Problem 40 - Calculus Early Transcendentals

Question

Suppose a curve is described by \(y=f(x)\) on the interval \([-b, b],\) where \(f^{\prime}\) is continuous on \([-b, b] .\) Show that if \(f\) is symmetric about the origin ( \(f\) is odd) or \(f\) is symmetric about the \(y\) -axis ( \(f\) is even), then the length of the curve \(y=f(x)\) from \(x=-b\) to \(x=b\) is twice the length of the curve from \(x=0\) to \(x=b .\) Use a geometric argument and prove it using integration.

Step-by-Step Solution

Answer

Question: Show that if a curve is described by y = f(x) on the interval [-b, b] and f′ is continuous on that interval, then if f is either odd (symmetric about the origin) or even (symmetric about the y-axis), the length of the curve from x = -b to x = b is twice the length of the curve from x = 0 to x = b. Answer: Using both geometric arguments and integration, we have proved that for a curve described by y = f(x) on the interval [-b, b], if f is either an odd function (symmetric about the origin) or an even function (symmetric about the y-axis), the length of the curve from x = -b to x = b is indeed twice the length of the curve from x = 0 to x = b.

Step 1: Geometric Argument for Odd Functions (Symmetric about the Origin)

Since f is an odd function, it satisfies the property that f(-x) = -f(x) for all x. For every point on the curve y = f(x) on the interval [0, b], there is a corresponding point on the curve on the interval [-b, 0], which is obtained by reflecting the point across the origin. As a result, the length of the curve from x = -b to x = b is simply twice the length of the curve from x = 0 to x = b for odd functions.

Step 2: Geometric Argument for Even Functions (Symmetric about the Y-axis)

Since f is an even function, it satisfies the property that f(-x) = f(x) for all x. For every point on the curve y = f(x) between 0 and b, there is a corresponding point on the curve on the interval [-b, 0], which is obtained by reflecting the point across the y-axis. As a result, the length of the curve from x = -b to x = b is simply twice the length of the curve from x = 0 to x = b for even functions.

Step 3: Proof Using Integration

To prove the result using integration, we first find the length of the curve between x = -b and x = b, and then the length of the curve between x = 0 and x = b. The length of a curve, L, can be found by integrating the square root of 1 plus the square of the derivative of the function with respect to x, over the given interval. For the length from x = -b to x = b: \( L_{-b \to b} = \int_{-b}^{b} \sqrt{1 + [f'(x)]^2} \space dx\) For the length from x = 0 to x = b: \( L_{0 \to b} = \int_{0}^{b} \sqrt{1 + [f'(x)]^2} \space dx\) Now, we will consider both the case of odd and even functions separately: 1. For odd functions, we have: \(f(-x) = -f(x) \Rightarrow f'(-x) = -f'(x)\) We can rewrite the length from x = -b to x = b as: \( L_{-b \to b} = \int_{-b}^{b} \sqrt{1 + [f'(-x)]^2} \space (-dx) = 2 \int_{0}^{b} \sqrt{1 + [f'(x)]^2} \space dx = 2 L_{0 \to b}\) 2. For even functions, we have: \(f(-x) = f(x) \Rightarrow f'(-x) = f'(x)\) We can rewrite the length from x = -b to x = b as: \( L_{-b \to b} = \int_{-b}^{b} \sqrt{1 + [f'(x)]^2} \space dx = 2 \int_{0}^{b} \sqrt{1 + [f'(x)]^2} \space dx = 2 L_{0 \to b}\) In both cases, we have proved that the length of the curve y = f(x) from x = -b to x = b is twice the length of the curve from x = 0 to x = b for odd and even functions.

Key Concepts

Odd and Even Functions

Understanding odd and even functions is essential in analyzing symmetry in calculus. Odd functions have the unique property that for every point on the curve, the function satisfies \(f(-x) = -f(x)\). An example of an odd function is \(f(x) = x^3\), because if you take any point \(x\), then \(-x\) gives you the negative of the value at \(x\). This reflection across the origin indicates that if you know the behavior of the function on one side of the axis, you can determine the behavior on the opposite side by mirroring it through the origin.
On the other hand, even functions satisfy \(f(-x) = f(x)\). An example of an even function is \(f(x) = x^2\). This means that the graph of an even function is symmetric about the y-axis, and the left half of the graph mirrors the right half. Recognizing odd and even properties helps in calculating certain integrals and understanding the symmetry of graphs.

Symmetry in Calculus

Symmetry plays a crucial role in calculus as it simplifies both computation and analysis. There are two primary types of symmetry relevant to functions in calculus: symmetry about the origin and symmetry about the y-axis.

• Symmetry about the origin occurs with odd functions. This type of symmetry suggests that any transformation on one side of the function is reversed on the opposite side.
• Symmetry about the y-axis involves even functions. Here, every point on the function graph can be mirrored around the y-axis.

The impact of these symmetries becomes evident when calculating properties like the length of a curve or areas under the curve. As demonstrated in the exercise, the curve's length from \(-b\) to \(b\) simplifies due to symmetry. For odd or even functions, the length can be calculated by focusing only on one side of the axis and doubling it, a powerful simplification to speed up calculations. This understanding essentially cuts the computational effort in half by leveraging the inherent symmetry.

Integration in Calculus

Integration helps in calculating areas under curves and more complex figures like determining the length of a curve. When a function is defined between \(-b\) and \(b\), and we want to find the curve's length, we integrate using the formula:
\[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \ dx \]
This formula considers the derivative of the function, \(f'(x)\), to account for the curve's slope at each point, ensuring that inclines and declines are measured accurately. For symmetric functions:

• With odd functions, we use integration and symmetry rules (i.e., symmetric about the origin) to conclude that the integral from \(-b\) to \(b\) equals twice the integral from \(0\) to \(b\).
• For even functions, the symmetry about the y-axis leads to similar results, with both sides of the curve contributing equally to the overall length.

Mastering how to set up and evaluate integrals in such symmetric scenarios greatly enhances efficiency and clarity in solving calculus problems.