81–88. Arc length Find the arc length of the following curves on the given interval.

x = 2t sin t - t² cos t, y = 2t cos t + t² sin t; 0 ≤ t ≤ π

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Recall the formula for the arc length of a parametric curve given by \(x = x(t)\) and \(y = y(t)\) over the interval \(a \leq t \leq b\): \[L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]

Find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) by differentiating each component with respect to \(t\). Use the product rule where necessary since both \(x\) and \(y\) are products of functions of \(t\).

Compute \(\left(\frac{dx}{dt}\right)^2\) and \(\left(\frac{dy}{dt}\right)^2\) and then add them together inside the square root to form the integrand: \[\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\]

Simplify the expression under the square root as much as possible to make the integral easier to evaluate.

Set up the definite integral for the arc length from \(t=0\) to \(t=\pi\): \[L = \int_0^{\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\] This integral can then be evaluated using appropriate techniques or numerical methods.

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Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves. Understanding how to work with parametric forms is essential for calculating properties like arc length.

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Parameterizing Equations

Arc Length Formula for Parametric Curves

The arc length of a curve defined parametrically by x(t) and y(t) from t = a to t = b is given by the integral of the square root of (dx/dt)² + (dy/dt)² dt. This formula measures the distance along the curve by summing infinitesimal line segments, requiring differentiation of both x and y with respect to t.

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Arc Length of Parametric Curves

Differentiation of Parametric Functions

To apply the arc length formula, one must compute the derivatives dx/dt and dy/dt accurately. This involves using standard differentiation rules, including the product and chain rules, especially when x(t) and y(t) are products or compositions of functions. Correct differentiation is crucial for evaluating the integral correctly.

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