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Chapter 6: Problem 55

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

Short Answer

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

Step by step solution

01

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:\[L_{\sin} = \int _{0}^{2\pi} \sqrt{1+ [\cos(x)]^{2}} dx \]
02

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:\[ L_{\cos} = \int _{\pi/2}^{5\pi/2} \sqrt{1+ [-\sin(x)]^{2}} dx \]
03

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.

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