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 Jamers328 August 6th, 2012 08:09 AM

Area of a Circle problem

The area of a circle
i. Let r be a positive real number and let O = (0, 0) and P = (r, 0) be points in the coordinate plane. Let ?? be the (half)-line that emanates from O and makes an angle ? with respect to the positive x-axis (see figure in attachment).
Let Q be the point on ?? whose distance to O is r and let Q' be the point on ?? whose perpendicular projection onto the positive x-axis is P (see figure in attachment).
Find the x and y coordinates of Q and Q'.

ii. Partition the angle 2? into N equal parts of size ? and let
Cr = area of a circle with radius r,
A = area of the triangle ?OPQ,
A' = area of the triangle ?OPQ',
where O,P,Q,Q' are as in the above figure with angle ? = ?.

(a) Show that
A = r^2 sin(?) / 2
and
A' = r^2 tan(?) / 2

(b) Explain why the inequalities
Nr2 sin(?) / 2 ? Cr ? Nr2 tan(?)/ 2

hold for any r > 0 and any natural number N ? 3.

(c) Find an expression for N in terms of ? and take the limit as ? --> 0 in (7)
to find Cr.

Can anyone help get me started on this? Thank you!

 mathbalarka August 6th, 2012 10:15 AM

Re: Area of a Circle problem

It is straightforoward that the x coordinate of Q' is simply r.

Now, $Q'P = OP \tan(\theta) = r \tan(\theta)$

So, the y coordinate of Q' is $r \tan(\theta)$

Can you proceed for Q now?

 Jamers328 August 6th, 2012 10:59 AM

Re: Area of a Circle problem

From there, I know that the coordinates of QP are (rcos x, rsinx). To find the area, I need to take the anti-derivative of the the function. However, even though we have the two coordinates, how do I know which formula to take the anti-derivative of?

 MarkFL August 6th, 2012 12:42 PM

Re: Area of a Circle problem

10.)

i) We may use the definition of sine and cosine to find:

$\cos$$\theta$$=\frac{Q_x}{r}\:\therefore\:Q_x=r\co s$$\theta$$$

$\sin$$\theta$$=\frac{Q_y}{r}\:\therefore\:Q_y=r\si n$$\theta$$$

Since $Q_x'=r$ we then find $\tan$$\theta$$=\frac{Q_y'}{r}\:\therefore\:Q_y #39;=r\tan$$\theta$$$

ii)

a) Using the formula for the area of a triangle $A=\frac{1}{2}bh$ we then find:

$A=\frac{1}{2}r\cdot r\sin$$\Delta\varphi$$=\frac{r^2\sin$$\Delta\varph i$$}{2}$

$A'=\frac{1}{2}r\cdot r\tan$$\Delta\varphi$$=\frac{r^2\tan$$\Delta\varph i$$}{2}$

b) For any natural number $N\ge3$ and $r>0$ the geometrical interpretation of the given inequality is that for a circle of radius r, its area is greater than an inscribed N-gon and less than that of a circumscribed N-gon.

c) $\Delta\varphi=\frac{2\pi}{N}\:\therefore\:N=\frac{ 2\pi}{\Delta\varphi}$ and hence:

$C_r=\lim_{\Delta\varphi\to0}\frac{2\pi r^2\sin$$\Delta\varphi$$}{2\Delta\varphi}=\lim_{\D elta\varphi\to0}\frac{\pi r^2\sin$$\Delta\varphi$$}{\Delta\varphi}=$

$\pi r^2\lim_{\Delta\varphi\to0}\frac{\sin$$\Delta\varp hi$$}{\Delta\varphi}=\pi r^2(1)=\pi r^2$

 Jamers328 August 6th, 2012 12:47 PM

Re: Area of a Circle problem

Thank you for the help! As you can see from my prior post, I was really off from what I needed to do, so I appreciate the help in preparation for my final. Thank you thank you thank you!

 MarkFL August 6th, 2012 01:02 PM

Re: Area of a Circle problem

To be more formal about the last part, we could state:

$\pi r^2\lim_{\Delta\varphi\to0}\frac{\sin$$\Delta\varp hi$$}{\Delta\varphi}\le C_r\le\pi r^2\lim_{\Delta\varphi\to0}\frac{\tan$$\Delta\varp hi$$}{\Delta\varphi}$

Apply L'Hôpital's rule to the limits (although this isn't necessary, just quicker):

$\pi r^2\lim_{\Delta\varphi\to0}\cos$$\Delta\varphi$$\l e C_r\le\pi r^2\lim_{\Delta\varphi\to0}\sec^2$$\Delta\varphi$$$

$\pi r^2\le C_r\le\pi r^2$

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