Question

Find the arc length of the following curves on the given interval. $$x=2 t \sin t-t^{2} \cos t, y=2 t \cos t+t^{2} \sin t ; 0 \leq t \leq \pi$$

Short answer

Question: Find the arc length of the curve described by the parametric equations \(x = 2t \sin t - t^2 \cos t\) and \(y = 2t \cos t + t^2 \sin t\) for \(t\) ranging from \(0\) to \(\pi\). Answer: The arc length of the curve is given by the integral \(s = \int_{0}^{\pi} \sqrt{(2 \sin t + 4t \cos t - t^2 \sin t)^2 + (-2 \sin t - 2t \sin t +2 \cos t + t^2 \cos t)^2} dt\). Due to the complexity of this integral, a numerical approximation is needed to obtain the value for \(s\).

Step-by-step solution

Differentiate the parametric equations with respect to \(t\).

We have \(x = 2t \sin t - t^2 \cos t\) and \(y = 2t \cos t + t^2 \sin t\). Differentiating \(x(t)\) and \(y(t)\) with respect to \(t\), we get: $$\frac{dx}{dt} = 2 \sin t + 2t \cos t + 2t \cos t + t^2(-\sin t)$$ $$\frac{dy}{dt} = -2 \sin t + 2t(-\sin t) + 2 \cos t + t^2(\cos t)$$

Calculate \((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2\).

We have: $$\left(\frac{dx}{dt}\right)^2 = \left(2 \sin t + 4t \cos t - t^2 \sin t\right)^2$$ $$\left(\frac{dy}{dt}\right)^2 = \left(-2 \sin t - 2t \sin t + 2 \cos t + t^2 \cos t\right)^2$$ Now, sum the two squares: $$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \left(2 \sin t + 4t \cos t - t^2 \sin t\right)^2 + \left(-2 \sin t - 2t \sin t +2 \cos t + t^2 \cos t\right)^2$$

Compute the square root and simplify.

Take the square root of the above expression: $$\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}= \sqrt{(2 \sin t + 4t \cos t - t^2 \sin t)^2 + (-2 \sin t - 2t \sin t +2 \cos t + t^2 \cos t)^2}$$

Integrate the simplified expression.

Finally, compute the definite integral between the limits \(0\) and \(\pi\): $$s = \int_{0}^{\pi} \sqrt{(2 \sin t + 4t \cos t - t^2 \sin t)^2 + (-2 \sin t - 2t \sin t +2 \cos t + t^2 \cos t)^2} dt$$ As this integral is quite complex and challenging to compute directly, a numerical approximation (e.g., using a calculator or computer software) can be employed to obtain the value for \(s\).

Key concepts

Understanding Parametric Equations

When dealing with parametric equations, we're exploring a set of functions that define both the x and y coordinates using another variable, typically denoted as "t". This variable often represents time, but it can represent any continuous value. Parametric equations are incredibly useful in mathematics because they allow us to describe more complex forms and motions that aren’t easily expressed by a single function.

• In our problem, we have two equations: - The x-coordinate is defined by: \( x = 2t \sin t - t^2 \cos t \) - The y-coordinate is expressed as: \( y = 2t \cos t + t^2 \sin t \)
• These equations trace a path in the plane as t varies from 0 to \( \pi \).

By understanding parametric equations, we gain the ability to analyze curves beyond simple lines and polygons. This is key for studying paths of moving objects or shapes changing over time.

The Role of Differentiation

Differentiation, a fundamental concept in calculus, helps us determine the rate of change of a variable concerning another. With parametric equations, differentiation is crucial to find out how the x and y coordinates change as the parameter changes.In our exercise, we differentiate both parametric equations with respect to \( t \).

• For \( x(t) = 2t \sin t - t^2 \cos t \), the derivative \( \frac{dx}{dt} \) calculates as: \( 2 \sin t + 2t \cos t + 2t \cos t - t^2 \sin t \).
• Similarly, for \( y(t) = 2t \cos t + t^2 \sin t \), \( \frac{dy}{dt} \) is: \( -2 \sin t + 2 \cos t - 2t \sin t + t^2 \cos t \).

Differentiating these equations sets the stage for computing the arc length, as it gives insights into how steeply the curve ascends or descends along different parts of its path.

Understanding the Definite Integral

The definite integral allows us to calculate precise joint totals between points on a curve, which in our case is the arc length of the path defined by the parametric equations. The integration process finds the total 'accumulation' of changes over an interval, which here is the interval \( 0 \le t \le \pi \).For arc length, we integrate the expression \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \) over the desired parameter interval. This gives:\[s = \int_{0}^{\pi} \sqrt{(2 \sin t + 4t \cos t - t^2 \sin t)^2 + (-2 \sin t - 2t \sin t +2 \cos t + t^2 \cos t)^2} \, dt\]This integral looks complicated at first, and indeed it often requires methods such as numerical approximation to solve practically. Calculators and software like MATLAB or Wolfram Alpha are helpful here as they can handle the heavy lifting, providing a numerical solution for the arc length. Using definite integrals, we can transform complex parameter-driven scenarios into straightforward calculations of distances or areas.