Problem 37 - Calculus

Question

About exponential \(v^{3}\) s and \(f^{\prime}\) s. Estimate the slope of \(f(t)=2^{t}\) at \(t=0\). Use a calculator to compute (increase in \(f)\) /(increase in \(t\) ) when \(t\) is small: \(\frac{f(t)-f(0)}{t}=\frac{2-1}{1}\) and \(\frac{2^{.1}-1}{.1}\) and \(\frac{2^{.01}-1}{.01}\) and \(\frac{2^{001}-1}{.001}\)

Step-by-Step Solution

Answer

The slope of \( f(t) = 2^t \) at \( t = 0 \) is approximately 0.7.

Step 1: Understanding the function

The function given is an exponential function \[ f(t) = 2^t \]. Exponential functions grow by multiplying the previous value by a constant (in this case, 2) each step. We are tasked with estimating the derivative or slope \( f'(t) \) at \( t = 0 \). The derivative is the rate of change or the slope of the tangent to the curve at a point.

Step 2: Definition of the Derivative

For a small increase in \( t \), the derivative at a point \( t \) can be approximated using the difference quotient: \[ f'(t) \approx \frac{f(t) - f(0)}{t}. \]We will use this approximation for various small values of \( t \) to estimate the slope at \( t = 0 \).

Step 3: Calculate for \( t = 1 \)

Substitute \( t = 1 \) into the difference quotient. \[ \frac{f(1) - f(0)}{1} = \frac{2^1 - 2^0}{1} = \frac{2 - 1}{1} = 1. \]

Step 4: Compute for \( t = 0.1 \)

Using a calculator, compute \( 2^{0.1} \):\[ \frac{f(0.1) - f(0)}{0.1} = \frac{2^{0.1} - 1}{0.1}. \]Approximate by calculator:\[ 2^{0.1} \approx 1.0718, \]\[ \frac{1.0718 - 1}{0.1} = \frac{0.0718}{0.1} = 0.718. \]

Step 5: Calculate for \( t = 0.01 \)

Again, using a calculator, compute \( 2^{0.01} \):\[ \frac{f(0.01) - f(0)}{0.01} = \frac{2^{0.01} - 1}{0.01}. \]Approximate by calculator:\[ 2^{0.01} \approx 1.0069, \]\[ \frac{1.0069 - 1}{0.01} = \frac{0.0069}{0.01} = 0.69. \]

Step 6: Calculate for \( t = 0.001 \)

Calculate for a very small \( t \):\[ \frac{f(0.001) - f(0)}{0.001} = \frac{2^{0.001} - 1}{0.001}. \]Approximate by calculator:\[ 2^{0.001} \approx 1.0007, \]\[ \frac{1.0007 - 1}{0.001} = \frac{0.0007}{0.001} = 0.7. \]

Step 7: Conclude the Slope at \( t = 0 \)

As \( t \) gets smaller, the results are converging to approximately 0.7. Thus, the slope or derivative \( f'(0) \) of \( f(t) = 2^t \) at \( t = 0 \) is estimated to be 0.7.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus, representing the rate at which a function changes at any given point. In simpler terms, it measures the instantaneous rate of change or the slope of the tangent line to the curve of a function at a specific point. For exponential functions like \( f(t) = 2^t \), derivatives help determine how quickly the function grows or decreases. Calculating the derivative provides insight into the behavior of the function, especially around points of interest like \( t = 0 \). This process involves using the difference quotient, which is a way of estimating the derivative by considering very small changes in \( t \). Understanding derivatives will allow you to analyze and describe the growth dynamics of functions in fields like physics, biology, and economics.

Difference Quotient

The difference quotient is a formula used to estimate the derivative of a function at a particular point. It's expressed as \( \frac{f(t+h) - f(t)}{h} \), where \( h \) is a small increase or difference in \( t \). In the exercise, this concept allows us to approximate the slope of the function \( f(t) = 2^t \) at \( t = 0 \).
The smaller the value of \( h \), the closer the approximation is to the actual derivative. This method helps to understand how the function changes over extremely small intervals, bringing us closer to calculating the exact rate of change. Employing the difference quotient consistently helps to simplify complex problems involving changes in exponential functions.

Slope Estimation

Slope estimation refers to the process of approximating the steepness or incline of a function at a given point. In mathematical terms, it's about estimating the rate of change using available data points. For exponential functions, slope estimation is significant because it illustrates how swiftly the function increases or decreases.
The exercise involves estimating the slope at \( t = 0 \) for the function \( f(t) = 2^t \). By calculating the difference quotient for several small values of \( t \), we determine that the slope converges toward 0.7. This approximation uses a practical approach to grasp how the exponential function behaves without directly computing the exact derivative. Using tools like calculators or software continues to enhance accuracy in these estimations.

Rate of Change

The rate of change describes how one quantity changes in relation to another. In the context of functions, it indicates how fast the function's output changes as the input varies. For our exponential function \( f(t) = 2^t \), understanding the rate of change is crucial in identifying how quickly the function's values grow.
The concept of the rate of change is closely tied to the derivative, with both offering insights into the dynamics of various functions. By examining our function using the difference quotient, we ascertained that around \( t=0 \), the rate of change is approximately 0.7. This estimation reflects the behavior of \( f(t) \) at very small increments, showcasing the exponential growth characteristic of such functions. Understanding this allows students to apply similar methods to a wide range of real-world problems where exponential growth or decay occurs.