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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given curves are $r_1 = \sin 	heta + \cos 	heta$ and $r_2 = 2 \sin 	heta$.At the point of intersection,$r_1 = r_2$,so $\sin 	heta + \cos 	het

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given curves are $r_1 = \sin 	heta + \cos 	heta$ and $r_2 = 2 \sin 	heta$.At the point of intersection,$r_1 = r_2$,so $\sin 	heta + \cos 	heta = 2 \sin 	heta$,which implies $\cos 	heta = \sin 	heta$,or $	an 	heta = 1$. Thus,$	heta = \frac{\pi}{4}$.For $r_1 = \sin 	heta + \cos 	heta$,$\frac{dr_1}{d	heta} = \cos 	heta - \sin 	heta$. At $	heta = \frac{\pi}{4}$,$\frac{dr_1}{d	heta} = 0$. The angle $\phi_1$ is given by $	an \phi_1 = \frac{r_1}{dr_1/d	heta} = \frac{\sqrt{2}}{0} = \infty$,so $\phi_1 = \frac{\pi}{2}$.For $r_2 = 2 \sin 	heta$,$\frac{dr_2}{d	heta} = 2 \cos 	heta$. At $	heta = \frac{\pi}{4}$,$\frac{dr_2}{d	heta} = 2(\frac{1}{\sqrt{2}}) = \sqrt{2}$. The angle $\phi_2$ is given by $	an \phi_2 = \frac{r_2}{dr_2/d	heta} = \frac{\sqrt{2}}{\sqrt{2}} = 1$,so $\phi_2 = \frac{\pi}{4}$.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:

Similar Questions

Find the equation of the normal lines to the curve $3x^{2}-y^{2}=8$ which are parallel to the line $x+3y=4$.

The values of parameter $a$ such that the line $(\log_{2}(1 + 5a - a^{2}))x - 5y - (a^{2} - 5) = 0$ is a normal to the curve $xy = 1$ may lie in the interval

Find the equations of all lines having slope $0$ that are tangent to the curve $y = \frac{1}{x^2 - 2x + 3}$.

The curves $y=x^2-1$ and $y=8x-x^2-9$:

Find the equation of all the tangents to the curve $y=\cos(x+y)$ for $-2\pi \leq x \leq 2\pi$ that are parallel to the line $x+2y=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module