Problem 100 - Calculus Single Variable

Question

Plot \(y=\ln (x)\) for \(1 / 2 \leq x \leq 3\). Let \(P(c)\) denote the point \((c, \ln (c))\) on the graph. The purpose of this exercise is to graphically explore the relationship between \(1 /\) \(\mathrm{c}\) and the slope of the tangent line at \(P(c) .\) For \(c=1,3 / 2,\) and \(2,\) calculate the slope \(m(c)\) of the secant line that passes through the pair of points \(P(c-0.001)\) and \(P(c+0.001) .\) For each \(c,\) calculate \(|1 / c-m(c)|\) to see that \(m(c)\) is a good approximation of \(1 / \mathrm{c}\). Add the three secant lines to your viewing window. For each of \(c=1\), \(3 / 2,\) and \(2,\) add to the viewing window the line through \(P(c)\) with slope \(1 / \mathrm{c}\). As we will see in Chapter \(3,\) these are the tanget lines at \(P(1), P(3 / 2),\) and \(P(2) .\) It is likely that they cannot be distinguished from the secant lines in your plot.

Step-by-Step Solution

Answer

The secant slopes \( m(c) \) closely approximate the tangent slopes \( 1/c \).

Step 1: Understand the Function

We are working with the natural logarithm function, \( y = \ln(x) \), where \( \ln \) denotes the natural logarithm. This function is defined for all \( x > 0 \). We need to focus on the interval \( 1/2 \leq x \leq 3 \), and specifically analyze points at \( x = 1, \frac{3}{2}, \text{ and } 2 \).

Step 2: Plot the Function

To start, plot the function \( y = \ln(x) \) from \( x = 1/2 \) to \( x = 3 \). This will give us a visual representation of the logarithm function over the desired interval. Note the important points: \( P(1), P(3/2), \text{ and } P(2) \) correspond to \( (1, \ln(1)), (3/2, \ln(3/2)), \text{ and } (2, \ln(2)) \).

Step 3: Define the Slope of the Secant Lines

For each value of \( c = 1, 3/2, 2 \), calculate the slope \( m(c) \) of the secant line using the formula for points \( P(c-0.001) \) and \( P(c+0.001) \). The slope \( m(c) \) is given by the difference quotient: \[ m(c) = \frac{\ln(c+0.001) - \ln(c-0.001)}{(c+0.001) - (c-0.001)} \]

Step 4: Calculate Slope m(c) for Specific Points

Compute \( m(c) \) for each \( c \):- **At \( c = 1: \)** \[ m(1) \approx \frac{\ln(1.001) - \ln(0.999)}{1.001 - 0.999} \]- **At \( c = 3/2: \)** \[ m(3/2) \approx \frac{\ln(1.501) - \ln(1.499)}{1.501 - 1.499} \]- **At \( c = 2: \)** \[ m(2) \approx \frac{\ln(2.001) - \ln(1.999)}{2.001 - 1.999} \]Compute these to get specific numerical values for each point.

Step 5: Compare Secant and Tangent Slopes

Determine how each \( m(c) \) compares to \( 1/c \) by calculating \[ |1/c - m(c)| \] for each \( c = 1, 3/2, 2 \). Observe that \( m(c) \) should closely approximate \( 1/c \).

Step 6: Plot Secant and Tangent Lines

Add the secant lines through the points \( P(c-0.001) \) and \( P(c+0.001) \) for each \( c \) to the plot. Additionally, plot the tangent lines at each \( P(c) \) using the slope \( 1/c \). The tangent line at point \( P(c) \) can be expressed as: \[ y - \ln(c) = \frac{1}{c}(x - c) \] Plot these alongside the function and secant lines.

Key Concepts

Natural Logarithm Function

The natural logarithm function is an essential part of calculus and mathematics in general. It is denoted by \( y = \ln(x) \) and is defined for all positive values of \( x \). This function represents the power to which the base of the natural logarithm (Euler's number \( e \)) must be raised to obtain the number \( x \).
The natural logarithm is uniquely important because it simplifies the process of differentiating and integrating exponential functions. In addition to its smooth and continuous nature, it has some notable properties:

• The natural logarithm of 1 is 0 (\( \ln(1) = 0 \)).
• The function is logarithmically increasing: as \( x \) increases, \( \ln(x) \) increases but at a decreasing rate.
• It is undefined for non-positive values of \( x \), which means the domain is \( x > 0 \).

In this specific exercise, we're examining \( \ln(x) \) within the range \( 1/2 \leq x \leq 3 \). This gives us a window into how the function behaves over this interval and allows us to calculate slopes at specific points.

Secant Line Slope

In calculus, the secant line provides an approximation of a tangent line by connecting two points on a curve. The slope of a secant line is a crucial concept when transitioning from an average rate of change to an instantaneous rate of change.
In this exercise, for the function \( y = \ln(x) \), the secant line slope \( m(c) \) can be found using the points \( P(c-0.001) \) and \( P(c+0.001) \). The formula used is known as the difference quotient:

• \( m(c) = \frac{\ln(c+0.001) - \ln(c-0.001)}{0.002} \)

This calculation tells us the average rate of change of the function over a small interval around \( c \). Essentially, by narrowing the interval (making it approach zero), the secant line slope approximates the tangent line slope more closely. The exercise involved computing this slope at specific points \( c = 1, \frac{3}{2}, \text{ and } 2 \), giving us a numerical approximation to be compared with their corresponding tangent line slopes.

Tangent Line Slope

The tangent line slope is a fundamental aspect of understanding derivatives in calculus. Unlike the secant line, which connects two distinct points, the tangent line touches the curve at exactly one point, providing an instantaneous rate of change.
For the function \( y = \ln(x) \), the derivative tells us the slope of the tangent line at a point \( P(c) = (c, \ln(c)) \). The derivative \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \) reveals the slope of the tangent line at any given \( x = c \). Therefore, the tangent slope is \( \frac{1}{c} \).
In this example, we evaluated the tangent line slopes at points \( c = 1, \frac{3}{2}, ext{ and } 2 \), where:

• For \( c = 1, \text{ the slope is } \frac{1}{1} = 1 \).
• For \( c = \frac{3}{2}, \text{ the slope is } \frac{2}{3} \approx 0.6667 \).
• For \( c = 2, \text{ the slope is } \frac{1}{2} = 0.5 \).

Plotting these tangent lines along with the secant lines can visually demonstrate how closely the secant line slope, calculated using small intervals, approximates the true instantaneous rate of change, \( \frac{1}{c} \).

Difference Quotient

The difference quotient is a central concept for understanding how derivatives work. It is used to estimate the slope of a function over an interval and forms the basis for defining the derivative as the interval approaches zero.
Mathematically, the difference quotient for a function \( f(x) \) at a point \( x = c \) is:

• \( \frac{f(c+h) - f(c-h)}{2h} \)

In this exercise with \( y = \ln(x) \), \( h \) was set to a small value \( 0.001 \), giving us specific points \( P(c-0.001) \) and \( P(c+0.001) \). By computing:

• \( \frac{\ln(c+0.001) - \ln(c-0.001)}{0.002} \)

we apply the difference quotient to get \( m(c) \), which serves as an approximation for the derivative at \( c \), i.e., \( \frac{1}{c} \).
The closer \( h \) is to zero, the more accurate \( m(c) \) will be in approximating the tangent slope. This exercise shows that even with a very small \( h \), the secant line provides a close estimate, illustrating the utility and practicality of the difference quotient in calculus.