Question

find the angle between the radius vector and tangent to the curve r=a(1+cos theta) and also find the slop of the curve at theta=phi÷6

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{r=a\left(1+cos(\theta)\right)}$

Now,

$\rm{\implies\dfrac{dr}{d\theta}=-a\cdot\,sin(\theta)}$

So, the angle between radius vector and tangent is given by,

$\sf{tan(\phi)=r\cdot\dfrac{d\theta}{dr}}$

$\sf{\implies\,tan(\phi)=a\left(1+cos(\theta)\right)\cdot\dfrac{-1}{a\cdot\,sin(\theta)}}$

$\sf{\implies\,tan(\phi)=-\dfrac{1+cos(\theta)}{sin(\theta)}}$

$\sf{\implies\,tan(\phi)=-\dfrac{2\,cos^{2}\left(\dfrac{\theta}{2}\right)}{2\,sin\left(\dfrac{\theta}{2}\right)\,cos\left(\dfrac{\theta}{2}\right)}}$

$\sf{\implies\,tan(\phi)=-\dfrac{cos\left(\dfrac{\theta}{2}\right)}{sin\left(\dfrac{\theta}{2}\right)}}$

$\sf{\implies\,tan(\phi)=-\,cot\left(\dfrac{\theta}{2}\right)}$

$\sf{\implies\,tan(\phi)=tan\left(\dfrac{\pi}{2}+\dfrac{\theta}{2}\right)}$

$\sf{\implies\,\phi=\dfrac{\pi}{2}+\dfrac{\theta}{2}}$

Hence, the angle between the radius vector and the tangent is  $\dfrac{\pi}{2}+\dfrac{\theta}{2}$

Now, the slope of the tangent is given by,

$\rm{\dfrac{dy}{dx}=\dfrac{\dfrac{dr}{d\theta}\cdot\,sin(\theta)+r\cdot\,\cos(\theta)}{\dfrac{dr}{d\theta}\cdot\,\cos(\theta)-r\cdot\,\sin(\theta)}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{-a\,\sin(\theta)\cdot\,sin(\theta)+a\left(1+\cos(\theta)\right)\cdot\,\cos(\theta)}{-a\,\sin(\theta)\cdot\,\cos(\theta)-a\left(1+\cos(\theta)\right)\cdot\,\sin(\theta)}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{-a\,\sin^{2}(\theta)+a\,\cos(\theta)+a\,\cos^{2}(\theta)}{-a\,\sin(\theta)\,\cos(\theta)-a\,\sin(\theta)-a\,\sin(\theta)\,\cos(\theta)}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{\cos(\theta)+\cos^{2}(\theta)-\sin^{2}(\theta)}{-2\,\sin(\theta)\,\cos(\theta)-\sin(\theta)}}$

$\rm{\implies\,\dfrac{dy}{dx}=\dfrac{\cos(\theta)+\cos(2\theta)}{-\sin(2\theta)-\sin(\theta)}}$

$\rm{\implies\,\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{6}}=\dfrac{\cos\left(\dfrac{\pi}{6}\right)+\cos\left(\dfrac{\pi}{3}\right)}{-\sin\left(\dfrac{\pi}{3}\right)-\sin\left(\dfrac{\pi}{6}\right)}}$

$\rm{\implies\,\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{6}}=\dfrac{\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}}{-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}}}$

$\rm{\implies\,\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{6}}=\dfrac{\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}}{-\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)}}$

$\rm{\implies\,\dfrac{dy}{dx}\bigg|_{\theta=\frac{\pi}{6}}=-1}$