Problem 33 - Calculus Early Transcendentals

Question

Suppose the graph of \(f\) on the interval \([a, b]\) has length \(L,\) where \(f^{\prime}\) is continuous on \([a, b] .\) Evaluate the following integrals in terms of \(L\) a. \(\int_{a / 2}^{b / 2} \sqrt{1+f^{\prime}(2 x)^{2}} d x \quad\) b. \(\int_{a / c}^{b / c} \sqrt{1+f^{\prime}(c x)^{2}} d x\) if \(c \neq 0.\)

Step-by-Step Solution

Answer

Answer: The ratio of the arc lengths can be obtained by dividing the integral expression of Part (a) by the integral expression of Part (b). $$\frac{\int_{a/2}^{b/2} \sqrt{1+f^{\prime}(2x)^{2}} dx}{\int_{a/c}^{b/c} \sqrt{1+f^{\prime}(cx)^{2}} dx} = \frac{\frac{1}{2} L}{\frac{1}{c} L}$$ After canceling out the \(L\) terms, we get the ratio: $$\frac{\frac{1}{2}}{\frac{1}{c}} = \frac{c}{2}$$ So, the ratio of the arc lengths is \(\frac{c}{2}\).

Step 1: Part (a) - Substitution

To solve the integral \(\int_{a/2}^{b/2} \sqrt{1+f^{\prime}(2x)^{2}} dx\), substitute \(2x=u\). We need to find \(du\) in terms of \(dx\) as well as new limits of integration. Since \(2x = u\), we get \(dx = \frac{1}{2} du\). When \(x=a/2\), we have \(u=2(a/2)=a\), and when \(x=b/2\), we have \(u=2(b/2)=b\). So, the limits of integration become \(a\) and \(b\) after substitution.

Step 2: Part (a) - Evaluate the Integral

Using the substitution \(2x=u\) and the new limits of integration, we rewrite the integral: $$\int_{a/2}^{b/2} \sqrt{1+f^{\prime}(2x)^{2}} dx = \int_{a}^{b} \sqrt{1+f^{\prime}(u)^{2}}\frac{1}{2} du$$ We recognize that the integral is half the arc length on the interval \([a, b]\), which is given by \(L\). Thus, we get: $$\int_{a/2}^{b/2} \sqrt{1+f^{\prime}(2x)^{2}} dx = \frac{1}{2} L$$

Step 1: Part (b) - Substitution

Similarly, for the second part, substitute \(cx=v\). We need to find \(dv\) in terms of \(dx\) as well as new limits of integration. Since \(cx = v\), we get \(dx = \frac{1}{c} dv\). When \(x=a/c\), we have \(v=c(a/c)=a\), and when \(x=b/c\), we have \(v=c(b/c)=b\). So, the limits of integration become \(a\) and \(b\) after substitution.

Step 2: Part (b) - Evaluate the Integral

Using the substitution \(cx=v\) and the new limits of integration, we rewrite the integral: $$\int_{a/c}^{b/c} \sqrt{1+f^{\prime}(cx)^{2}} dx = \int_{a}^{b} \sqrt{1+f^{\prime}(v)^{2}}\frac{1}{c} dv$$ We recognize that the integral is \(\frac{1}{c}\) times the arc length on the interval \([a, b]\), which is given by \(L\). Thus, we get: $$\int_{a/c}^{b/c} \sqrt{1+f^{\prime}(cx)^{2}} dx = \frac{1}{c} L$$

Key Concepts

Arc Length

When discussing calculus, the concept of arc length is vital to understanding the curvature and overall distance along a curve. Imagine tracing a path along a winding river. The total path you travel, following every bend and curve, is akin to the arc length in mathematics. Mathematical formulas allow us to calculate this length precisely for a variety of functions and scenarios.

The arc length of the function's curve between two points is calculated using an integral. The formula is:

• For a function \( y = f(x) \), the arc length \( L \) is given by: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \ dx \]
• Where \( \frac{dy}{dx} \) is the derivative of the function, representing the slope at any given point.

To compute the arc length, one needs an integral that evaluates the path length between two points on a curve, accounting for the function's slope. Understanding these basics helps when applying careful calculations to solve more complex integrals emphasizing length.

Integrals

Integrals are foundational tools in calculus used for finding areas under curves and accumulating quantities. Imagine filling a container with water—the integral calculates the total amount of water required, considering every tiny bit along the way.

The definite integral is expressed as \( \int_{a}^{b} f(x) \ dx \), where \( a \) and \( b \) are the limits of integration. This process requires calculating the sum of infinitely many tiny areas between specific bounds. Key aspects include:

• **Limits of Integration:** They establish the range over which the function is evaluated.
• **Integrand:** The function \( f(x) \) to be integrated, indicative of change.
• **Area Calculation:** Integrals determine the total change or area that the function \( f(x) \) outlines from point \( a \) to point \( b \).

Critical in solving integrals is understanding the substitution method, which simplifies integrals by changing variables, making them easier to solve.

Substitution Method

The substitution method allows for simplification of complex integrals by changing variables, facilitating easier computation. Visualize rearranging pieces in a puzzle for smoother assembly—substitution in integrals does something similar by altering how the integral is approached.

When performing substitution:

• Identify a substitution such as \( u = g(x) \), which effectively transforms the integrand and simplifies integration.
• Compute the differential, \( du \), in terms of \( dx \), forming the foundation of the substitution.
• Change the limits of integration according to the new variable, adjusting them to fit the substitution context.
• Rewrite the integral with new expressions, focusing on developing a simpler integrand.

This method streamlines the process, allowing complicated integrals to become accessible, highlighting their relationships with standard functions easily solved through basic integration techniques. It is essential for students to practice this method to gain proficiency in solving a wide variety of integral problems effectively.