Problem 4 - Calculus: Early Transcendentals

Question

(a) Find the slope of the tangent line to the curve \( y = x - x^3 \) at the point \( (1, 0) \) (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at \( (1, 0) \) until the curve and the line appear to coincide.

Step-by-Step Solution

Answer

The slope of the tangent line at (1, 0) is -2, and the equation is \( y = -2x + 2 \).

Step 1: Differentiate the function

First, we find the derivative of the function \( y = x - x^3 \). The derivative \( y' \) is calculated as follows:\[y' = \frac{d}{dx}(x - x^3) = 1 - 3x^2.\]

Step 2: Compute the slope using Definition 1

Definition 1 involves the derivative from first principles, i.e., using the limit definition:\[m = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]Substituting \( f(x) = x - x^3 \) and evaluating at \( x = 1 \):\[m = \lim_{h \to 0} \frac{(1+h) - (1+h)^3 - (1-1^3)}{h}\]After simplification, this leads to \( m = 1 - 3(1)^2 = -2 \).

Step 3: Verify slope using Equation 2

Substitute \( x = 1 \) into the derivative obtained in Step 1:\[m = 1 - 3(1)^2 = 1 - 3 = -2\]Thus the slope is \( -2 \).

Step 4: Find the tangent line equation

The equation of the tangent line is \( y = mx + c \), where \( m = -2 \). We know the point \( (1,0) \) lies on this line.Using the point-slope form \( y - y_1 = m(x - x_1) \), we have:\[y - 0 = -2(x-1)\].Simplifying gives \( y = -2x + 2 \).

Step 5: Graph the curve and tangent

To graph the curve \( y = x - x^3 \) and the line \( y = -2x + 2 \), center the window at \( (1, 0) \) and adjust the viewing rectangle size smaller until the curve and the line appear to coincide. This involves using a graphing calculator or software, progressively zooming in while focusing around the point \( (1,0) \).

Key Concepts

Slope of Tangent Line

The slope of a tangent line at a specific point on a curve represents how steep the curve is at that point. To find this slope, you evaluate the derivative of the function at the given point. In our example, the slope of the tangent line for the curve defined by the equation \( y = x - x^3 \) is evaluated at the point \( (1, 0) \). To determine the slope, we first find the derivative of the function, \( y' = 1 - 3x^2 \), and then substitute \( x = 1 \) to find the slope: \( m = 1 - 3 \times 1^2 = -2 \). This slope tells us the rate of change of the function at that specific point, and hence, the slope of the tangent line is \(-2\).

Derivative

The concept of a derivative is fundamental in calculus. It quantifies the rate at which a function is changing at any given point. In other words, derivatives provide us with the slope of the tangent line to the curve at any chosen point.To compute a derivative, you often use differentiation rules. For our function \( y = x - x^3 \), the derivative \( y' \) is found by differentiating each term separately:

• Derivative of \( x \) is \( 1 \)
• Derivative of \(-x^3\) is \(-3x^2\)

This results in a derivative \( y' = 1 - 3x^2 \), which provides a general formula for the slope of the tangent at any value of \( x \).

Limit Definition of Derivative

The limit definition of a derivative is a formal way of defining the derivative of a function. It expresses the derivative as a limit:\[ m = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This formula calculates the slope of the tangent by considering the limit as \( h \), a small increment in \( x \), approaches zero. It captures the precise instantaneous rate of change of the function.In our exercise, this method was used to find the slope of the tangent line at \( x = 1 \). After substituting the function \( f(x) = x - x^3 \) into the limit definition and simplifying, we are led to the same derivative slope as before: \(-2\). While often more complex to work through manually, this approach confirms the accuracy of derivative rules.

Point-Slope Form

Once the slope of the tangent line is known and a point on this line is given, the equation of the tangent line can be derived using the point-slope form:\[ y - y_1 = m(x - x_1) \]Here, \( (x_1, y_1) \) is a known point on the line, and \( m \) is the slope of the line. Using our example, the slope \( m = -2 \) and the point \( (1, 0) \) gives the equation:\[ y - 0 = -2(x - 1) \]Simplifying this, we get:\[ y = -2x + 2 \]This equation describes the tangent line to the curve \( y = x - x^3 \) at the point \( (1, 0) \), closely representing the behavior of the curve at this precise location.