Problem 45 - Calculus Single Variable

Question

In each of Exercises \(45-52,\) calculate the length \(L\) of the given parametric curve. $$ x=t \cos (t) \quad y=t \sin (t) \quad 0 \leq t \leq 3 \pi $$

Step-by-Step Solution

Answer

The length of the curve is \( \int_0^{3\pi} \sqrt{1 + t^2} \, dt \), evaluated numerically.

Step 1: Understand the Formula for Arc Length

The length of a parametric curve defined by \( x = f(t) \) and \( y = g(t) \) from \( t = a \) to \( t = b \) is given by \( L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). This formula ensures that we account for changes in both the x and y coordinates.

Step 2: Compute Derivatives

Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Given \( x = t \cos(t) \), we differentiate to get \( \frac{dx}{dt} = \cos(t) - t \sin(t) \). Similarly, for \( y = t \sin(t) \), we differentiate to get \( \frac{dy}{dt} = \sin(t) + t \cos(t) \).

Step 3: Plug Derivatives into Arc Length Formula

Substitute \( \frac{dx}{dt} = \cos(t) - t \sin(t) \) and \( \frac{dy}{dt} = \sin(t) + t \cos(t) \) into the arc length formula: \[ L = \int_0^{3\pi} \sqrt{(\cos(t) - t \sin(t))^2 + (\sin(t) + t \cos(t))^2} \, dt \].

Step 4: Simplify the Integrand

Simplify the integrand: \((\cos(t) - t \sin(t))^2 + (\sin(t) + t \cos(t))^2 = \cos^2(t) + 2t \cos(t) \sin(t) + t^2 \cos^2(t) + \sin^2(t) - 2t \cos(t) \sin(t) + t^2 \sin^2(t) \). Combine terms to get \(1 + t^2\).

Step 5: Evaluate the Integral

The integral becomes \( L = \int_0^{3\pi} \sqrt{1 + t^2} \, dt \). This is a standard integral that can be evaluated using numerical methods or tables.

Step 6: Calculate the Numerical Value

Using a numerical method or integrator, evaluate \( \int_0^{3\pi} \sqrt{1 + t^2} \, dt \) to find the length of the curve. This gives an approximate value based on calculations.

Key Concepts

Parametric Curves

Parametric curves are a fascinating concept in calculus as they allow us to describe a curve in the coordinate plane using a parameter, typically denoted as \( t \). Instead of expressing \( y \) solely as a function of \( x \), we express both \( x \) and \( y \) in terms of a third variable. This way, both horizontal and vertical positions are tied together by the parameter \( t \).
To put it simply, a parametric equation for a curve will have two components:

• \( x = f(t) \)
• \( y = g(t) \)

This flexibility allows more complex shapes to be represented, including loops and paths that double back over themselves. In our example, the parametric equations \( x = t\cos(t) \) and \( y = t\sin(t) \) trace a spiral-like path as \( t \) ranges from 0 to \( 3\pi \). Parametric curves drop the restrictions of traditional functions, offering a broader scope of shapes to explore in geometry and physics.

Derivatives

Derivatives play a vital role when dealing with parametric equations, especially in calculating the arc length. The arc length formula for parametric curves necessitates the derivatives of both \( x \) and \( y \) with respect to the parameter \( t \). This is because we need to consider how both \( x \) and \( y \) change as \( t \) progresses.
This is why **calculating derivatives** for each component is essential:

• For \( x = t\cos(t) \), the derivative \( \frac{dx}{dt} = \cos(t) - t\sin(t) \) captures how \( x \) changes.
• For \( y = t\sin(t) \), the derivative \( \frac{dy}{dt} = \sin(t) + t\cos(t) \) describes the change in \( y \).

These derivatives measure the instantaneous rate of change or the slope of the tangent to the curve at any point \( t \). Understanding these rate of changes is crucial, as they are used to plug into the arc length formula.

Integral Calculus

Integral calculus provides us with the necessary tools to find the arc length of parametric curves. In our exercise, we make use of the arc length formula, which integrates the square root of the sum of the squares of the derivatives. The integral we evaluate is:\[L = \int_0^{3\pi} \sqrt{1 + t^2} \, dt\]
The expression inside the square root derives from simplifying the squares of the derivatives from the parametric equations. The challenge of calculating this integral lies in its complexity, often requiring **numerical methods** or tables for evaluation. When dealing with challenging integrals like this, methods such as trapezoidal rule or Simpson's rule are often employed.
This integral captures the total distance traveled along the curve from point \( t = 0 \) to \( t = 3\pi \). Integral calculus is indispensable in issues involving total change or accumulation, making it essential for calculating everything from areas under curves to distances traveled.