Problem 30 - Calculus Early Transcendentals

Question

Find the arc length of the following curves by integrating with respect to \(y\) $$x=2 e^{\sqrt{2} y}+\frac{1}{16} e^{-\sqrt{2} y}, \text { for } 0 \leq y \leq \frac{\ln 2}{\sqrt{2}}$$

Step-by-Step Solution

Answer

Answer: The arc length is given by the expression: $$\sin^{-1}\left(\frac{1}{8\sqrt{2}}\right)- \sin^{-1}\left(\frac{45}{8\sqrt{2}}\right)$$

Step 1: Find \(\frac{dx}{dy}\)

To find the derivative of \(x\) with respect to \(y\), we calculate: $$\frac{dx}{dy}=\frac{d}{dy} \left(2 e^{\sqrt{2} y}+\frac{1}{16} e^{-\sqrt{2} y}\right)$$ Using the chain rule on this expression gives: $$\frac{dx}{dy} = 2\sqrt{2}e^{\sqrt{2} y}-\frac{1}{8\sqrt{2}}e^{-\sqrt{2} y}$$

Step 2: Calculate \(1+(\frac{dx}{dy})^2\)

Next, we need to calculate the expression inside the square root. $$1+\left(\frac{dx}{dy}\right)^2 = 1 + \left(2\sqrt{2}e^{\sqrt{2} y}-\frac{1}{8\sqrt{2}}e^{-\sqrt{2} y}\right)^2 $$

Step 3: Finding the Arc Length

The arc length is given by the following integral: $$\int_{0}^{\frac{\ln 2}{\sqrt{2}}} \sqrt{1+(\frac{dx}{dy})^2} dy= \int_{0}^{\frac{\ln 2}{\sqrt{2}}} \sqrt{1+\left(2\sqrt{2}e^{\sqrt{2}y}-\frac{1}{8\sqrt{2}}e^{-\sqrt{2}y}\right)^2} dy$$ We can use a substitution to simplify the integral. Let $$u = 2\sqrt{2}e^{\sqrt{2}y}-\frac{1}{8\sqrt{2}}e^{-\sqrt{2}y}$$ Therefore, $$du = \left(2\sqrt{2}e^{\sqrt{2}y}+\frac{1}{8\sqrt{2}}e^{-\sqrt{2}y}\right)dy = \left(\frac{dx}{dy}\right)dy$$ Now, the integral becomes: $$\int_{0}^{\frac{\ln 2}{\sqrt{2}}} \sqrt{1+(\frac{dx}{dy})^2} dy = \int_{u(0)}^{u\left(\frac{\ln 2}{\sqrt{2}}\right)} \sqrt{1+u^2} \frac{du}{\frac{dx}{dy}}$$ Notice that $$u\left(0\right)= 2\sqrt{2}e^{\sqrt{2}(0)}-\frac{1}{8\sqrt{2}}e^{-\sqrt{2}(0)} = 2\sqrt{2}-\frac{1}{8\sqrt{2}} = \frac{45}{8\sqrt{2}}$$ and $$u\left(\frac{\ln 2}{\sqrt{2}}\right) = 2\sqrt{2}e^{\ln \frac{\sqrt{2}}{2}}-\frac{1}{8\sqrt{2}}e^{-\ln \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{8}-\frac{1}{8\sqrt{2}} = \frac{1}{8\sqrt{2}}$$ So the integral becomes: $$\int_{\frac{45}{8\sqrt{2}}}^{\frac{1}{8\sqrt{2}}} \sqrt{1+u^2} \frac{1}{\sqrt(1-u^2)}du = \int_{\frac{45}{8\sqrt{2}}}^{\frac{1}{8\sqrt{2}}} \frac{du}{\sqrt{1-u^2}}$$ This integral is the difference of inverse sines: $$\int_{\frac{45}{8\sqrt{2}}}^{\frac{1}{8\sqrt{2}}} \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}\left(\frac{1}{8\sqrt{2}}\right)- \sin^{-1}\left(\frac{45}{8\sqrt{2}}\right)$$ Thus, the arc length is $$\sin^{-1}\left(\frac{1}{8\sqrt{2}}\right)- \sin^{-1}\left(\frac{45}{8\sqrt{2}}\right)$$

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus focused on the process of integration, which, in a geometrical context, can be seen as finding the area under a curve. While differential calculus focuses on rates of change and slopes of curves, integral calculus deals with the accumulation of quantities and the areas between curves.

In the context of arc length, integral calculus plays a key role. To find the arc length of a curve, we integrate a function that represents the length of an infinitesimally small segment of the curve. This segment's length is typically found using the Pythagorean theorem in the infinitesimal form, resulting in the integrand \(\sqrt{1 + (\frac{dx}{dy})^2}\) for integration with respect to \(y\).

In our exercise, the arc length of a given curve is determined by integrating this function over the interval specified for \(y\), which requires evaluating a somewhat complex square root function. Algebraic manipulation and substitution are often used to simplify the integral before solving it.

Chain Rule

The chain rule is a fundamental theorem in calculus used to differentiate compositions of functions. It is essential when dealing with complex functions that are made up of simpler functions, particularly when these simpler functions are nested within each other.

The chain rule can be expressed as \(\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\), where \(z\) is a function of \(y\), and \(y\) is a function of \(x\). When differentiating a function like \(2e^{\sqrt{2} y} + \frac{1}{16} e^{-\sqrt{2} y}\), the chain rule allows us to calculate the derivative \(\frac{dx}{dy}\) by recognizing the composition of the exponential and square root functions.

The steps provided in the solution show how to apply the chain rule to find the derivative of each term in the function \(x\) with respect to \(y\). The first term, \(2e^{\sqrt{2} y}\), is the composition of an exponential function and the function \(\sqrt{2} y\), while the second term, \(\frac{1}{16} e^{-\sqrt{2} y}\), is similar but involves a negative exponent. The chain rule is crucial in finding \(\frac{dx}{dy}\) correctly for use in the arc length formula.

Exponential Functions

Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a constant called the base, and \(x\) is the exponent. These functions are characterized by a constant rate of growth or decay and appear frequently in various scientific disciplines, including physics, finance, and biology.

In our example, \(2e^{\sqrt{2} y}\) and \(\frac{1}{16} e^{-\sqrt{2} y}\) are exponential functions with the base \(e\), which is the natural logarithm base approximately equal to 2.71828. Exponential functions with this base are particularly important in calculus because they have a unique property: the rate of change (derivative) of \(e^x\) with respect to \(x\) is equal to \(e^x\) itself.

The use of exponential functions in the arc length formula complicates the integrand, but it also provides a connection to real-world problems, where natural growth and decay processes can be modeled with these functions.

Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arcus functions or anti-trigonometric functions, are the inverse operations of the basic trigonometric functions. They are used to find the angles when the value of the trigonometric function is known.

There are six primary inverse trigonometric functions: \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), \(\tan^{-1}(x)\), \(\csc^{-1}(x)\), \(\sec^{-1}(x)\), and \(\cot^{-1}(x)\). In the context of arc length, we come across the inverse sine function, represented as \(\sin^{-1}(x)\) or \(\arcsin(x)\), which gives us the angle whose sine value is \(x\).

The final step of the provided solution involves evaluating the integral to find the difference between inverse sines. The limits of integration are obtained by substituting the bounds of \(y\) into our substituted function \(u\), after which the evaluation yields the length of the curve in terms of these inverse sine values. The inverse trigonometric functions, in this case, make it possible to express a complex-looking integral in terms of more familiar trigonometric terms.