Find the equation(s) of the tangent line(s) to the curve given $(x,y)$ point.
$$r=1-2\sin(\theta )$$ at $(0,0)$. I am not sure how to go about find the the tangent line. Do I need to convert from polar to rectangular?
Thanks!
Find the equation(s) of the tangent line(s) to the curve given $(x,y)$ point.
$$r=1-2\sin(\theta )$$ at $(0,0)$. I am not sure how to go about find the the tangent line. Do I need to convert from polar to rectangular?
Thanks!
For polar coordinates , $x=r\cos\theta\implies dx=\cos\theta dr-r\sin\theta d\theta$
$y=r\sin\theta\implies dy=\sin\theta dr+r\cos\theta d\theta$
So, $$\frac{dy}{dx}=\frac{\sin\theta dr+r\cos\theta d\theta}{\cos\theta dr-r\sin\theta d\theta}$$
Now, at $(0,0)$, $r=0\implies \sin\theta=1/2\implies \tan\theta=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$ which gives $$\frac{dy}{dx}=\tan\theta=\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$$
So, equations of tangents are $y=\frac{1}{\sqrt{3}}x,y=-\frac{1}{\sqrt{3}}x$
Hint: You know by using the Chain rule and that $y=r\sin(\theta),~~x=r\cos(\theta)$, we have $$m_{\text{tangent}}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}$$ So find the coordinates of point in polar coordinates and then write the line equation using $m$ above.
HINT:
$x = r(\theta)\cdot\cos(\theta)$
$y = r(\theta)\cdot\sin(\theta)$
$ m = \dfrac{dy}{dx}$ , where $m$ is the slope of the tangent line.
You can probably take it from here.
Let's use the polar form for the slope, which can be derived using the product rule as follows: $$\frac{dy}{dx}=\frac{d(r\sin\theta)}{d(r\cos\theta)}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta}$$
Now since $r(\theta)=1-2\sin\theta$, we have $r'(\theta)=-2\cos\theta$, so our slope becomes: $$\frac{dy}{dx}=\frac{\cos\theta-4\sin\theta\cos\theta}{2\sin^2\theta-2\cos^2\theta-\sin\theta}=\frac{\cos\theta-2\sin2\theta}{4\sin^2\theta-\sin\theta-2}$$ Now $(0,0)$ corresponds to angles $\theta=\frac{\pi}{6},\frac{5\pi}{6}$, and plugging these in gives two slopes: $$m=\frac{\sqrt{3}}{3},\;\frac{-\sqrt{3}}{3}$$ corresponding to the lines $$y=\pm\frac{\sqrt{3}}{3}x$$