Problem 9 - 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. \(x=\left(y^{3} / 3\right)+1 /(4 y) \quad\) from \(\quad y=1\) 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 9 units.

Step 1: Find the Derivative of x with Respect to y

Given that \(x = \frac{y^3}{3} + \frac{1}{4y}\), we need to find \(\frac{dx}{dy}\). The derivative of \(\frac{y^3}{3}\) with respect to \(y\) is \(y^2\), and the derivative of \(\frac{1}{4y}\) with respect to \(y\) is \(-\frac{1}{4y^2}\). Thus, \(\frac{dx}{dy} = y^2 - \frac{1}{4y^2}\).

Step 2: Set Up the Arc Length Formula

The arc length of a curve defined by \(x = f(y)\) is calculated using the formula \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\), where \(a\) and \(b\) are the limits from 1 to 3 in this case.

Step 3: Simplify the Expression Inside the Square Root

Given the hint, \(1 + \left(\frac{dx}{dy}\right)^2\) is a perfect square. Substitute \(\frac{dx}{dy} = y^2 - \frac{1}{4y^2}\) and simplify: \[1 + \left(y^2 - \frac{1}{4y^2}\right)^2 = \left(y^2 + \frac{1}{2y^2}\right)^2\].

Step 4: Solve the Integral

Since \(\sqrt{1 + \left(\frac{dx}{dy}\right)^2} = y^2 + \frac{1}{2y^2}\), we compute the integral: \(L = \int_{1}^{3} \left(y^2 + \frac{1}{2y^2}\right) \, dy\). This integral can be separated into two separate integrals: \(L = \int_{1}^{3} y^2 \, dy + \frac{1}{2} \int_{1}^{3} \frac{1}{y^2} \, dy\).

Step 5: Evaluate the Integral

The integral \(\int y^2 \, dy\) gives \(\frac{y^3}{3}\) and \(\int \frac{1}{y^2} \, dy\) gives \(-\frac{1}{y}\). So, \[\int_{1}^{3} y^2 \, dy = \left[\frac{y^3}{3}\right]_{1}^{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}\], and \[\frac{1}{2} \int_{1}^{3} \frac{1}{y^2} \, dy = \frac{1}{2} \left[-\frac{1}{y}\right]_{1}^{3} = \frac{1}{2} \left(-\frac{1}{3} + 1\right) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}\].

Step 6: Calculate the Total Length

Add the two results from the integrals to find the total arc length: \[L = \frac{26}{3} + \frac{1}{3} = \frac{27}{3} = 9\].

Key Concepts

Derivative Calculation

Calculating the derivative is a foundational skill in calculus. For a given function that describes a curve, like our function for x in terms of y, the derivative \(\frac{dx}{dy}\) tells us the rate at which x changes with respect to y. This is crucial when determining how steep a curve is at any given point. In our exercise, the function given is \(x = \frac{y^3}{3} + \frac{1}{4y}\). To find the derivative, we need to apply the basic rules of differentiation separately to each term:

• The derivative of \(\frac{y^3}{3}\) is \(y^2\), because using the power rule, the exponent reduces by one and gets multiplied by the original exponent.
• The derivative of \(\frac{1}{4y}\) involves first rewriting the function as \(\frac{1}{4}y^{-1}\) and then applying the power rule again, resulting in \(-\frac{1}{4y^2}\).

Thus, the overall derivative of x with respect to y becomes \(y^2 - \frac{1}{4y^2}\). By understanding how to derive each term in the function, students can learn to break down complex problems into manageable steps.

Integral Evaluation

Integrals help us calculate many useful things, such as areas and arc lengths. In this exercise, once we have the derivative, we use an integral to find the total length of the curve. The formula to find the arc length given a function x = f(y) involves the integral \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy, \]where \(a\) and \(b\) are the limits from \(y=1\) to \(y=3\). This requires evaluating the expression inside the square root.

Our simplification of \(1 + \left(\frac{dx}{dy}\right)^2\) as a perfect square makes the evaluation straightforward: \(1 + \left(y^2 - \frac{1}{4y^2}\right)^2 = \left(y^2 + \frac{1}{2y^2}\right)^2\). Hence the integral simplifies to \(L = \int_{1}^{3} \left(y^2 + \frac{1}{2y^2}\right) \, dy\). This breaks into two simpler integrals:

• \(\int_{1}^{3} y^2 \, dy\)
• \(\frac{1}{2} \int_{1}^{3} \frac{1}{y^2} \, dy\)

Integrals allow us to sum both these contributions over the interval, giving the total arc length.

Perfect Square Form

Perfect squares are expressions that can be rewritten as the square of another expression. Recognizing a perfect square can greatly simplify calculations, as it often does in exercises involving integrals. In our problem, we see that the expression \(1 + \left(\frac{dx}{dy}\right)^2\) can be simplified to a perfect square, \((y^2 + \frac{1}{2y^2})^2\).

This revelation is powerful because it allows us to replace the complex expression inside the square root with a simpler one. For our integral calculation, it meant we could bypass the square root directly, resulting in simpler integrals to evaluate.

• Recognizing that an expression is a perfect square helps in reducing potentially complicated calculations.
• It converts an intimidating square-rooted expression into a neatly squared form, easing the integration process.

Understanding how to identify and utilize perfect square forms can be invaluable for efficiently solving math problems.