9.3 Finding Arc Lengths of Curves Given by Parametric Equations

Learning objectives

—CHA-6.B—Determine the length of a curve in the plane defined by parametric functions, using a definite integral• CHA-6.B Determine the length of a curve in the plane defined by parametric functions, using a definite integral.• CHA-6.B.1 The length of a parametrically defined curve can be calculated using a definite integral.• Enduring understanding CHA-6: Definite integrals allow us to solve problems involving the accumulation of change in length over an interval.

For $x=x(t)$ and $y=y(t)$ , a tiny parameter change produces horizontal change $(dx/dt)dt$ and vertical change $(dy/dt)dt$ . The Pythagorean magnitude of these components is the nonnegative rate at which length is traced.

L=\int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},dt

Differentiate $x(t)$ and $y(t)$ .\n2. Square and add the two component derivatives.\n3. Take the nonnegative square root.\n4. Integrate over the stated $t$ -interval.\n5. Check the interval describes the intended part of the curve.

Example: $x=3\cos t$ and $y=3\sin t$ for $0\le t\le\pi/2$ trace one quarter of a circle of radius 3. Here $dx/dt=-3\sin t$ and $dy/dt=3\cos t$ , so the speed is $\sqrt{9\sin^2t+9\cos^2t}=3$ . Thus $L=\int_0^{\pi/2}3\,dt=3\pi/2$ units.

Use the parameter bounds, not automatically the curve's $x$ -coordinates. The square root is the magnitude of both component rates, so neither sign can make length negative. If the parameterization retraces a segment, this integral counts the retraced distance again.

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—9.4 Defining and Differentiating Vector-Valued FunctionsRead