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Mitch's question at Yahoo! Answers regarding a polar equation

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MarkFL

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Feb 24, 2012
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Here is the question:

Calculus II Polar Coordinates?

Find the slope of the tangent line to the given polar curve at the point specified by the value of Ɵ using the exact answer: r = 2 - sinƟ, Ɵ = π/3
Here is a link to the question:

Calculus II Polar Coordinates?

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
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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:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{d\theta}\cdot\frac{d\theta}{dx}\)

and we need the conversions from polar to Cartesian coordinate systems:

\(\displaystyle x=r\cos(\theta)=(2-\sin(\theta))\cos(\theta)\)

\(\displaystyle y=r\sin(\theta)=(2-\sin(\theta))\sin(\theta)\)

Hence, using the product rule, we find:

\(\displaystyle \frac{dx}{d\theta}=(2-\sin(\theta))(-\sin(\theta))+(-\cos(\theta))\cos(\theta)=-2\sin(\theta)+\sin^2(\theta)-\cos^2(\theta)=-(2\sin(\theta)+\cos(2\theta))\)

\(\displaystyle \frac{dy}{d\theta}=(2-\sin(\theta))\cos(\theta)+(-\cos(\theta))\sin(\theta)=2\cos(\theta)(1-\sin(\theta))\)

Thus, we have:

\(\displaystyle \frac{dy}{dx}=\frac{2\cos(\theta)(\sin(\theta)-1)}{2\sin(\theta)+\cos(2\theta)}\)

\(\displaystyle \frac{dy}{dx}|_{\theta=\frac{\pi}{3}}=\frac{2\cdot\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1 \right)}{2\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}-2}{2\sqrt{3}-1}=\frac{4-3\sqrt{3}}{11}\)

We now have the slope, now the point:

\(\displaystyle (x,y)=(r\cos(\theta),r\sin(\theta))=\left(\frac{4-\sqrt{3}}{4},\frac{4\sqrt{3}-3}{4} \right)\)

And so, using the point-slope formula, we find the equation of the tangent line is:

\(\displaystyle y-\frac{4\sqrt{3}-3}{4}=\frac{4-3\sqrt{3}}{11}\left(x-\frac{4-\sqrt{3}}{4} \right)\)

Writing this in slope-intercept form, we obtain:

\(\displaystyle y=\frac{4-3\sqrt{3}}{11}x+\frac{30\sqrt{3}-29}{22}\)

To Mitch and any other guests viewing this topic, I invite and encourage you to register and post other calculus problems here in our Calculus forum.

Best Regards,

Mark.
 
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MarkFL

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Feb 24, 2012
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I notice only now that the OP is only asked for the slope of the tangent line...but I suppose is is better to provide too much rather than not enough information. (Giggle)

Once we have the slope, and can find the point, it only seems natural to go ahead and find the line. (Malthe)