implies a general solution.
- Particular solutions are often derived once initial conditions are supplied, tailoring the equation to real-world scenarios.
