Question

Lengths of symmetric curves Suppose a curve is described by y=f(x) on the interval [−b, b], where f′ is continuous on [−b, b]. Show that if f is odd or f is even, then the length of the curve y=f(x) from x=−b to x=b is twice the length of the curve from x=0 to x=b. Use a geometric argument and prove it using integration.

Accepted Answer

Recall the formula for the length of a curve $$y = f(x) on $$the interval $$[a, c] is $$given by the integral $$L = \int_a^c \sqrt{1 + (f'(x))^2$} \, dx. $$Since the curve is defined on $$[-b, b]$$, the total length is $$L = \int_{-b}^b \sqrt{1 + (f'(x))^2$} \, dx. $$Consider the geometric symmetry: if $$f is $$even, then $$f(-x) = f(x)$$, and if $$f is $$odd, then $$f(-x) = -f(x). In $$both cases, the shape of the curve on $$[-b, 0] is a $$mirror image (possibly reflected) of the curve on $$[0, b]. $$Use the substitution $$u = -x to $$rewrite the integral over $$[-b, 0]$$: $$\int_{-b}^0 \sqrt{1 + (f'(x))^2$} \, dx = \int_b^0$ \sqrt{1 + (f'(-u))^2$} (-du) = \$$int_0$$^b \sqrt{1 + (f'(-u))^2$} \, du. $$Since $$f is $$even or odd, analyze $$f'(-u)$$: for even $$f$$, $$f' is $$odd, so $$f'(-u) = -f'(u)$$; for odd $$f$$, $$f' is $$even, so $$f'(-u) = f'(u). In $$both cases, $$(f'(-u))^2$ = (f'(u))^2$, so the integrand is the same on $$[0, b]$$. Therefore, the length from -b$ to 0$ equals the length from 0$ to b$, and the total length is twice the length from 0$ to b$.

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