show that area =

Joel · #1 $Old$ November 3rd, 2009, 06:35 AM

Hi All,

OP and OQ are radii of length r cm of a circle centred at O.
The arc PQ of the circle subtends an angle of theta radians at O and the perimeter of the sector OPQ is 12 cm.

Show that the are A cm squared of the sector is given by

A = 72 theta / (2 + theta)^2

Any help would be appreciated

alexmahone · #2 $Old$ November 3rd, 2009, 06:54 AM

Quote:

Originally Posted by Joel $View Post$

Hi All,

OP and OQ are radii of length r cm of a circle centred at O.
The arc PQ of the circle subtends an angle of theta radians at O and the perimeter of the sector OPQ is 12 cm.

Show that the are A cm squared of the sector is given by

A = 72 theta / (2 + theta)^2

Any help would be appreciated

$r+r+r\theta=12$

$2r+r\theta=12$

$r=\frac{12}{2+\theta}$

$A=\frac{1}{2}{r^2}{\theta}$

$=\frac{1}{2}(\frac{12}{2+\theta})^2*\theta$

$=\frac{72\theta}{(2+\theta)^2}$

Joel · #3 $Old$ November 3rd, 2009, 06:33 PM

Thanks,

I am now asked to find the maximumarea of this sector. Answer in the book gives me 9cm^2.

How do I show this??? Do I differentiate or am I on the wrong path?

alexmahone · #4 $Old$ November 3rd, 2009, 06:45 PM

Quote:

Originally Posted by Joel $View Post$

Thanks,

I am now asked to find the maximumarea of this sector. Answer in the book gives me 9cm^2.

How do I show this??? Do I differentiate or am I on the wrong path?

For maximum area of the sector, $\frac{dA}{d\theta}=0$ .

Joel · #5 $Old$ November 3rd, 2009, 06:54 PM

to find dy / dx of (72 theta) / (2 + theta)^2

do i get rid of the fraction by multiplying 72 theta by (2 + theta)^2??

This may be completly wrong

I get 288theta + 288theta^2 + 78 theta^3

dy dx = 576theta + 234theta

I think this is just flat out WRONG.

Any help appreciated

alexmahone · #6 $Old$ November 3rd, 2009, 07:12 PM

Joel · #7 $Old$ November 3rd, 2009, 09:56 PM

to make this = 0

2 + theta would have to be 0 = theta = -2

is this correct?

alexmahone · #8 $Old$ November 3rd, 2009, 10:01 PM

Quote:

Originally Posted by Joel $View Post$

to make this = 0

2 + theta would have to be 0 = theta = -2

is this correct?

Incorrect!

Factorise the numerator and equate it to zero.

Joel · #9 $Old$ November 4th, 2009, 01:29 AM

thanks heaps for your patience,

factorise the numerator,

[IMG]file:///C:/DOCUME%7E1/Nate/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]

(2 + theta)^2 . 72(1 - theta) . 2(2 + theta)

am I on the right track ?

alexmahone · #10 $Old$ November 4th, 2009, 01:50 AM

$\frac{dA}{d\theta}=\frac{(2+\theta)^2*72-72\theta*2(2+\theta)}{(2+\theta)^4}$

= $\frac{72}{(2+\theta)^4}\{(2+\theta)^2-2\theta(2+\theta)\}$

= $\frac{72}{(2+\theta)^4}\{(2+\theta)(2+\theta-2\theta)\}$

= $\frac{72}{(2+\theta)^3}(2-\theta)$

$\frac{dA}{d\theta}=0\implies\theta=2$

Joel · #11 $Old$ November 4th, 2009, 04:25 AM

so putting 2 back into equation

= 144 / 16

= 9

Thats the answer.... woo hoo.

Thanks very much for your patience and guidance. Very much appreciated.