Problem 30 - Calculus: Early Transcendentals

Question

(a) If \( G(x) = 4x^2 - x^3 \), find \( G'(a) \) and use it to find equations of the tangent lines to the curve \( y = 4x^2 - x^3 \) at the points \( (2, 8) \) and \( (3, 9) \). (b) Illustrate part (a) by graphing the curve and the tangent lines on the same screen.

Step-by-Step Solution

Answer

The tangent lines are \( y = 4x \) at \( (2, 8) \) and \( y = -3x + 18 \) at \( (3, 9) \).

Step 1: Differentiate G(x)

First, we need to find the derivative of the function \( G(x) = 4x^2 - x^3 \). Applying the power rule, we differentiate term by term: \[ G'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(x^3) = 8x - 3x^2. \]

Step 2: Find the derivative at specific points

Now, we evaluate the derivative at the given points to find the slopes of the tangent lines. For \( x = 2 \), calculate \( G'(2) = 8(2) - 3(2)^2 = 16 - 12 = 4. \) For \( x = 3 \), calculate \( G'(3) = 8(3) - 3(3)^2 = 24 - 27 = -3. \)

Step 3: Write equations of tangent lines

We will use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope of the line.For the point \( (2, 8) \), use \( m = 4 \): \[ y - 8 = 4(x - 2) \]which simplifies to \[ y = 4x - 8 + 8 = 4x. \]For the point \( (3, 9) \), use \( m = -3 \): \[ y - 9 = -3(x - 3) \]which simplifies to \[ y = -3x + 9 + 9 = -3x + 18. \]

Step 4: Graph the curve and tangent lines

To visualize part (a), plot the original curve \( y = 4x^2 - x^3 \) along with the tangent lines \( y = 4x \) and \( y = -3x + 18 \). The curve represents the function, and the lines represent the tangent lines at the points \( (2, 8) \) and \( (3, 9) \). The slope of the tangent line at \( (2, 8) \) is positive, indicating an upwards slope, while at \( (3, 9) \), it is negative, indicating a decline.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which involves finding the rate at which a function is changing at any given point. This rate of change is known as the derivative. In our problem, we are asked to find the derivative of the function \( G(x) = 4x^2 - x^3 \). Differentiation gives us the slope of the tangent line to the curve at any point \( x \).
To find the derivative of \( G(x) \), we apply differentiation rules to each term of the function. This involves techniques like the power rule, which is especially powerful when dealing with polynomial functions. Differentiation helps in understanding how changes in \( x \) affect the function \( G(x) \).

• The derivative, \( G'(x) \), represents the slope of the tangent line at a particular point on the curve.
• Through differentiation, we can find critical points and analyze the behavior of the function around those points.

In our exercise, after differentiating, we obtain \( G'(x) = 8x - 3x^2 \), which we further evaluate at specific points to find the slope of tangent lines.

Power Rule

The power rule is a simple yet crucial rule in calculus that assists in finding the derivative of a function that consists of terms of the form \( x^n \). The rule states that if \( f(x) = x^n \), then the derivative \( f'(x) = n \cdot x^{n-1} \).

Using the power rule can greatly simplify the process of differentiation for functions, particularly polynomial ones. In this exercise, the function \( G(x) = 4x^2 - x^3 \) is a polynomial, and we can apply the power rule to differentiate it term by term.

• The term \( 4x^2 \) becomes \( 8x \) after differentiation, using the rule that \( n \cdot 4x^{2-1} = 8x \).
• The term \( -x^3 \) becomes \( -3x^2 \) after applying the power rule article, where \( n \cdot x^{2} = -3x^2 \).

This efficient method allows us to attain \( G'(x) = 8x - 3x^2 \), which tells us how the function \( G(x) \) changes with respect to \( x \), and helps in determining the slopes at specific points.

Point-Slope Form

The point-slope form is an essential formula in analytical geometry used to write the equation of a line given its slope and one point on the line. The point-slope equation is formulated as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the line.

It is particularly useful when dealing with tangent lines, as in our exercise, where the derivative gives us the slope, and the function provides the point. Once we determine the slope using differentiation, we employ the point-slope formula to find the equation of the tangent line.

• For the point \( (2, 8) \) with slope 4, the equation becomes \( y - 8 = 4(x - 2) \), simplifying to \( y = 4x \).
• For the point \( (3, 9) \) with slope -3, it becomes \( y - 9 = -3(x - 3) \), simplifying to \( y = -3x + 18 \).

This approach helps us easily derive the equations for tangent lines, which are paramount to understanding the instantaneous rate of change at specific points on the curve.