# Thread: What's theta??

1. ## What's theta??

Stuck on what's next......Find the sector angle (where cone volume is maximised)when a sector is removed and the circle circumference is joined (point to point) to make a cone.
I came up with this equation:
Volume of cone – sector=V=1/3h(pir^2-(theta/360xpir^2))
Unknowns…V, theta, r, h.
I know I need derivative but I need to put equation in terms one variable.
Any tips on how to proceed??

2. Originally Posted by Neverquit
Stuck on what's next......Find the sector angle (where cone volume is maximised)when a sector is removed and the circle circumference is joined (point to point) to make a cone.
I came up with this equation:
Volume of cone – sector=V=1/3h(pir^2-(theta/360xpir^2))
Unknowns…V, theta, r, h.
I know I need derivative but I need to put equation in terms one variable.
Any tips on how to proceed??
Area of the circle = π*R^2
Area of the sector = 1/2*R^2*θ
Therefore surface area of the cone = π*R^2 - 1/2*R^2*θ = πR*r,....(1) where r is the radius of the cone.
Now volume of the cone V = 1/3*π*r^2*h.
But r^2 = (R^2 - h^2)
So V = 1/3*π*h*(R^2 - h^2)
dV/dh = 0
Find R in terms of h and the in terms of r. And put it in eq(1) to find θ.

3. Originally Posted by Neverquit
Stuck on what's next......Find the sector angle (where cone volume is maximised)when a sector is removed and the circle circumference is joined (point to point) to make a cone.
I came up with this equation:
Volume of cone – sector=V=1/3h(pir^2-(theta/360xpir^2))
Unknowns…V, theta, r, h.
I know I need derivative but I need to put equation in terms one variable.
Any tips on how to proceed??
Hi Neverquit,

$V_{cone}=\frac{{\pi}r^2h}{3}$

Use Pythagoras' theorem to obtain one variable instead of 2

$R^2=r^2+h^2\ \Rightarrow\ r^2=R^2-h^2$

$V_{cone}=\frac{{\pi}\left(R^2-h^2\right)h}{3}=\frac{{\pi}R^2h-{\pi}h^3}{3}$

differentiate wrt "h"

$\frac{{\pi}R^2}{3}-\frac{{\pi}3h^2}{3}=0$

$3h^2=R^2$

$h^2=\frac{R^2}{3}$

If we find "r", we can find the remaining arc length of the circle from which the sector was cut from, since

$2{\pi}r=cone\ circular\ circumference=arc\ length\ from\ circle$

$r^2=R^2-h^2=R^2-\frac{R^2}{3}=\frac{2R^2}{3}$

$r=\sqrt{\frac{2}{3}}R$

$\frac{2{\pi}r}{2{\pi}R}=\frac{\alpha}{360^o}=\frac {r}{R}=\sqrt{\frac{2}{3}}$

$\theta=360^o-\alpha$

4. Hello, Neverquit!

This is an intricate (and badly worded) problem . . .

Find the sector angle (where cone volume is maximized)
when a sector is removed and the remainder is joined to form a cone.
Code:
              * * *
*           *
*               *
*                 *

*         O         *
*         o         *
*       * θ *  R    *
*       *
*  *           *  *
A o               o B
*           *
* * *
Rθ

The sector angle is $\theta \,=\,\angle AOB.$
The radius of the circle is $R.$

The arc length of $AB$ is: . $R\theta.$
This is the circumference of the circular base of our cone.
. . $2\pi r \:=\:R\theta \quad\Rightarrow\quad r \:=\:\frac{R\theta}{2\pi}$ .[1]

A cross-section of our cone looks like this:

Code:
            *
/|\
/ | \
/ h|  \ R
/   |   \
/    |    \
* - - + - - *
r
We have: . $h^2 \;=\;R^2 - r^2 \;=\;R^2 - \left(\frac{R\theta}{2\pi}\right)^2 \;=\;\frac{R^2(4\pi^2 - \theta^2)}{4\pi^2}$

. . Hence: . $h \;=\;\frac{R}{2\pi}\sqrt{4\pi^2 - \theta^2}$ .[2]

The volume of a cone is: . $V \;=\;\frac{\pi}{3}r^2h$

Substitute [1] and [2]: . $V \;=\;\frac{\pi}{3}\left(\frac{R\theta}{2\pi}\right )^2\cdot\frac{R}{2\pi}\sqrt{4\pi^2 - \theta^2}$

. . Therefore: . $V \;=\;\frac{R^3}{24\pi^2}\theta^2\sqrt{4\pi^2-\theta^2}$

And that is the function we must maximize.

5. ## Updated

Hi Soroban,

The sector angle is
The radius of the circle is
The arc length is .

The cone is created when the sector is removed and AB are joined together.

I wanted to make the question less wordy.

6. Hi soroban,
"The arc length of is: .
This is the circumference of the circular base of our cone.
. . .[1]
"
This is not correct. The cone is not formed from the sector removed from the circle. It is formed from the remaining part of the circle.
Your calculation of r is correct. But θ is not correct. For that you have to equate the area of the circle to sum of the area of the sector and surface area of the cone.
Therefore surface area of the cone = π*R^2 - 1/2*R^2*θ = πR*r
Substitute the value of r and find θ.

7. my calculation is fine,
i was wasn't working with the cone surface area,
i worked with the original circle.

8. Originally Posted by Archie Meade
my calculation is fine,
i was wasn't working with the cone surface area,
i worked with the original circle.
Sorry. You are right.

9. ## A few more questions.....

I followed Archie Meads methodology and I don't understand a few parts:

How.. r^2=2R^2/3

and

how I get..theta=360deg-alpha

Theta should =66.1deg

10. Originally Posted by Neverquit
I followed Archie Meads methodology and I don't understand a few parts:

How.. r^2=2R^2/3

and

how I get..theta=360deg-alpha

Theta should =66.1deg
Hi Neverquit,

in solving for $h^2$ of the cone corresponding to max volume,

it is $\frac{R^2}{3}$

where $R$ is the radius of the original circle.

When the sector is cut from that, we are left with the "pac-man" shaped remainder of the circle,
that looks like a pizza with a slice removed.
The cone is formed from this.

I calculated the angle of that and subtracted the answer from 360 degrees,
since that then gives the angle of the sector that was removed from the original circle.

The outside arc length of the "pac-man" equals the circumference of the circular opening of the cone.

Also, using Pythagoras' theorem, we can express r in terms of R
using our result for the height of the cone corresponding to maximum volume,
since R is the slant length of the cone.

Evaluating that gives

$r^2=R^2-h^2=R^2-\frac{R^2}{3}=\frac{3R^2-R^2}{3}=\frac{2R^2}{3}$

The cone circumference length is the "pac-man" arc length.

The ratio of this length to the original circle circumference
is the ratio of alpha to 360 degrees, where alpha is the angle of the "pac-man".

Theta is then alpha subtracted from 360 degrees.

Alpha is about 294 degrees or so, giving theta approximately 66 degrees.

$\alpha=\frac{\sqrt{2}360^o}{\sqrt{3}}=293.93^o$

Then $\theta=360^o-293.938^o=66.062^o$