Problem 71 - Calculus - AP Edition

Question

-Show that for any constants \(A\) and \(k\), the function \(y=A e^{k t}\) satisfies the equation \(d y / d t=k y\).

Step-by-Step Solution

Answer

The function satisfies the equation because both expressions for \( \frac{d y}{d t} \) and \( k y \) are identical.

Step 1: Understand the Given Function

The given function is \( y = A e^{k t} \), where \( A \) and \( k \) are constants. We need to differentiate this function with respect to \( t \) to find \( \frac{d y}{d t} \).

Step 2: Differentiate the Function

To find the derivative \( \frac{d y}{d t} \), apply the derivative rule for exponential functions. \(\frac{d}{d t} (e^{k t}) = k e^{k t}\),\so the derivative of \( y = A e^{k t} \) is: \[\frac{d y}{d t} = A \cdot k \cdot e^{k t} = k A e^{k t}\].

Step 3: Compare with the Given Equation

The differential equation given in the problem is \( \frac{d y}{d t} = k y \). Substitute \( y = A e^{k t} \) into the equation to obtain \( k y = k (A e^{k t}) = k A e^{k t} \).

Step 4: Confirmation

The expression obtained for \( \frac{d y}{d t} = k A e^{k t} \) is exactly the same as the expression \( k y = k A e^{k t} \). Thus, this confirms that the function satisfies the equation.

Key Concepts

Exponential Functions

Exponential functions are a fundamental part of mathematics, especially in calculus and differential equations. These functions have the form \( y = A e^{k t} \), where \( A \) and \( k \) are constants, \( e \) is the base of natural logarithms, and \( t \) is the variable. They grow or decay at a rate proportional to their current value.

The constant \( A \) is known as the initial value of the function. It determines the starting point of the function when \( t = 0 \).

• If \( A > 0 \), the function's graph starts above the t-axis.
• If \( A < 0 \), it starts below the t-axis.

The constant \( k \) is the rate of growth or decay. It influences how quickly the function increases or decreases over time.

• When \( k > 0 \), the function grows exponentially.
• When \( k < 0 \), it decays exponentially.

These properties make exponential functions ideal for modeling phenomena such as population growth, radioactive decay, and continuous compounding interest.

Derivatives

Derivatives are a core concept in calculus, representing the rate of change of a function with respect to a variable. The derivative of a function tells us how the function's output changes as we change its input.

Mathematically, the derivative of a function \( y = f(t) \) with respect to \( t \) is denoted as \( \frac{d y}{d t} \). For exponential functions like \( y = A e^{k t} \), taking the derivative reveals how the quantity changes over time.

Finding derivatives is essential for solving differential equations. In our exercise, we calculated \( \frac{d y}{d t} \) to confirm a relationship between the function and its rate of change. This process helps understand how systems modeled by the function behave dynamically.

Differentiation Rules

Differentiation rules are guidelines that help us calculate derivatives easily. These rules apply to various kinds of functions, making the process systematic and reliable.

For exponential functions, the differentiation rule is particularly straightforward: the derivative of \( e^{k t} \) with respect to \( t \) is \( k e^{k t} \). This rule reflects the unique property of the exponential function where the function itself also appears in its derivative.

• The constant multiplier rule states that the derivative of a constant times a function is the constant times the derivative of the function. So, \( \frac{d}{d t} (A e^{k t}) = A \cdot \frac{d}{d t} (e^{k t}) \).
• For our function \( y = A e^{k t} \), applying these rules confirms the derivative \( \frac{d y}{d t} = k A e^{k t} \).

Understanding these rules helps in accurately finding derivatives across a range of functions, simplifying the process of solving complex equations.