Problem 61 - Calculus of a Single Variable

Question

Using a tangent Line In Exercises \(61-64,\) set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point. $$ f(x)=x^{3}, \quad(1,1) $$

Step-by-Step Solution

Answer

The area of the region bounded by the graph of the function \(f(x) = x^{3}\) and the tangent line to the graph at the given point (1,1) is 0.75 square units.

Step 1: Find the Derivative of the Function

Find the derivative of \(f(x) = x^{3}\). The derivative is \(f'(x) = 3x^{2}\).

Step 2: Calculate the slope of tangent line

The slope of the tangent line will be given by the derivative at the point we're interested in. For the point (1,1), the derivative of \(f(x) = x^{3}\) is \(f'(1) = 3(1)^{2} = 3\). This is the slope of the tangent line.

Step 3: Find the Equation of Tangent Line

To find the equation of the tangent line, use the slope obtained in Step 2 along with the given point. Apply the point-slope form of a linear equation, \(y-y_{1}=m(x-x_{1})\), where \(m\) is the slope and \((x_{1},y_{1})\) is the point. Substituting, the equation becomes \(y-1=3(x-1)\). Simplified, this becomes \(y = 3x - 2\). This is the equation of the tangent line.

Step 4: Set up the definite integral

To find the area between the curve and the tangent line, calculate the integral of the absolute difference of the function and the tangent line equation from \(x=0\) to \(x=1\). This gives: \(\int_{0}^{1}|x^3 - (3x - 2)| dx\).

Step 5: Evaluate the Definite Integral

Now we can evaluate the integral from Step 4. As \(x^3 > 3x - 2\) for \(x\) in the interval \([0, 1]\), the absolute value is not required. Therefore, we have: \(\int_{0}^{1} (x^3 - (3x - 2)) dx = [0.25x^4 - 1.5x^2 + 2x]_{0}^{1} = 0.25 - 1.5 + 2 = 0.75\).

Key Concepts

Tangent Line Equation

Understanding the tangent line equation is fundamental in calculus, especially when studying the behavior of curves. At a given point, the tangent line represents the instantaneous direction of the function. To work out the equation of a tangent line, we start by finding the derivative of the function at the desired point. This derivative gives us the slope.

In the exercise discussed, the derivative of the function at the point (1,1) yields a slope of 3. Using this information, along with the point-slope form, gives us a linear equation that characterizes our tangent line. Thus, through a tangible application, we illustrate the interplay between a function and its tangent, bridging the derivative and the geometric manifestation.

Derivative of a Function

The derivative of a function represents the rate at which the function's output changes with respect to its input. In our exercise, the derivative is found by applying the power rule to the function f(x) = x³, resulting in f'(x) = 3x². Through this process, we can uncover the slope of the tangent line at any point along the function's curve. For those seeking to deepen their comprehension, consider how the derivative visually traces the steepness of the tangent line as one slides along the curve, reinforcing the concept that the derivative epitomizes the function's instantaneous rate of change.

Point-Slope Form

The point-slope form is a straightforward way of writing the equation for a line when you have a point on that line and know its slope. The formula is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) are the coordinates of the given point. After substituting the known values, you transform this formula into the slope-intercept form, making it easier to graph. In our case, knowing the point (1,1) and slope 3, we were able to quickly derive the linear equation that graphically represents the tangent line.

Evaluate Definite Integral

Calculating the definite integral of a function provides the area under the curve between two specific points. This concept is quintessential when determining the space enclosed between curves or a curve and a straight line, such as a tangent. During the problem-solving process, we start by setting the integral up, usually based on the limits determined by the points of intersection or relevance. Through integration, we then distill this to a concrete value representing the measure of the area sought. As seen in the exercise, after integrating, we calculate the difference at the boundary points leading to a final value of 0.75, which is the area under the curve and above the tangent line from x=0 to x=1.

By conceptually digesting these steps, students can better appreciate the power of definite integrals beyond memorization, recognizing them as tools to quantify real-life scenarios.