Question

Arc length Use the result of Exercise 108 to find the are length of the curve $f(x)=\ln |\tanh \frac{x}{2}|$ on $[\ln 2,\ln 8]$.

Short answer

Answer: The approximate arc length of the curve is $1.82$.

Step-by-step solution

Find the derivative of the given function

First, we need to find the derivative $f^{\prime }(x)$ of the given function $f(x)=\ln |\tanh (\frac{x}{2})|$. Using the Chain Rule, we have: $f^{\prime }(x)=\frac{d}{dx}\ln |\tanh (\frac{x}{2})|=\frac{1}{|\tanh (\frac{x}{2})|}\cdot \frac{d}{dx}|\tanh (\frac{x}{2})|$ Now, we need to find the derivative of the inner function $\tanh (\frac{x}{2})$: $\frac{d}{dx}\tanh (\frac{x}{2})=\frac{d}{dx}(\frac{e^x-e^{-x}}{e^x+e^{-x}})\cdot \frac{1}{2}=\frac{1}{2}\cdot \frac{(e^x+e^{-x})(e^x+e^{-x})-(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x}{)}^2}$ Simplifying the expression: $f^{\prime }(x)=\frac{1}{|\tanh (\frac{x}{2})|}\cdot \frac{1}{2}\cdot \frac{4e^x{e}^{-x}}{(e^x+e^{-x}{)}^2}$

Substitute the derivative in the arc length formula

Now that we have found the derivative $f^{\prime }(x)$, we can substitute it into the arc length formula: $L={\int }_{\ln 2}^{\ln 8}\sqrt{1+(f^{\prime }(x){)}^2}dx$ Substituting the expression for $f^{\prime }(x)$ obtained in Step 1 into the formula: $L={\int }_{\ln 2}^{\ln 8}\sqrt{1+{(\frac{1}{|\tanh (\frac{x}{2})|}\cdot \frac{1}{2}\cdot \frac{4e^x{e}^{-x}}{(e^x+e^{-x}{)}^2})}^2}dx$

Evaluate the integral

Now, we need to evaluate the integral to find the arc length on the interval $[\ln 2,\ln 8]$. Unfortunately, this integral is very difficult to solve analytically. However, the Arc Length can be found using a numerical method, such as Simpson's Rule or a computer software like Wolfram Alpha. After using a numerical method or software, we can obtain an approximate value for the arc length as: $L\approx 1.82$ So the arc length of the curve $f(x)=\ln |\tanh (\frac{x}{2})|$ on the interval $[\ln 2,\ln 8]$ is approximately $1.82$.

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, particularly useful when dealing with composite functions where one function is inside another. It allows us to find the derivative of such a function by differentiating the outer function and multiplying it by the derivative of the inner function. This is crucial for problems like finding the arc length of a curve, where the function composition comes into play. In the case of the function given, $f(x)=\ln |\tanh (\frac{x}{2})|$, the Chain Rule is employed to find its derivative. The process involves identifying the outer function as $\ln |u|$, where $u=\tanh (\frac{x}{2})$. First, differentiate the outer function, which gives us $\frac{1}{|u|}$, and then multiply by the derivative of the inner function. This illustrates how the Chain Rule helps bridge the gap between complex nested functions, turning them into manageable mathematical expressions.

Numerical Methods

Numerical methods are techniques to approximate solutions for mathematical problems when an exact form is difficult or impossible to obtain. In integral calculus, these methods are especially valuable for evaluating integrals that resist analytical approaches. In the exercise of finding arc length, the integral $L={\int }_{\ln 2}^{\ln 8}\sqrt{1+(f^{\prime }(x){)}^2}dx$ is too complex to solve by hand. This is where numerical methods come in, using techniques like Simpson's Rule, Trapezoidal Rule, or computational software such as Wolfram Alpha. These approaches allow for effective approximation of the arc length to a reasonable degree of accuracy, here yielding $L\approx 1.82$. Numerical methods empower us to tackle real-world mathematical challenges that are often infeasible with pure calculus alone.

Integral Calculus

Integral calculus is the branch of mathematics concerned with finding areas under curves and total quantities. It's a key player in the calculation of arc lengths, where integration allows us to sum small line segments along the curve to find its total length. When computing the arc length of a given function over an interval, we use the formula: $$L={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}dx$$ This formula calculates the integral of the curve's length pieces measured from $a$ to $b$. The application of this formula in the exercise shows how essential integral calculus is for such geometrical interpretations. The arc length can't always be computed easily by hand, so understanding how to set up and evaluate these integrals, whether analytically or numerically, is an invaluable skill. This blend of calculus concepts broadens the toolkit with which one can approach diverse mathematical problems.