Question

find the angle between the polar curves r=a(1-costheta) and r=(1+costheta)​

Answer 1

Answer:

diff loga=0

pi1=theta /2

$pi2 |\pi \div 2 + theta \div 2|$

$|pi1 - pi2|$

ans =

$\pi \div 2$

Answer 2

Answer:

The angle between the polar curves r=a(1-costheta) and r=(1+costheta)​ is $\pi / 2$

Explanation:

Step 1: A form created using the polar coordinate system is called a polar curve. Points on polar curves have varying distances from the origin (the pole), depending on the angle taken off the positive x-axis to calculate distance. The cardioids, limaons, lemniscates, rose curves, and Archimedes spirals are the five traditional polar curves.

Step 2: The polar equation, r =  a cos nθ, looks like a rose curve. The rose will have 2n petals since n is an even number. With the help of the rose curve's symmetry, these points will provide us a sufficient number of points to finish the rest of the graph.

Step 3: Taking $\log$ on both sides of 1 st equation, $\log r=\log a(1-\cos \theta)$

On differentiation,$1 / r . d r / d \theta=\sin \theta / 1-\cos \theta=\cot \theta / 2$

$\therefore \tan \varphi_1=r d \theta / d r=\tan \theta / 2$ or $\varphi_1=\theta / 2$

Similarly from 2 nd equation, $\tan \varphi_2=1+\cos \theta /-\sin \theta=-\cot \theta / 2$

or $\varphi_2=(\pi / 2+\theta / 2)$

Thus the angle of intersection ' $\alpha$ ' between the two curves is given by

$\tan \alpha=\tan \left(\varphi_1 \sim \varphi_2\right)$ or $\alpha=\varphi_1 \sim \varphi_2=\pi / 2$

To learn more about similar questions visit:

https://brainly.in/question/34435823?referrer=searchResults

https://brainly.in/question/40669509?referrer=searchResults

#SPJ3