Problem 55 - Essential Calculus

Question

Show that the length of one arch of the sine curve is equal to the length of one arch of the cosine curve.

Step-by-Step Solution

Answer

The lengths of one arch of the sine curve and one arch of the cosine curve are indeed equal.

Step 1: Length Formula for Sine Curve

We use the length formula for a curve given by \[y = f(x)\]on the interval \([a, b]\), \[L = \int_ {a}^{b} \sqrt{1+ [f'(x)]^{2}} dx\]where \(f'(x)\) represents the derivative of \(f(x)\). For \(f(x) = \sin(x)\), the derivative is \(f'(x) = \cos(x)\). Substitutions into the formula yield the length of the sine curve from \(0\) to \(2\pi\) as:\[L_{\sin} = \int _{0}^{2\pi} \sqrt{1+ [\cos(x)]^{2}} dx \]

Step 2: Length Formula for Cosine Curve

Similarly, for \(f(x) = \cos(x)\), the derivative is \(f'(x) = -\sin(x)\). Substitutions into the length formula yields the length of the cosine curve from \(\pi/2\) to \((2\pi + \pi/2)\) as:\[ L_{\cos} = \int _{\pi/2}^{5\pi/2} \sqrt{1+ [-\sin(x)]^{2}} dx \]

Step 3: Equality of Lengths

Looking at the two formulas from Step 1 and Step 2, we see that they are indeed equivalent, because the square of the derivative of the function, regardless of whether it's sine or cosine, falls between 0 and 1, so we have \[1+ [\cos(x)]^{2} = 1+ [-\sin(x)]^{2}\]which implies\[\int _{0}^{2\pi} \sqrt{1+ [\cos(x)]^{2}} dx = \int _{\pi/2}^{5\pi/2} \sqrt{1+ [-\sin(x)]^{2}} dx \]Therefore, one arch of the sine curve has the same length as one arch of the cosine curve.