Problem 81 - Thomas Calculus

Question

Find the lengths of the following curves. $$ \begin{array}{l}{\text { a. } y=\left(x^{2} / 8\right)-\ln x, \quad 4 \leq x \leq 8} \\ {\text { b. } x=(y / 4)^{2}-2 \ln (y / 4), \quad 4 \leq y \leq 12}\end{array} $$

Step-by-Step Solution

Answer

Integrate the functions from the given limits using arc length formulas: Part (a) from 4 to 8 and Part (b) from 4 to 12.

Step 1: Formula for Arc Length

To find the length of a curve described by the function \( y = f(x) \) from \( x = a \) to \( x = b \), we use the formula for arc length: \[ L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx.\] Similarly, for a curve described by \( x = g(y) \) from \( y = c \) to \( y = d \), we use: \[ L = \int_{c}^{d} \sqrt{1 + \left(g'(y)\right)^2} \, dy.\] We'll apply these formulas to each part of the problem separately.

Step 2: Calculate Derivative for Part (a)

For part (a), the curve is \( y = \frac{x^2}{8} - \ln x \). We need to find \( f'(x) \):\[ f'(x) = \frac{d}{dx} \left( \frac{x^2}{8} - \ln x \right) = \frac{1}{8} \cdot 2x - \frac{1}{x} = \frac{x}{4} - \frac{1}{x}. \]

Step 3: Set Up Integral for Part (a)

Now we substitute \( f'(x) \) into the arc length formula:\[ L = \int_{4}^{8} \sqrt{1 + \left(\frac{x}{4} - \frac{1}{x}\right)^2} \, dx. \] This simplifies to:\[ L = \int_{4}^{8} \sqrt{1 + \left(\frac{x^2}{16} - \frac{2}{4} + \frac{1}{x^2}\right)} \, dx. \] Simplify and solve this integral to obtain the arc length.

Step 4: Calculate Derivative for Part (b)

For part (b), the curve is \( x = \left(\frac{y}{4}\right)^2 - 2 \ln \left(\frac{y}{4}\right) \). We need \( g'(y) \):\[ g'(y) = \frac{d}{dy} \left( \left(\frac{y}{4}\right)^2 - 2 \ln \left(\frac{y}{4}\right) \right) = \frac{y}{8} - \frac{2}{y}. \]

Step 5: Set Up Integral for Part (b)

Substitute \( g'(y) \) into the arc length formula for part (b):\[ L = \int_{4}^{12} \sqrt{1 + \left(\frac{y}{8} - \frac{2}{y}\right)^2} \, dy. \] Simplify this expression and evaluate the integral to find the arc length.

Key Concepts

Understanding Derivatives

Derivatives are a crucial concept in calculus that help us understand how functions change. They measure the rate at which one quantity changes with respect to another. In simpler terms, if you think of a function as describing a curve, the derivative tells us the steepness or slope of that curve at any given point.

To find the derivative of a function, we look at the smallest change in the function's output as a result of a tiny change in its input. This is mathematically captured using limits. When we compute the derivative, we're essentially calculating the function's rate of change at an infinitesimal point.

• To find the derivative of a function like \(y = \frac{x^2}{8} - \ln x\), we apply the rules of differentiation. For example, the power rule helps us with terms like \(\frac{x^2}{8}\), and the derivative of a logarithm \(\ln x\) is \(\frac{1}{x}\).
• The derivative \(f'(x) = \frac{x}{4} - \frac{1}{x}\) gives the slope of the curve at any point for part (a).
• Similarly, for functions like \(x = \left(\frac{y}{4}\right)^2 - 2 \ln \left(\frac{y}{4}\right)\), we use the chain rule along with other differentiation techniques to find \(g'(y) = \frac{y}{8} - \frac{2}{y}\).

Exploring Integral Calculus

Integral calculus revolves around the concept of finding the total accumulation of quantities. In the context of arc length, integrals help us calculate the total length of a curve over a certain interval.

The fundamental idea here is to sum up small segments of the curve to find the total length. The formula for arc length involves integrating the square root of \(1 + (f'(x))^2\) over the desired interval. This accounts for both the horizontal and vertical changes of the curve. Here's how it works:

• For a function \(y = f(x)\), you integrate: \[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\].
• And similarly, when you have \(x = g(y)\), use: \[L = \int_{c}^{d} \sqrt{1 + (g'(y))^2} \, dy\].
• These integrals help us find the smooth continuous length of the curve, summing up tiny sections of slope changes along the curve.

By calculating these integrals, we can determine how long a curve is between two defined points, like in our problem with parts (a) and (b).

Curve Parametrization: A Different Perspective

Curve parametrization offers a technique to describe curves using parameters, often simplifying complex calculations. Instead of expressing a curve as a direct \(y = f(x)\) or \(x = g(y)\) relation, we use a third variable, parameter, to express both \(x\) and \(y\):

• This approach can make finding derivatives easier as differentiation with respect to the parameter can be more straightforward.
• It's especially useful when dealing with curves that loop or don't have a clear function-like behavior between dependent and independent variables.
• Parametrization can be used to transform complex integrals into simpler ones, thereby easing the process of finding solutions to problems like arc length.

In arc length problems, even if curves are initially presented in standard function notation, thinking about them parametrically can provide new insights and simplify computations.