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Math Help - orthogonal trajectaries

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    orthogonal trajectaries

    Find the orthogonal trajectory of the system r^2cos^2 2\theta=c^2 sin 2 \theta.

    ans: r^4(3-cos \theta)=k.
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  2. #2
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    Quote Originally Posted by dimuk View Post
    Find the orthogonal trajectory of the system r^2\cos^2 2\theta=c^2 \sin 2 \theta.

    ans: r^4(3-\cos \theta)=k. I'm getting \color{red}r^4(3-\cos4\theta)=k.
    In polar coordinates, the quantity \tfrac1r\tfrac{dr}{d\theta} represents the slope of the tangent of a curve with respect to the radial direction. (Informally, a point on the curve moves an infinitesimal distance dr in the radial direction and rdθ in the transverse direction.) So the orthogonal system should have "radial slope" -r/\tfrac{dr}{d\theta} at that point.

    Write r^2\cos^2 2\theta=c^2 \sin 2 \theta as  r^2 = c^2\sec2\theta\tan2\theta, and differentiate with respect to θ: 2rr' = 2c^2(\sec2\theta\tan^22\theta + \sec^32\theta), where r' = \tfrac{dr}{d\theta}. Substitute c^2 = \frac{r^2\cos^22\theta}{\sin2\theta} and simplify a bit, to get \frac{r'}r = \frac{1+\sin^22\theta}{\sin2\theta\cos2\theta}.

    The orthogonal trajectory will therefore have \frac{r'}r =-\frac{\sin2\theta\cos2\theta}{1+\sin^22\theta}. Integrate this: \int\frac{dr}r = -\int\frac{\sin2\theta\cos2\theta}{1+\sin^22\theta}  d\theta.

    To find the integral on the right-hand side, substitute u = \sin2\theta, and it becomes -\int\frac{u}{2(1+u^2)}du = -\tfrac14\ln(1+u^2).

    That gives the orthogonal trajectory as \ln r = -\tfrac14\ln(1+\sin^22\theta)\; + const., or r^4(1+\sin^22\theta) = const.

    Finally, the trig identity \sin^2\phi = \tfrac12(1-\cos2\phi) can be used to write the answer as r^4(3-\cos4\theta)= const.
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