Question

Find the angle between the radius vector and the tangent line of the curve r=a(1+cos theta)

Answer 1

Answer:

We have tan ϕ = r dθ/dr = r / dr/dθ

Step-by-step explanation:

r = a (1 - cos θ) .

So dr/dθ = a sin θ .

Hence, tan ϕ = a (1 - cos θ) / (a sin θ)

= 2 sin^2 (θ/2) / (2 sin (θ/2) cos (θ/2) )

= tan (θ/2) .

So ϕ = θ/2 .

Answer 2

Answer:

The angle between radius vector and tangent is  $\frac{\pi }{2} +\frac{\theta}{2}$

Step-by-step explanation:

The given curve is r = a(1 + cos$\theta$)

differentiate r with respect to $\theta$, we get

$\frac{dr}{d\theta} =-asin\theta$

We know that

tanФ =  $\frac{r}{\frac{dr}{d\theta} }$

Here Ф is the angle between radius vector and tangent

tanФ = r ×$\frac{d\theta}{dr}$

        $=\frac{r}{\frac{dr}{d\theta} }$

        $=\frac{a(1+cos\theta)}{-asin\theta}$

       $=-\frac{2cos^{2}\frac{\theta}{2} }{2sin\frac{\theta}{2}cos\frac{\theta}{2} }$

       $=-\frac{cos\frac{\theta}{2} }{sin\frac{\theta}{2} }$

      $=-cot(\frac{\theta}{2} )$

       $=tan(\frac{\pi }{2} +\frac{\theta}{2} )$

Ф $=\frac{\pi }{2} +\frac{\theta}{2}$

Therefore the angle between radius vector and tangent is $\frac{\pi }{2} +\frac{\theta}{2}$