Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider \(y=x^{2}+1\). (a) Sketch its graph as carefully as you can. (b) Draw the tangent line at \((1,2)\). (c) Estimate the slope of this tangent line. (d) Calculate the slope of the secant line through \((1,2)\) and \(\left(1.01,(1.01)^{2}+1.0\right)\) (e) Find by the limit process (see Example 1) the slope of the tangent line at \((1,2)\).

Step-by-Step Solution

Answer

The tangent line at \((1,2)\) has a slope of 2.

Step 1: Graph Sketching

The graph of the function \(y = x^2 + 1\) is a parabola that opens upwards. The vertex of the parabola is at \((0, 1)\), and the parabola is symmetric about the y-axis. To sketch, plot some points like \((0, 1)\), \((1, 2)\), \((-1, 2)\), \((2, 5)\), and \((-2, 5)\) and then draw a smooth curve through these points.

Step 2: Drawing the Tangent Line at (1, 2)

A tangent line at a point on a curve touches the curve at exactly one point without crossing it. Draw a line that touches the parabola at \((1,2)\) and closely follows the curve near this point.

Step 3: Estimating the Slope of the Tangent Line

To estimate the slope of the tangent line at \((1, 2)\), visually inspect the steepness of the line drawn in the previous step. The line appears to rise over run smoothly and could be estimated from the curvature or changes in adjacent points.

Step 4: Calculating the Slope of the Secant Line

A secant line through points \((1, 2)\) and \((1.01, (1.01)^2 + 1.0)\) will have a slope given by the formula \( m_{secant} = \frac{(f(x_2) - f(x_1))}{(x_2 - x_1)} \). Calculate \( (1.01)^2 + 1 = 1.0201 + 1 = 2.0201\). The slope is then \( \frac{2.0201 - 2}{1.01 - 1} = 2.01 \).

Step 5: Finding the Slope Using the Limit Process

To calculate the exact slope of the tangent, use derivatives. First, take the derivative of \(y = x^2 + 1\), resulting in \(y' = 2x\). At \(x = 1\), the slope \(m = 2 \times 1 = 2\). Alternatively, consider the limit definition: \( \lim_{{h \to 0}} \frac{(f(1+h) - f(1))}{h} = \lim_{{h \to 0}} \frac{(1+h)^2 + 1 - 2}{h} \). Simplifying gives \( \lim_{{h \to 0}} \frac{2h + h^2}{h} = \lim_{{h \to 0}} (2 + h) = 2\).

Key Concepts

Differentiation: Unraveling the Tangent Slope

Differentiation is a fundamental tool in calculus that helps us find the slope of a tangent line to a curve at any given point. When we differentiate a function, we get another function that tells us the slope at any point along the original curve. Let’s break this down using the example function given:

• The original function is \[ f(x) = x^2 + 1 \].
• To find its derivative, apply the differentiation rules for power functions. The derivative, denoted as \( f'(x) \), is calculated by the formula \( f'(x) = 2x \).

This means at each point \( x \), the slope of the tangent line to the graph of \( y = x^2 + 1 \) is given by \( 2x \).
In the exercise, at the point \( (1, 2) \), simply substitute \( x = 1 \) into the derivative: \[ f'(1) = 2 imes 1 = 2 \].So, the slope of the tangent line at this point is exactly 2.
Differentiation thus allows us to pinpoint precisely how steep a function is at any given value of \( x \), using just a simple calculation.

Secant Lines: A Step Between Points on a Curve

A secant line is a line that intersects a curve at two distinct points. It’s useful for approximating the slope of a curve when you don’t have the tangent line information readily available. Let's delve into how secant lines work using the exercise example.

• To find the slope of a secant line through the points \( (1, 2) \) and \( (1.01, (1.01)^2 + 1) \), use the formula for the slope: \[ m_{secant} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \].

Calculate each point’s function value:
- At \( x = 1 \), \( f(1) = 2 \).- At \( x = 1.01 \), \( f(1.01) = 2.0201 \).
Now, plug these into the secant slope formula:

• \[ m_{secant} = \frac{2.0201 - 2}{1.01 - 1} = 2.01 \].

This value, 2.01, is the average rate of change in y per unit change in x between the points.
Understanding secant lines and their slopes gives a practical means of measuring and comparing changes between two points on a curve.

Slopes of Curves: From Approximation to Precision

Slopes tell us how steep a curve is at a given point. For curves, the slope isn't constant, as it would be for straight lines. Let’s explore these concepts a bit more:
Tangent lines provide the exact slope at a single point. They touch the curve without crossing it,giving us the direction the curve is taking at that point.

• In the previous sections we discussed how the derivative \( f'(x) = 2x \) gives the exact slope at any point \( x \) on the curve \( y = x^2 + 1 \).

Secant lines, in contrast, connect two points on the curve, providing an average slope between those points. They are useful to approximate the curve’s behaviour over small intervals.
The crucial step of finding the tangent slope using the limit process involves approximating the slope of the secant lines as they get infinitely close.For example, as the secant line from \( x = 1 \) to \( x = 1.01 \) gets closer, using the formula

• \[ m = \lim_{{h \to 0}} \frac{f(1+h) - f(1)}{h} \]

allows us to approach the true slope of 2 at \( (1, 2) \).
Slopes of curves thus reveal deeper insights into the function's behaviour, giving both a local (tangent slopes) and broader picture (secant slopes) of how the function operates.