Problem 36 - Calculus Early Transcendentals

Question

Bernoulli's “parabolas" Johann Bernoulli (1667-1748) evaluated the arc length of curves of the form \(y=x^{(2 n+1) / 2 n},\) where \(n\) is a positive integer, on the interval \([0, a].\) a. Write the arc length integral. b. Make the change of variables \(u^{2}=1+\left(\frac{2 n+1}{2 n}\right)^{2} x^{1 / n}\) to obtain a new integral with respect to \(u.\) c. Use the Binomial Theorem to expand this integrand and evaluate the integral. d. The case \(n=1\left(y=x^{3 / 2}\right)\) was done in Example \(1 .\) With \(a=1,\) compute the arc length in the cases \(n=2\) and \(n=3\) Does the arc length increase or decrease with \(n\) ? e. Graph the arc length of the curves for \(a=1\) as a function of \(n.\)

Step-by-Step Solution

Answer

Answer: The arc length increases as n increases.

Step 1: Write the arc length integral

The first step is to determine the arc length integral for the curve given by \(y=x^{(2n+1)/2n}\). The formula for the arc length of a curve \(y = f(x)\) on an interval \([a, b]\) is given by: $$L=\int_{a}^{b} \sqrt{1+(f'(x))^2} dx$$ First, we need to compute the derivative of \(y(x)\): $$y'(x) = \frac{d}{dx} \left(x^{(2n+1)/2n}\right) = \frac{2n+1}{2n} x^{(2n+1)/2n-1}.$$ Now, we can insert the function's derivative into the arc length equation and integrate it over the given interval [0, a]: $$L=\int_{0}^{a} \sqrt{1+\left( \frac{2n+1}{2n}x^{\frac{2n+1}{2n}-1} \right)^2} dx.$$

Step 2: Change of variables

Next, we will make the change of variables \(u^2=1+\left(\frac{2n+1}{2n}\right)^2 x^{\frac{1}{n}}\). First, we need to differentiate this equation with respect to \(x\) to obtain \(du\): $$2u\frac{du}{dx} = \frac{1}{n}\left(\frac{2n+1}{2n}\right)^2 x^{\frac{1}{n}-1}$$ Now, we can solve for \(dx\): $$dx=\frac{2nu}{\frac{2n+1}{2n}\left(\frac{2n+1}{2n}\right)^2 x^{\frac{1}{n}-1}}du$$ Substitute this into the integral from Step 1: $$L=\int \sqrt{1+\left( \frac{2n+1}{2n}\right)^2 x^{\frac{2n+1}{2n}-1}}\frac{2nu}{\frac{2n+1}{2n}\left(\frac{2n+1}{2n}\right)^2 x^{\frac{1}{n}-1}}du$$ After cancelling terms, we obtain: $$L=\int 2n u \sqrt{u^{2}-1} du$$

Step 3: Expand the integrand using the Binomial Theorem

To evaluate the integral, we will expand the integrand using the Binomial Theorem. The Binomial theorem is: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ Although in this context, we apply the formula for the square root of \((u^2-1)\), so we have: $$\sqrt{u^2-1} = (u^2-1)^{1/2} = \sum_{k=0}^{\infty} \binom{1/2}{k} (u^2)^{1-2k} (-1)^k$$ Now, we can substitute this into our integral from Step 2: $$L=\int 2n u \sum_{k=0}^{\infty} \binom{1/2}{k} (u^2)^{1-2k} (-1)^k du$$ We can then change the order of summation and integration: $$L = 2n \sum_{k=0}^{\infty} \binom{1/2}{k} (-1)^k \int u^{3-4k} du$$ Finally, we can evaluate the integral: $$L = 2n \sum_{k=0}^{\infty} \binom{1/2}{k} (-1)^k \frac{u^{4-4k}}{4-4k}$$

Step 4: Investigate the cases n=1, n=2, and n=3

First, let's compute the arc length for n=2 and n=3 when a=1. For n=2, we'll substitute this value into the expression of L and then substitute the limits of integration [1, 1]: $$L_{n=2} = 5 \sum_{k=0}^{\infty} \binom{1/2}{k} (-1)^k \frac{u^{4-4k}}{4-4k} \Big|_{1}^{1} = 5 \sum_{k=0}^{\infty} \binom{1/2}{k} \frac{(-1)^k}{4-4k}$$ For n=3, we do the same: $$L_{n=3} = \frac{9}{2} \sum_{k=0}^{\infty} \binom{1/2}{k} (-1)^k \frac{u^{4-4k}}{4-4k} \Big|_{1}^{1} = \frac{9}{2} \sum_{k=0}^{\infty} \binom{1/2}{k} \frac{(-1)^k}{4-4k}$$ To determine if the arc length increases or decreases with n, we can compare the expressions for L. We can see that the arc length increases as n increases.

Step 5: Graph the arc length of the curves

We are asked to graph the arc length of the curves for a=1 as a function of n. To do this, we can use the expression for L that we derived in Step 3: $$L = 2n \sum_{k=0}^{\infty} \binom{1/2}{k} (-1)^k \frac{u^{4-4k}}{4-4k}$$ Evaluate the expression for different values of n and plot the resulting points. Join them with a smooth curve to visualize the relationship between the arc length and n.

Key Concepts

Bernoulli's Parabolas

The concept of Bernoulli's parabolas involves investigating a family of curves given by the formula \(y = x^{(2n+1)/2n}\). These curves are named after Johann Bernoulli, a prominent Swiss mathematician from the late 17th and early 18th century. Bernoulli's work focused on a variety of forms, one of which included exploring the arc length of these parabolas. This type of curve is particularly interesting because its form changes with the integer \(n\), influencing the curvature and overall shape of the parabola.

• The curves are defined over the interval \([0, a]\)
• The shape depends on whether \(n\) is small or large

For each curve, finding the arc length involves calculus techniques like differentiation and integration, often necessitating advanced methods due to the complexity of the equations involved.

Change of Variables

To solve the arc length problem for Bernoulli's parabolas, a common method employed is the "change of variables" technique. In calculus, changing variables is a strategy to simplify an integral into an easier form to evaluate.
For Bernoulli's parabolas, we use the substitution \(u^2 = 1 + \left(\frac{2n+1}{2n}\right)^2 x^{1/n}\) to transform the arc length integral.

• This substitution helps to rearrange the complex expression into a manageable form
• Differentiation of \(u^2\) is necessary to replace \(dx\) in the integral with terms of \(du\)

By using this substitution, the structure of the integral changes, often making it possible to apply other mathematical techniques such as the Binomial Theorem, which aids in evaluation.

Binomial Theorem

The Binomial Theorem is a versatile mathematical tool that is extremely useful in expanding expressions that have exponents. It's given by \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In the context of Bernoulli's parabolas, it aids in expanding the term \((u^2 - 1)^{1/2}\), which arises from the change of variables substitution in the arc length formula.
The power series expansion converts this expression into a series summing a sequence of terms which then can be integrated individually. This approach simplifies the integral by addressing each term in the series separately:

• Reduces a complex problem into a sequence of simpler calculations
• Each term can be integrated over the given interval

Using the Binomial Theorem thus streamlines the process of solving integrals, particularly for complex polynomials or irrational functions as seen in this exercise.

Calculus Integration

Integration is a core component of calculus, used here to find the arc length of Bernoulli's parabolas. Integrating a function involves calculating the area under its curve on a defined interval, answering questions about lengths, areas, and volumes.
In this exercise, once we have transformed the integral and expanded it using the Binomial Theorem, calculating the integral involves evaluating each term of the expanded series. For the integral calculations:

• Apply integration techniques to each term separately
• The limits of integration provide the bounds for calculating the total area

Once solved, the sum of these integrals gives the arc length of the curve over the specific interval \([0, a]\). The resulting integral expression indicates how the arc length changes with parameters, like \(n\), in Bernoulli's parabolas, providing insights into the geometric properties of these mathematical constructs.