Re: Mitch's questions at Yahoo! Answers regarding a polar equation
Hello Mitch,
We need to use the chain rule as follows to find the slope of the tangent line:
[math]\frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}[/math]
and we need the conversions from polar to Cartesian coordinate systems:
[math]x=r\cos(\theta)=(2-\sin(\theta))\cos(\theta)[/math]
[math]y=r\sin(\theta)=(2-\sin(\theta))\sin(\theta)[/math]
Hence, using the product rule, we find:
[math]\frac{dx}{d\theta}=(2-\sin(\theta))(-\sin(\theta))+(-\cos(\theta))\cos(\theta)=-2\sin(\theta)+\sin^2(\theta)-\cos^2(\theta)=-(2\sin(\theta)+\cos(2\theta))[/math]
[math]\frac{dy}{d\theta}=(2-\sin(\theta))\cos(\theta)+(-\cos(\theta))\sin(\theta)=2\cos(\theta)(1-\sin(\theta))[/math]
Thus, we have:
[math]\frac{dy}{dx}=\frac{2\cos(\theta)(\sin(\theta)-1)}{2\sin(\theta)+\cos(2\theta)}[/math]
[math]\frac{dy}{dx}|_{\theta=\frac{\pi}{3}}=\frac{2\cdot\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1 \right)}{2\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}-2}{2\sqrt{3}-1}=\frac{4-3\sqrt{3}}{11}[/math]
We now have the slope, now the point:
[math](x,y)=(r\cos(\theta),r\sin(\theta))=\left(\frac{4-\sqrt{3}}{4},\frac{4\sqrt{3}-3}{4} \right)[/math]
And so, using the point-slope formula, we find the equation of the tangent line is:
[math]y-\frac{4\sqrt{3}-3}{4}=\frac{4-3\sqrt{3}}{11}\left(x-\frac{4-\sqrt{3}}{4} \right)[/math]
Writing this in slope-intercept form, we obtain:
[math]y=\frac{4-3\sqrt{3}}{11}x+\frac{30\sqrt{3}-29}{22}[/math]
To Mitch and any other guests viewing this topic, I invite and encourage you to register and post other calculus problems here in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.
