Question

2. Find the angle between the radius vector and the tangent for the curve r2 = a2 cos 2θ at θ = π/6.

Answer 1

Answer:

Yoshi

Step-by-step explanation:

thanks flush out of the day

Answer 2

The angle between the radius vector and the tangent for the curve $r^{2}$ = $a^{2}$ cos 2θ at θ = π/6  is  5π/6.

Step-by-step explanation:

Given:

The curve $r^{2}$ = $a^{2}$ cos 2θ

To Find:

The angle between the radius vector and the tangent at  θ = π/6 for the curve $r^{2}$ = $a^{2}$ cos 2θ .

Formula Used:

  Cot A = 1/r. dr/dθ       --------------------------- formula no. 01

The angle between the radius vector and the tangent for the curve is A.

The polar coordinates are r and θ.

Solution:

$r^{2}$ = $a^{2}$ cos 2θ

Taking Log of both sides.

Log $r^{2}$ = $a^{2}$ Log( cos 2θ)

2 Log r = Log $a^{2}$  + Log cos 2θ

Differentiating  r with respect to θ.

2.1/r. dr/dθ = 0+ (1/ cos 2θ).(-2 sin 2 θ)

2.1/r. dr/d θ = - 2 tan 2θ

1/r .dr/d θ =- - tan 2θ

Value of  dr/dθ   at θ= π/6

1/r .dr/d θ =  - tan (2 π/6)

Applying formula  no 01 ,  Cot A = 1/r. dr/dθ      

Cot A   =  - tan π/3

Cot A    = - $\sqrt{3}$

 A= 5 π/6

                                                                                                         

Thus, The angle between the radius vector and the tangent for the curve $r^{2}$ = $a^{2}$ cos 2θ at θ = π/6  is  5 π/6.