Question

Find the arc length of the following curves on the given interval. $$x=\cos t-\sin t, y=\cos t+\sin t ; 0 \leq t \leq \pi$$

Short answer

Answer: The arc length of the curve on the given interval is $\pi\sqrt{2}$.

Step-by-step solution

1. Differentiate x(t) and y(t) with respect to t

We first find the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\): $$\frac{dx}{dt} = \frac{d(\cos t - \sin t)}{dt} = -\sin t - \cos t$$ $$\frac{dy}{dt} = \frac{d(\cos t + \sin t)}{dt} = -\sin t + \cos t$$

2. Square the derivatives and add them together

Now we square the derivatives of \(x(t)\) and \(y(t)\) and add them together: $$(\frac{dx}{dt})^2 = (-\sin t - \cos t)^2 = \sin^2 t + 2\sin t\cos t + \cos^2 t$$ $$(\frac{dy}{dt})^2 = (-\sin t + \cos t)^2 = \sin^2 t - 2\sin t\cos t + \cos^2 t$$ Adding them together yields: $$(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = 2\sin^2 t + 2\cos^2 t = 2(\sin^2 t +\cos^2 t) = 2$$

3. Take the square root and write down the integrand

We now take the square root of the sum \((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2\) and write down the resulting integrand: $$\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{2}$$

4. Integrate the integrand and evaluate the result over the interval [0, pi]

Finally, we integrate the integrand over the interval \(0\leq t\leq \pi\) to find the arc length: $$L = \int_0^\pi \sqrt{2} \, dt = \sqrt{2} \int_0^\pi dt = \sqrt{2}(t\big|_0^\pi) = \sqrt{2}(\pi - 0) = \pi\sqrt{2}$$ Hence, the arc length of the curve on the given interval is: $$L = \pi\sqrt{2}$$

Key concepts

Parametric Equations

When tackling parametric equations in mathematics, they provide a powerful way of describing curves in a plane. Unlike regular equations that relate variables such as \( x \) and \( y \) directly, parametric equations define each variable in terms of a third one, often denoted as \( t \). This parameter \( t \) usually represents time or some other changeable quantity.
For example, in our exercise, we have:

• \( x(t) = \cos t - \sin t \)
• \( y(t) = \cos t + \sin t \)

These equations trace out the path of a curve as \( t \) changes. The range given, \(0 \leq t \leq \pi\), tells us the specific part of the curve we're examining. Analyzing parametric equations gives us insight into the curve's trajectory and orientation.

Differentiation

Differentiation is a pivotal concept in calculus that enables us to determine the rate of change of a function with respect to a variable. When working with parametric equations, we differentiate both components, \( x(t) \) and \( y(t) \), separately with respect to \( t \).
In this problem, we found:

• \( \frac{dx}{dt} = -\sin t - \cos t \)
• \( \frac{dy}{dt} = -\sin t + \cos t \)

These derivatives describe how each component of our parametric equation changes as \( t \) varies. By understanding these changes, we can move into further analysis of the curve, such as finding its arc length or determining the slope of a tangent at any point.

Integration

Integration is the process of finding the accumulated sum over a continuous range, and it is the inverse operation of differentiation. In the context of arc length, integration helps us calculate the distance along a curve over an interval.
To begin with, we find the square of the derivatives of the functions. By adding these squares, we discover a simplified expression:

• \( (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = 2 \)

Taking the square root gives the integrand needed for our integration. As seen in the solution, integrating over the interval \( [0, \pi] \) involves considering the constant integrand \( \sqrt{2} \), which simplifies the integration process considerably.
Ultimately, the result \( \int_0^\pi \sqrt{2} \, dt = \pi\sqrt{2} \) reveals the total arc length of the curve between the specified \( t \) values.

Trigonometric Functions

Understanding trigonometric functions such as sine and cosine is essential, as they represent fundamental wave patterns in mathematics. They frequently appear in physics and engineering alongside parametric equations.
In our exercise:

• \( x(t) = \cos t - \sin t \)
• \( y(t) = \cos t + \sin t \)

The combination of trigonometric functions here forms intriguing patterns. They utilize known identities like \( \sin^2 t + \cos^2 t = 1 \) to simplify expressions and calculations. This identity is crucial when working with derivatives and calculating arc lengths using the Pythagorean Theorem.
Furthermore, using trigonometric functions helps visualize the curve's oscillations and rotations, providing geometric insights essential to solving problems involving curves on the plane.