Problem 9 - Calculus: Early Transcendentals

Question

(a) Find the slope of the tangent to the curve \( y = 3 + 4x^2 - 2x^3 \) at the point where \( x = a \). (b) Find equations of the tangent lines at the points \( (1, 5) \) and \( (2, 3) \). (c) Graph the curve and both tangents on a common screen.

Step-by-Step Solution

Answer

The slope at \( x = a \) is \( 8a - 6a^2 \); tangents are \( y = 2x + 3 \) at (1,5) and \( y = -8x + 19 \) at (2,3).

Step 1: Differentiate the Function

To find the slope of the tangent, we need to first differentiate the given function with respect to \( x \). The function is \( y = 3 + 4x^2 - 2x^3 \). Differentiate term by term. \[ \frac{dy}{dx} = \frac{d}{dx}(3) + \frac{d}{dx}(4x^2) - \frac{d}{dx}(2x^3) \] This simplifies to \[ \frac{dy}{dx} = 0 + 8x - 6x^2 \] Thus, the derivative function is \( \frac{dy}{dx} = 8x - 6x^2 \).

Step 2: Find the Slope at x = a

The slope of the tangent at any point \( x = a \) is given by the derivative \( \frac{dy}{dx} \) evaluated at \( x = a \). Using the result from Step 1:\[ \text{Slope at } x = a = 8a - 6a^2 \].

Step 3: Compute the Slope at Specific Points

Now compute the slope using \( 8x - 6x^2 \) at the given points:- For \( x = 1 \): \[ \frac{dy}{dx} = 8(1) - 6(1)^2 = 8 - 6 = 2 \]- For \( x = 2 \): \[ \frac{dy}{dx} = 8(2) - 6(2)^2 = 16 - 24 = -8 \]

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line.- At point \( (1,5) \), the slope is 2: \[ y - 5 = 2(x - 1) \] Simplifying gives: \[ y = 2x + 3 \]- At point \( (2,3) \), the slope is -8: \[ y - 3 = -8(x - 2) \] Simplifying gives: \[ y = -8x + 19 \]

Step 5: Graph the Curve and Tangent Lines

To graph the function \( y = 3 + 4x^2 - 2x^3 \) and the tangents \( y = 2x + 3 \) and \( y = -8x + 19 \), plot them on the same set of axes. The curve will be a cubic polynomial, and the lines will be straight lines tangent to this curve at the specified points.

Key Concepts

Derivative

A derivative represents the rate of change of a function with respect to a variable. When we find the derivative of a function, we're essentially calculating the function's slope at any given point. For a simple function like a straight line, the slope is constant. However, for curves like cubic functions, the slope can vary at different points on the curve.

To determine the derivative of a function such as \( y = 3 + 4x^2 - 2x^3 \), we differentiate each term:

• For the constant \( 3 \), the derivative is \( 0 \) since constants do not change.
• The derivative of \( 4x^2 \) is \( 8x \), found by multiplying the power of \( x \) by its coefficient and reducing the power by one.
• Similarly, the derivative of \( 2x^3 \) is \(-6x^2 \), calculated in the same way.

Putting these derivatives together, we have \( \frac{dy}{dx} = 8x - 6x^2 \), which represents the slope of the tangent line at any point "a" on the curve.

The slope, \( 8a - 6a^2 \), can then be used to find how steep the curve is at any specific \( x \)-value.

Tangent Line Equation

To determine the equation of a tangent line to a curve, we use the point-slope form of a line \( y - y_1 = m(x - x_1) \). Here, \( m \) is the slope of the tangent, while \( (x_1, y_1) \) is a point on the curve.

Let's break it down with the examples given:

• At the point \((1, 5)\), the slope \( m \) from our previous calculation is 2. So the equation becomes: \( y - 5 = 2(x - 1) \). Simplifying this gives the tangent line: \( y = 2x + 3 \).
• At the point \((2, 3)\), the slope is \(-8\). Thus, the equation turns into: \( y - 3 = -8(x - 2) \), which simplifies to \( y = -8x + 19 \).

These equations show how the characteristics of cubic functions produce varying slopes and, hence, different tangent lines at different points.

Understanding and working with tangent lines is crucial because they provide instantaneous rates of change and are foundational in calculus.

Graphing Cubic Functions

Cubic functions, like \( y = 3 + 4x^2 - 2x^3 \), create interesting curves that can have turning points, and they often appear in the form of a "S" shape or inverted forms. Graphing them visually highlights their behavior over an interval.

To graph a cubic function:

• Begin by identifying key features such as intercepts, turning points, and asymptotic behavior if applicable.
• Use a graphing tool or plot manually by calculating a few key values for \( x \) to see how \( y \) changes.
• Place your calculated tangent lines on the graph to see how they intersect the curve at the specific points.

In our exercise, you'd graph the cubic curve and then add the tangent lines \( y = 2x + 3 \) and \( y = -8x + 19 \) as straight lines that touch the curve at the points \((1, 5)\) and \((2, 3)\) respectively.

Visualizing these elements on a graph enhances the understanding of cubic functions and their properties, making it easier to see the dynamics between the curve and its tangent lines.