Problem 19 - Thomas Calculus

Question

Find the lengths of the curves. The spiral \(r=\theta^{2}, \quad 0 \leq \theta \leq \sqrt{5}\)

Step-by-Step Solution

Answer

The length of the curve is found by integrating the expression from step 4 over the given limits.

Step 1: Understand the formula for arc length in polar coordinates

To find the length of a curve given in polar coordinates, use the formula: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \] where \( r = f(\theta) \) is the polar equation, and \( \theta \) ranges from \( a \) to \( b \).

Step 2: Identify the function and its derivative

Given the spiral function \( r = \theta^2 \) we need to find its derivative: \[ \frac{dr}{d\theta} = 2\theta \] Now we have both \( r = \theta^2 \) and \( \frac{dr}{d\theta} = 2\theta \).

Step 3: Set up the integral

Substitute \( r = \theta^2 \) and \( \frac{dr}{d\theta} = 2\theta \) into the arc length formula: \[ L = \int_{0}^{\sqrt{5}} \sqrt{ (2\theta)^2 + (\theta^2)^2 } \, d\theta = \int_{0}^{\sqrt{5}} \sqrt{ 4\theta^2 + \theta^4 } \, d\theta = \int_{0}^{\sqrt{5}} \sqrt{ \theta^4 + 4\theta^2 } \, d\theta \]

Step 4: Simplify the integrand

Factor \( \theta^2 \) out of the expression inside the square root: \[ L = \int_{0}^{\sqrt{5}} \sqrt{ \theta^2 (\theta^2 + 4)} \, d\theta = \int_{0}^{\sqrt{5}} \theta \sqrt{ \theta^2 + 4 } \, d\theta \]

Step 5: Solve the integral

To integrate \( \int_{0}^{\sqrt{5}} \theta \sqrt{ \theta^2 + 4 } \, d\theta \), use substitution. Let \( u = \theta^2 + 4 \), then \( du = 2\theta \, d\theta \). Replace the limits accordingly and solve the resulting integral. The step involves some standard integration techniques.

Step 6: Evaluate the definite integral

After solving, evaluate the integral from step 5 between \( \theta = 0 \) and \( \theta = \sqrt{5} \). This provides the arc length of the given spiral curve. The calculation will involve computing the definite integral using the derived expressions from performing the antiderivative.

Key Concepts

Polar Coordinates

Polar coordinates are a fascinating way to represent points in the plane, differing from the traditional Cartesian coordinates. Instead of using an x and y to determine a point's position, polar coordinates use a distance from the origin, called the radius (r), and an angle (\(\theta\)) from a fixed direction, usually the positive x-axis.

This system is particularly useful for problems involving curves and spirals, like the one in the exercise. When dealing with objects that naturally extend outward in circular forms, such as spirals, using polar coordinates simplifies the math significantly.

• Coordinates expressed as \( (r, \theta) \)
• \( r = f(\theta) \) describes the function in polar form.
• Aids in understanding curves that wrap around a central point.

For the curve in the exercise, the equation \( r = \theta^2 \) describes a spiral as \(\theta\) increases. This spiral radiates outwards as \(\theta\) increases from 0 to \(\sqrt{5}\). Polar coordinates not only make the process of setting up the integral more intuitive but also streamline the path to finding the arc length.

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration, which is, in a sense, the reverse process of differentiation. Integration is about finding the total accumulated quantity, for instance, the area under a curve, or in this case, the length of a curve described in polar coordinates.

The problem involves calculating the arc length, which requires setting up an integral based on the derivative of the function and the function itself.

• Integration can determine lengths, areas, volumes, and other cumulative measures.
• Utilizes the integral formula for arc length in polar form, \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]
• In our exercise, \( r = \theta^2 \) leads to \( \frac{dr}{d\theta} = 2\theta \)

Understanding integral calculus equips you to solve equations that describe a wide range of physical curves and shapes, granting insight into the total length of such curves. Learning to set up and solve integrals like this is a key skill in mathematics and many applied sciences.

Definite Integrals

Definite integrals are a key concept in integral calculus, representing the integration of a function over a specific interval. When calculating the arc length of a curve, definite integrals come into play, covering the specific range of the parameter that describes the curve.

For the spiral described in our exercise, the integral \( \int_{0}^{\sqrt{5}} \theta \sqrt{ \theta^2 + 4 } \, d\theta \) computes the length of the spiral from \(\theta = 0\) to \(\theta = \sqrt{5}\).

• A definite integral has limits of integration, indicating the start and end points.
• \( \int_{a}^{b} f(x) \, dx \) with the limits \(a\) and \(b\).
• Evaluating it entails calculating the function's antiderivative and applying the limits.

Definite integrals provide the machinery to not just evaluate functions over a whole domain, but also allow us to conclude real, quantifiable measures specific to a curve segment, like exact arc length. This is an essential calculation in geometry and applied physics, giving practical meaning to the mathematics.