Sine vs Cosine

Calculus Level 5

What is the maximum ratio of the length of the cosine curve to that of the sine curve as measured from x = 0 ? x=0?

If the answer is R R , submit your answer as 1000 R \left \lfloor {1000R} \right \rfloor .


The answer is 1076.

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1 solution

Miles Koumouris
Dec 13, 2017

The arc length L L of a curve y = f ( x ) y=f(x) between x = a x=a and x = b x=b is given by L = a b 1 + ( d y d x ) 2 d x . L=\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\; dx. Thus, we seek the maximum possible value of 0 t 1 + sin 2 ( x ) d x 0 t 1 + cos 2 ( x ) d x , \dfrac{\int_0^t\sqrt{1+\sin^2(x)}\; dx}{\int_0^t\sqrt{1+\cos^2(x)}\; dx}, for some value of t t . By inspecting the graphs of y = 1 + cos 2 ( x ) y=\sqrt{1+\cos^2(x)} and y = 1 + sin 2 ( x ) y=\sqrt{1+\sin^2(x)} , it is easy to see that the maximum must occur when π 4 < t < 3 π 4 \tfrac{\pi }{4}<t<\tfrac{3\pi }{4} , but other than this, I found no way to evaluate the derivative of this ratio analytically (anyone else find a way?)

Solving numerically gives t 2.2449126574055164726 t\approx 2.2449126574055164726 and R 1.0764947166963004306 R\approx 1.0764947166963004306 , meaning that 1000 R = 1076 \lfloor 1000R\rfloor =\boxed{1076} .

@Digvijay Singh , is there a non-numerical way to solve this question?

Pi Han Goh - 3 years, 5 months ago

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I couldn't find it, which is why I posted this question.

Digvijay Singh - 3 years, 5 months ago

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