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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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.