Chapter 2: Problem 9
(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.
Short Answer
Expert verified
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 by step solution
01
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 \( \frac{dy}{dx} = 8x - 6x^2 \).
02
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:\[ \text{Slope at } x = a = 8a - 6a^2 \].
03
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 \( x = 2 \): \[ \frac{dy}{dx} = 8(2) - 6(2)^2 = 16 - 24 = -8 \]
04
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: \[ y = -8x + 19 \]
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
The slope, \( 8a - 6a^2 \), can then be used to find how steep the curve is at any specific \( x \)-value.
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.
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:
Understanding and working with tangent lines is crucial because they provide instantaneous rates of change and are foundational in calculus.
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 \( y = -8x + 19 \).
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:
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.
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.
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.