Problem 56 - Calculus: Early Transcendentals

Question

Find the area of the region bounded by the parabola \( y = x^2 \), the tangent line to this parabola at \( (1, 1) \), and the x-axis.

Step-by-Step Solution

Answer

The area is \( \frac{5}{24} \) square units.

Step 1: Find the Equation of the Tangent Line

To find the equation of the tangent line, first calculate the derivative of the parabola \( y = x^2 \), which is \( \frac{dy}{dx} = 2x \). At the point \((1, 1)\), the slope is \( 2 \cdot 1 = 2 \). Using the point-slope form of a line, the tangent line is \( y - 1 = 2(x - 1) \), simplifying to \( y = 2x - 1 \).

Step 2: Determine Points of Intersection

Set the equation of the parabola \( y = x^2 \) equal to the tangent line \( y = 2x - 1 \):\[ x^2 = 2x - 1 \]Rearrange to form a quadratic equation:\[ x^2 - 2x + 1 = 0 \]Factoring gives \((x-1)^2 = 0\), indicating they intersect at \( x = 1 \), \( y = 1 \). Check intersection with the x-axis by solving \( x^2 = 0 \) and \( 2x - 1 = 0 \) separately, giving \( x = 0 \) for the parabola and \( x = \frac{1}{2} \) for the tangent.

Step 3: Set Up Integrals for Area Calculation

The area between curves can be found using definite integrals over the appropriate intervals. The region of interest is bounded between \( x = 0 \) and \( x = \frac{1}{2} \). Set up the integral for the area between the tangent line and the x-axis (top function):\[ \int_{0}^{\frac{1}{2}} (2x - 1)\, dx \]And the integral between the parabola and the x-axis (bottom function):\[ \int_{0}^{\frac{1}{2}} x^2\, dx \].

Step 4: Evaluate the Integrals

Calculate the integral of the tangent line:\[ \int_{0}^{\frac{1}{2}} (2x - 1) \, dx = \left[x^2 - x\right]_{0}^{\frac{1}{2}} = \left(\frac{1}{4} - \frac{1}{2}\right) - (0) = -\frac{1}{4} \]Next, calculate the integral of the parabola:\[ \int_{0}^{\frac{1}{2}} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{\frac{1}{2}} = \frac{1}{24} \].

Step 5: Calculate the Enclosed Area

The area of the region is the difference between the area under the tangent line and the area under the parabola from \( x = 0 \) to \( x = \frac{1}{2} \):\[ \left(-\frac{1}{4}\right) - \left(\frac{1}{24}\right) = -\frac{6}{24} + \frac{1}{24} = -\frac{5}{24} \].Since area must be positive, take the absolute value: \( \frac{5}{24} \) square units.

Key Concepts

Parabola

A parabola is a U-shaped curve in a plane, represented by its equation, usually of the form \( y = ax^2 + bx + c \). In this exercise, the parabola is defined by the equation \( y = x^2 \). This specific parabola opens upwards, as indicated by the positive coefficient of \( x^2 \). Parabolas have distinctive properties that make them interesting in geometry and calculus:

• Vertex: The point \((0,0)\) serves as the vertex of the parabola \( y = x^2 \).
• Axis of Symmetry: It's a vertical line through the vertex, in this case, the y-axis itself.
• Directrix and Focus: Though not integral to this specific problem, these define another geometric relationship involving parabolas.

Understanding how a parabola behaves helps us to accurately set bounds when evaluating the area under or between curve(s). In this problem, the parabola provides one of the boundaries of the region for which we are calculating the area.

Tangent line

A tangent line is a straight line that just 'touches' a curve at a point, having the same slope as the curve at that point. For the parabola \( y = x^2 \), the tangent at the point \((1, 1)\) is determined by:
* **Finding the Derivative**: Derivative of \( y = x^2 \) is \( \frac{dy}{dx} = 2x \), giving the slope of the tangent line at any point \( x \). * **Slope at Point of Tangency**: At \((1, 1)\), the slope is 2, calculated by \( 2 \times 1 \). * **Equation of Tangent Line**: Using the point-slope form for a linear equation, we find \( y - 1 = 2(x - 1) \), simplifying to \( y = 2x - 1 \).
The tangent line \( y = 2x - 1 \) intersects the parabola at one point, and the x-axis serves as another boundary when considering the region between these entities. Knowing how to determine the equation of a tangent line is vital for solving problems involving curves and lines in calculus.

Definite integral

In calculus, a definite integral is used to calculate the exact area under a curve between two points. This becomes especially important when determining areas bounded by distinct curves, such as in this exercise. The formula for a definite integral \( \int_{a}^{b} f(x) \, dx \) represents the signed area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \). For calculating the area of the region, we:

• Identify the curves: \( y = 2x - 1 \) and \( y = x^2 \).
• Determine the bounds, such as \( x = 0 \) to \( x = \frac{1}{2} \).
• Calculate the integrals: Perform separate integrations for each curve over the specified intervals.
• Subtract the integral of the parabola from the integral of the tangent to find the area between them.

The use of definite integrals allows precise computation of areas that would otherwise be difficult to measure, especially when the boundaries are defined by curves rather than straight lines.

Intersection points

Intersection points are where two or more graphs meet. For this exercise, we find the intersection points to establish the boundaries of the region whose area we want to compute. An intersection point is determined mathematically by setting the equations of the curves equal and solving them.
To find intersection points between \( y = x^2 \) and \( y = 2x - 1 \), we equate both:

• \( x^2 = 2x - 1 \)
• Rearranging, we get \( x^2 - 2x + 1 = 0 \)
• This simplifies and factors to \((x-1)^2 = 0\).
• The solution here is \( x = 1 \), with corresponding \( y = 1 \).

This exercise also seeks intersection points with the x-axis, yielding \( x = 0 \) for the parabola and \( x = \frac{1}{2} \) for the tangent line. Understanding where each figure intersects is fundamental in understanding the shape of the area being calculated. It allows us to accurately define the limits for integration when determining the area bounded by these curves.