Problem 12 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(7-16\) . If you have a grapher, you may want to graph these curves to see what they look like. \(x=\left(y^{3} / 6\right)+1 /(2 y)\) from \(y=2\) to \(y=3\) (Hint: \(1+(d x / d y)^{2}\) is a perfect square.)

Step-by-Step Solution

Answer

The length of the curve is 3.75.

Step 1: Understand the Formula for Arc Length

To find the length of a curve defined parametrically or by functions, we use the formula for arc length. In this case, it is given by \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]where \(a\) and \(b\) are the limits for \(y\). In this problem, \(a = 2\) and \(b = 3\).

Step 2: Differentiate the Function x(y) with Respect to y

Given the function \[ x = \frac{y^3}{6} + \frac{1}{2y} \]differentiate with respect to \(y\): \[ \frac{dx}{dy} = \frac{d}{dy}\left(\frac{y^3}{6}\right) + \frac{d}{dy}\left(\frac{1}{2y}\right) \]Calculating separately,\[ \frac{d}{dy}\left(\frac{y^3}{6}\right) = \frac{1}{6} \cdot 3y^2 = \frac{y^2}{2} \]and\[ \frac{d}{dy}\left(\frac{1}{2y}\right) = \frac{-1}{2y^2} \].Thus,\[ \frac{dx}{dy} = \frac{y^2}{2} - \frac{1}{2y^2} \].

Step 3: Simplify the Expression under the Square Root

Compute \[ 1 + \left( \frac{dx}{dy} \right)^2 \].Substitute the derivative,\[ \frac{dx}{dy} = \frac{y^2}{2} - \frac{1}{2y^2} \].This gives,\[ \left( \frac{y^2}{2} - \frac{1}{2y^2} \right)^2 = \left( \frac{y^4 - 1}{4y^2} \right)^2 = \frac{(y^4 - 1)^2}{16y^4} \].Now, replacing and simplifying within \[ 1 + \left( \frac{dx}{dy} \right)^2 = 1 + \frac{(y^4 - 1)^2}{16y^4} \].By the hint, this expression should become a perfect square.

Step 4: Use the Hint to Form the Perfect Square

Following the hint, assume that the sum is indeed a perfect square:\[ 1 + \left( \frac{y^2}{2} - \frac{1}{2y^2} \right)^2 = \left( \frac{y^2}{2} + \frac{1}{2y^2} \right)^2 \].Verify:\[ \left( \frac{y^2}{2} + \frac{1}{2y^2} \right)^2 = \frac{y^4}{4} + \frac{1}{4y^4} + \frac{y^2}{2y^2} \].This gives:\[ \frac{y^4 + 2 + \frac{1}{y^4}}{4} = 1 + \frac{(y^4 - 1)^2}{16y^4} \] as required.

Step 5: Compute the Integral for Arc Length

Now that the expression is a perfect square, the integral simplifies. Using:\[ L = \int_{2}^{3} \frac{y^2}{2} + \frac{1}{2y^2} \, dy \]compute separately:The integral of \(\frac{y^2}{2}\) is:\[ \frac{1}{2} \cdot \frac{y^3}{3} = \frac{y^3}{6} \].The integral of \(\frac{1}{2y^2}\) is:\[ -\frac{1}{2y} \].Now sum over the intervals:\[ \int_{2}^{3} \frac{y^2}{2} + \frac{1}{2y^2} \, dy = \left[ \frac{y^3}{6} - \frac{1}{2y} \right]_{2}^{3} \].

Step 6: Evaluate the Definite Integral

Substitute the upper and lower bounds into the evaluated integral:\[ \left[ \frac{y^3}{6} - \frac{1}{2y} \right]_{2}^{3} = \left( \frac{27}{6} - \frac{1}{6} \right) - \left( \frac{8}{6} - \frac{1}{4} \right) \].Make the calculations:First, calculate at \(y=3\):\[ \frac{27}{6} - \frac{1}{6} = \frac{26}{6} \].Then at \(y=2\):\[ \frac{8}{6} - \frac{1}{4} = \frac{10}{12} - \frac{3}{12} = \frac{7}{12} \].Thus, the total length is:\[ \frac{26}{6} - \frac{7}{12} = \frac{52}{12} - \frac{7}{12} = \frac{45}{12} = \frac{15}{4} = 3.75 \].

Key Concepts

Parametric Equations

Parametric equations describe a method of defining a mathematical function or a curve. Instead of expressing one variable (like \(y\)) directly in terms of another (like \(x\)), both are defined separately as functions of a third variable, often denoted as \(t\). This allows for more flexibility and creativity in representing curves, including those that cannot be easily expressed in standard form.

In our original exercise, the curve is defined by \(x\) as a function of \(y\): \(x = \frac{y^3}{6} + \frac{1}{2y}\). Although \(x\) is defined in terms of \(y\), which might seem unusual at first, it retains the essence of parametric definition because it does not straightforwardly express \(y\) in terms of \(x\) as in a simple algebraic equation. Instead, it gives a clear way to explore complex relations between variables.

Using parametric equations, you gain the ability to model natural phenomena like planetary orbits, track the paths of flying objects, and sketch intricate geometrical shapes. This versatility is one reason they are so valuable in fields such as physics, engineering, and computer graphics.

Integral Calculus

Integral calculus is a crucial branch of mathematics focused on the concept of integration. If differentiation can be seen as finding the rate of change, integration is its complementary process, often described as finding the total accumulation. It enables you to compute areas under curves, volumes, surface areas, and much more.

For arc length, as in our exercise, integration helps sum up infinitely small lengths along a curve to find the total length of a given curve. The general formula for the arc length, given the function \(x = f(y)\), is:

• \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\), where \(a\) and \(b\) are limits for \(y\).

This integral accumulates the infinitesimal lengths \(ds\) along the curve from \(y=a\) to \(y=b\). Each infinitesimal length is calculated using the Pythagorean theorem.

In the specific problem, we tackled the challenge of forming a perfect square under the integral to simplify calculations, making it easier to evaluate the integral directly. Integral calculus thus provides the toolkit to solve such mathematical problems elegantly, even when the curves described are complex.

Differentiation

Differentiation is another cornerstone of calculus and deals with understanding how functions change. By computing derivatives, you learn about the rates at which quantities evolve, influencing fields from physics to economics.

In the context of arc length, as explored in the exercise, differentiation is used to determine how the curve \(x(y)\) changes with \(y\). Specifically, you find \(\frac{dx}{dy}\) to understand the slope of the curve as \(y\) varies.

• For the function \(x = \frac{y^3}{6} + \frac{1}{2y}\), differentiating with respect to \(y\) yields \(\frac{dx}{dy} = \frac{y^2}{2} - \frac{1}{2y^2}\).

This derivative plays an integral role in the arc length formula, entering as \(\left(\frac{dx}{dy}\right)\) and directly influencing the calculation for each infinitesimal segment of the curve.

Mastering differentiation allows you to delve into how quantities shift, providing clarity on the behavior of functions and enabling applications such as optimization and curve sketching. Thus, a solid grasp of differentiation paves the way for solving various real-world challenges.