8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled

Learning objectives

—CHA-6.A—Determine the length of a curve in the plane defined by a function, using a definite integral• CHA-6.A Determine the length of a curve in the plane defined by a function, using a definite integral.• CHA-6.A.1 The length of a planar curve defined by a function 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.

A short piece of a smooth graph has horizontal change $dx$ and vertical change $dy$ , so its length behaves like the hypotenuse $ds=\sqrt{dx^2+dy^2}$ . Dividing by the chosen differential produces a nonnegative length rate that can be accumulated across the interval.

y=f(x):\ L=\int_a^b\sqrt{1+[f'(x)]^2},dx;\qquad x=g(y):\ L=\int_c^d\sqrt{1+[g'(y)]^2},dy

Express the curve as one differentiable function of the integration variable.\n2. Differentiate that function.\n3. Form $\sqrt{1+(\text{derivative})^2}$ .\n4. Integrate over the coordinate interval.\n5. Check that the result is at least the straight-line distance between the endpoints.

Example: for $y=\frac23x^{3/2}$ on $0\le x\le1$ , $f'(x)=\sqrt{x}$ . Therefore $L=\int_0^1\sqrt{1+x}\,dx=\left[\frac23(1+x)^{3/2}\right]_0^1=\frac23(2\sqrt2-1)$ units.

The integrand contains the square of the derivative, not the square of the original function. Use this single-function formula only where the chosen orientation represents the curve smoothly across the full interval; otherwise a different orientation or a split into smooth pieces is needed.

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—AP Calculus BCRead