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

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Polar
AI Thread Summary
To find the slope of the tangent line for the polar curve r = 2 - sin(θ) at θ = π/3, the chain rule and conversions to Cartesian coordinates are employed. The derivatives dx/dθ and dy/dθ are calculated, leading to the slope formula dy/dx = (2cos(θ)(sin(θ)-1))/(2sin(θ)+cos(2θ)). Substituting θ = π/3 yields a slope of (4 - 3√3)/11. Additionally, the coordinates of the tangent point are determined, and the tangent line equation is provided in slope-intercept form.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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.
 
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top