Question #283572

Find orthogonal trajectory to the curve given by 𝑟 = 𝑎(1 + cos 𝜃)

1
Expert's answer
2021-12-30T10:42:21-0500
r=a(1+cosθ)r = a(1 + \cosθ)

Differentiate


dr=asinθdθdr=-a\sin \theta d\theta

a=r1+cosθa=\dfrac{r}{1+\cos \theta}

dr=r1+cosθsinθdθdr=-\dfrac{r}{1+\cos \theta}\sin \theta d\theta

1rdrdθ=sinθ1+cosθ-\dfrac{1}{r}\dfrac{dr}{d\theta}=\dfrac{\sin \theta}{1+\cos \theta}

This is the differential equation of the direction field D1D_1 for the given family F1.F_1. To find the differential equation of the direction field orthological to D1,D_1, we replace drdθ\dfrac{dr}{d\theta} by r2dθdr.-r^2\dfrac{d\theta}{dr}.

The differential equation for the orthogonal trajectories becomes


rdθdr=sinθ1+cosθ=2sin(θ2)cos(θ2)2cos2(θ2)=tan(θ2)r\dfrac{d\theta}{dr}=\dfrac{\sin \theta}{1+\cos \theta}=\dfrac{2\sin (\dfrac{\theta}{2})\cos (\dfrac{\theta}{2})}{2\cos^2 (\dfrac{\theta}{2})}=\tan(\dfrac{\theta}{2})

which is a case of variable-separable, and on integration gives


lnr=2ln(sin(θ2))+ln(2c)\ln r=2\ln(\sin(\dfrac{\theta}{2}))+\ln(2c)

This is the equation of the orthogonal family F2.F_2.

Since r=c(1cosθ)r = c(1 – \cosθ) represents the same curve as r=c(1+cosθ),r = c(1 + \cosθ), the member of F2F_2 with label cc is the same as the member of F1F_1 with label a=c.a = –c. Thus the given family is selforthogonal.





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