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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!

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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 $y=\frac{1}{\sqrt{3}}x,y=-\frac{1}{\sqrt{3}}x$

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  • $\begingroup$ No, my answers are saying y=1/sqrt(x)*x and y=-1/sqrt(3)*x $\endgroup$ – EhBabay May 4 '13 at 6:38
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    $\begingroup$ I edited my answer, have a look $\endgroup$ – Aang May 4 '13 at 6:42
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    $\begingroup$ Aha thank you very much. This makes more sense! $\endgroup$ – EhBabay May 4 '13 at 6:44
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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.

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  • $\begingroup$ Great prompts here, my friend! ;-) +1 $\endgroup$ – Namaste May 5 '13 at 0:26
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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.

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  • $\begingroup$ $y=r(\theta).\sin\theta$ $\endgroup$ – Aang May 4 '13 at 6:33
  • $\begingroup$ Thanks @Avatar, slip of the hand $\endgroup$ – kvmu May 4 '13 at 6:34
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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 $$y=\pm\frac{\sqrt{3}}{3}x$$

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