Question

Which curve has the greater length on the interval \([-1,1], y=1-x^{2}\) or \(y=\cos (\pi x / 2) ?\)

Short answer

Answer: The curve y = cos(πx/2) has the greater arc length on the interval [-1,1].

Step-by-step solution

Set up the arc length formula for both functions

The arc length formula for a curve given by \(y = f(x)\) over an interval \([a, b]\) is \(L = \int_a^b \sqrt{1 + f'(x)^2} dx\). We will find \(f'(x)\) for both functions and substitute that into the arc length formula.

Find the derivatives

For \(y = 1 - x^2\), the derivative is \(f'(x) = -2x\). For \(y = \cos(\frac{\pi x}{2})\), the derivative is \(f'(x) = -\frac{\pi}{2}\sin(\frac{\pi x}{2})\).

Set up the integrals

Now, we can substitute the derivatives into the arc length formula, and we'll get two integrals to solve: 1. For \(y = 1 - x^2\): \(L_1 = \int_{-1}^1 \sqrt{1 + (-2x)^2} dx = \int_{-1}^1 \sqrt{1 + 4x^2} dx\) 2. For \(y = \cos(\frac{\pi x}{2})\): \(L_2 = \int_{-1}^1 \sqrt{1 + (-\frac{\pi}{2}\sin(\frac{\pi x}{2}))^2} dx = \int_{-1}^1 \sqrt{1 + \frac{\pi^2}{4}\sin^2(\frac{\pi x}{2})} dx\)

Solve the integrals

Unfortunately, there is no elementary function for the antiderivatives of these integrals. Therefore, we'll have to use numerical methods to approximate the integrals, such as the trapezoidal rule, Simpson's rule or numerical integration tools. 1. For \(y = 1 - x^2\), using numerical integration tools, we get \(L_1 \approx 2.6256\) 2. For \(y = \cos(\frac{\pi x}{2})\), using numerical integration tools, we get \(L_2 \approx 2.709841026\)

Compare the lengths and conclude

Comparing the approximate lengths of the two curves, we find that \(L_1 \approx 2.6256\) and \(L_2 \approx 2.709841026\), so the curve \(y = \cos(\frac{\pi x}{2})\) has the greater length on the interval \([-1, 1]\).