Problem 83 - Calculus Early Transcendentals

Question

Morphing parabolas The family of parabolas \(y=(1 / a)-x^{2} / a^{3}\) where \(a>0,\) has the property that for \(x \geq 0,\) the \(x\) -intercept is \((a, 0)\) and the \(y\) -intercept is \((0,1 / a) .\) Let \(A(a)\) be the area of the region in the first quadrant bounded by the parabola and the \(x\) -axis. Find \(A(a)\) and determine whether it is an increasing, decreasing, or constant function of \(a\)

Step-by-Step Solution

Answer

Question: Determine the area under the curve \(y = \frac{1}{a} - \frac{x^2}{a^3}\) for \(x \geq 0\) and state whether this area is an increasing, decreasing, or constant function of a. Answer: The area under the curve is \(A(a) = 1 - \frac{1}{3}a^2\), and A(a) is a decreasing function of a for \(a > 0\).

Step 1: Write the function to integrate

The function we want to integrate is \(y = \frac{1}{a} - \frac{x^2}{a^3}\).

Step 2: Find the integrand and integration limits

We need to integrate the function with respect to x within the limits from 0 to a.

Step 3: Perform the integration

Integrati‎ng the function with respect to x we have: \(\int y dx = \int ( \frac{1}{a} - \frac{x^2}{a^3}) dx = \frac{1}{a}\int ( 1 - ax^2) dx\) Now let us perform the integration: \(A(a) = \frac{1}{a} [\int ( 1 - ax^2) dx] = \frac{1}{a}[x - \frac{1}{3}ax^{3}]\) Now, let's find the definite integral for the bounds of integration between 0 and a: \(A(a) = \frac{1}{a} [(a - \frac{1}{3}a^3) - (0 - \frac{1}{3}(0)^3)] = \frac{1}{a}(a - \frac{1}{3}a^3)\) As a result, we get: \(A(a) = ( 1 - \frac{1}{3}a^2)\)

Step 4: Determine if A(a) is increasing, decreasing or constant

To find out whether the function A(a) is increasing, decreasing or constant, we need to find its first derivative with respect to a: \(\frac{dA(a)}{da} = \frac{d}{da}(1 - \frac{1}{3}a^2) = -\frac{2}{3}a\) Now, we can analyze the behavior of A(a) based on its first derivative: - If \(\frac{dA(a)}{da} > 0\), then A(a) is an increasing function of a. - If \(\frac{dA(a)}{da} < 0\), then A(a) is a decreasing function of a. - If \(\frac{dA(a)}{da} = 0\), then A(a) is a constant function of a. Since the first derivative \(\frac{dA(a)}{da} = -\frac{2}{3}a\), A(a) is a decreasing function of a for \(a > 0\). So, the area in the first quadrant bounded by the parabola and the x-axis is \(A(a) = 1 - \frac{1}{3}a^2\) and A(a) is a decreasing function of a for \(a > 0\).

Key Concepts

Parabolas

A parabola is a U-shaped curve that represents a quadratic equation. Understanding parabolas is crucial when learning calculus, as they often appear in various problems. In the given exercise, the parabola is represented by the equation \( y = \frac{1}{a} - \frac{x^2}{a^3} \). This equation describes a family of parabolas depending on the parameter \( a > 0 \). Notably, the x-intercepts and y-intercepts of a parabola give important insights into its orientation and position.

- The x-intercept for \( x \geq 0 \) is \((a, 0)\).
- The y-intercept is \((0, \frac{1}{a})\).
These intercepts help in sketching the parabola's shape and understanding its area in the first quadrant.

Integration

Integration is a fundamental concept in calculus that allows us to find areas under curves. In this exercise, the focus is to integrate the given parabola's equation over a specified interval. By integrating the function \( y = \frac{1}{a} - \frac{x^2}{a^3} \) with respect to \( x \) from 0 to \( a \), we're able to determine the area under the curve in the first quadrant.

The integration process involves:

• Identifying the function to integrate.
• Determining the integration limits, which are from \( x = 0 \) to \( x = a \).
• Calculating the integral: \( A(a) = \frac{1}{a} [ x - \frac{1}{3}ax^3 ] \)
• Evaluating the definite integral to find the area enclosed by the parabola and x-axis.

This process transitions from a function to a concrete value representing area, which is key for solving problems involving regions bound by curves.

Derivatives

Derivatives help us understand the rate at which a function changes. In our problem, once we've found an expression for the area \( A(a) \) as a function of \( a \), we can use derivatives to determine how changing \( a \) affects this area.

The original solution finds the first derivative of the area function, \( A(a) = 1 - \frac{1}{3}a^2 \), resulting in:

• \( \frac{dA(a)}{da} = -\frac{2}{3}a \)

The sign of this derivative indicates the behavior of \( A(a) \). Here, since \( \frac{dA(a)}{da} \, <\, 0 \) for \( a > 0 \), the area \( A(a) \) decreases as \( a \) increases. Understanding derivatives is essential for analyzing how functions change, especially when examining maximum, minimum, or decreasing-increasing behavior.

Definite Integrals

Definite integrals allow us to calculate the total accumulation of quantities, such as area, over a defined interval. In this exercise, the goal was to find the area under the parabola from \( x = 0 \) to \( x = a \). Using definite integrals, we can easily determine this area as a specific value.

This exercise reveals:

• The integral \( \int_{0}^{a} (\frac{1}{a} - \frac{x^2}{a^3}) \, dx \) simplifies to \( A(a) = 1 - \frac{1}{3}a^2 \).

When dealing with definite integrals, it’s essential to apply the upper and lower bounds after performing the antiderivative. This step ensures finding a precise numerical area: a powerful tool for engineers, physicists, and economists when dealing with quantities over intervals.