Chapter 4: Problem 48
Show that for any constants \(A\) and \(k\), the function \(y=A e^{k t}\) satisfies the equation \(d y / d t=k y\)
Short Answer
Expert verified
The function \( y = A e^{kt} \) satisfies \( \frac{dy}{dt} = ky \).
Step by step solution
01
Identify the Function and Its Components
We start with the given function \( y = A e^{kt} \), where \( A \) and \( k \) are constants and \( t \) is the variable.
02
Differentiate the Function
To find \( \frac{dy}{dt} \), we differentiate \( y = A e^{kt} \) with respect to \( t \). Using the chain rule, the derivative of \( e^{kt} \) with respect to \( t \) is \( ke^{kt} \). Hence, \( \frac{dy}{dt} = A \times ke^{kt} = kAe^{kt} \).
03
Express the Original Function
Recall that the original function is \( y = A e^{kt} \). This will help us simplify our derivative expression if needed.
04
Substitute and Verify the Equation
Substitute \( y = A e^{kt} \) back into the derivative expression to verify the equation: \( \frac{dy}{dt} = kAe^{kt} = ky \). This shows that \( \frac{dy}{dt} \) equals \( ky \), confirming the solution satisfies the given equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are mathematical expressions of the form \( y = A e^{kt} \), where \( A \) is a constant that represents the initial value, \( e \) is the base of the natural logarithm, and \( k \) is another constant that affects the rate of exponential growth or decay.
In this function, \( t \) is the independent variable, often representing time. Exponential functions are prevalent in modeling real-world phenomena, such as population growth, radioactive decay, interest calculations in finance, and more.
In this function, \( t \) is the independent variable, often representing time. Exponential functions are prevalent in modeling real-world phenomena, such as population growth, radioactive decay, interest calculations in finance, and more.
- When \( k > 0 \), the function models exponential growth as the value of \( y \) increases over time.
- When \( k < 0 \), the function exhibits exponential decay, which decreases over time.
Calculus Differentiation
Differentiation is a foundational concept in calculus that involves finding the rate at which a function is changing at any given point.
The process of differentiation provides us with a derivative, which represents the slope of the tangent line to the curve at that particular point. For functions involving \( e \), the differentiation has specific rules that make it relatively straightforward.
The process of differentiation provides us with a derivative, which represents the slope of the tangent line to the curve at that particular point. For functions involving \( e \), the differentiation has specific rules that make it relatively straightforward.
- For differentiation of \( e^{kt} \), the resulting derivative is \( ke^{kt} \).
- This process is used to determine the instantaneous rate of change of one quantity with respect to another.
Chain Rule
The chain rule is a fundamental technique in calculus used for differentiating composite functions.
It's the rule that tells us how to take the derivative of a function that is nested within another. Consider the function \( y = A e^{kt} \), which can be seen as a composition of \( A \cdot g(t) \) where \( g(t) = e^{kt} \).
It's the rule that tells us how to take the derivative of a function that is nested within another. Consider the function \( y = A e^{kt} \), which can be seen as a composition of \( A \cdot g(t) \) where \( g(t) = e^{kt} \).
- When differentiating \( e^{kt} \), the chain rule helps by recognizing that since \( t \) is multiplied by \( k \), we must also multiply the derivative by \( k \).
- The result of applying the chain rule here is \( ke^{kt} \), ensuring we account for the inner function's influence.
Derivatives
Derivatives are a core part of calculus, capturing the idea of how a function changes as its input changes.
They quantify this change by providing us with a precise mathematical expression, representing the slope of the function at any point. For the function \( y = A e^{kt} \), the derivative \( \frac{dy}{dt} \) was shown to be \( kAe^{kt} \).
They quantify this change by providing us with a precise mathematical expression, representing the slope of the function at any point. For the function \( y = A e^{kt} \), the derivative \( \frac{dy}{dt} \) was shown to be \( kAe^{kt} \).
- This shows the derivative maintains the structure of the original function, scaled by the constant \( k \).
- The derivative \( \frac{dy}{dt} = ky \) perfectly aligns with our equation, emphasizing the nature of exponential functions where the rate of change is proportional to the function itself.