The angle of intersection of the curves r = s | vedclass

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(C) Given curves are $r_1 = \sin 	heta + \cos 	heta$ and $r_2 = 2 \sin 	heta$.To find the intersection,set $r_1 = r_2$:$\sin 	heta + \cos 	heta =

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$\frac{\pi}{4}$

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(C) Given curves are $r_1 = \sin 	heta + \cos 	heta$ and $r_2 = 2 \sin 	heta$.To find the intersection,set $r_1 = r_2$:$\sin 	heta + \cos 	heta = 2 \sin 	heta \Rightarrow \cos 	heta = \sin 	heta \Rightarrow 	an 	heta = 1 \Rightarrow 	heta = \frac{\pi}{4}$.For $r_1 = \sin 	heta + \cos 	heta$,$\frac{dr_1}{d	heta} = \cos 	heta - \sin 	heta$. Let $\phi_1$ be the angle between the tangent and the radius vector,then $	an \phi_1 = \frac{r_1}{dr_1/d	heta} = \frac{\sin 	heta + \cos 	heta}{\cos 	heta - \sin 	heta}$.At $	heta = \frac{\pi}{4}$,$	an \phi_1 = \frac{1/\sqrt{2} + 1/\sqrt{2}}{1/\sqrt{2} - 1/\sqrt{2}} = \frac{\sqrt{2}}{0} = \infty \Rightarrow \phi_1 = \frac{\pi}{2}$.For $r_2 = 2 \sin 	heta$,$\frac{dr_2}{d	heta} = 2 \cos 	heta$. Then $	an \phi_2 = \frac{r_2}{dr_2/d	heta} = \frac{2 \sin 	heta}{2 \cos 	heta} = 	an 	heta$.At $	heta = \frac{\pi}{4}$,$	an \phi_2 = 	an(\frac{\pi}{4}) = 1 \Rightarrow \phi_2 = \frac{\pi}{4}$.The angle of intersection $\psi = |\phi_1 - \phi_2| = |\frac{\pi}{2} - \frac{\pi}{4}| = \frac{\pi}{4}$.

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to

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