Cones

[tony3272]

Hey guys. Just wondering, if you had a circle and you cut out a sector of arc length x, and radius r in order to make a cone, how would you express the volume just in terms of x? The equation i've got at the moment still has an r in it and it's not helping my cause at all 

[Shark 774]

Hey guys. Just wondering, if you had a circle and you cut out a sector of arc length x, and radius r in order to make a cone, how would you express the volume just in terms of x? The equation i've got at the moment still has an r in it and it's not helping my cause at all 

I don't think it's possible. Think about it: "r" is the variable that determines the size of the circle, and hence you need "r" to determine the volume of the cone.

[b^3]

couldn't you just use since that is the arc length, sub it in and then find the volume?

[Shark 774]

couldn't you just use since that is the arc length, sub it in and then find the volume?

We don't have a theta.

[tony3272]

Hey guys. Just wondering, if you had a circle and you cut out a sector of arc length x, and radius r in order to make a cone, how would you express the volume just in terms of x? The equation i've got at the moment still has an r in it and it's not helping my cause at all 

I don't think it's possible. Think about it: "r" is the variable that determines the size of the circle, and hence you need "r" to determine the volume of the cone.

Yea that's what ive been thinking. I've punched in a bunch of random equations and ive found no way to express r in terms of x 

[xZero]

you can, the equations required are 'x=r(theta), circumference = 2*pi*r and a relationship between the circle part of the cone and the old circle'. You dont need to know theta, just leave it as a variable. I suggest you to get a piece of paper and do some experiment with it.

BTW is this question from essential text book?

[gossamer]

Hey guys. Just wondering, if you had a circle and you cut out a sector of arc length x, and radius r in order to make a cone, how would you express the volume just in terms of x? The equation i've got at the moment still has an r in it and it's not helping my cause at all 

I'm not sure that there's a way to express just in terms of x - the volume would also be dependent upon the angle of the sector.

Anyway: (Sorry, it's a bit messy and probably hard to read because I haven't used latex)

x = (2 * pi * r * theta) / 360, where theta is the angle of the sector in degrees
Rearrage:
r = (360 * x) / (2 * pi * theta)
  = (180 * x) / (pi * theta)

x = 2 * pi * r1 as well, where r1 is the radius of the base circle of the cone
r1 = x / (2 * pi)

If you look at the cross-section of the cone, the height of the cone can be found using pythag:
r1^2+h^2=r^2
h=sqrt(r^2-r1^2)
that is,
(x / (2 * pi))^2 + h^2 = ((180*x) / (pi*theta))^2
h = sqrt( ((180 * x) / (pi * theta))^2 - (x / (2*pi))^2 )
h = sqrt( (32400x^2 / pi^2 * theta^2) - ( x^2 / 4 pi^2 ) )
Make the thing inside the sqrt one fraction:
h = sqrt( (129600 x^2 - x^2*pi*theta^2) / (4*pi^2*theta^2) )
h = sqrt( (x^2(129600 -pi*theta^2)) / (4*pi^2*theta^2) ) --> factor x^2 out of the numerator
h = (x * sqrt(129600 -pi*theta^2) ) / 2*pi*theta

Volume of cone = 1/3 * pi*r^2*h

The r used for the formula is what we've called r1 (radius of the base circle of the cone) = x/2pi

So, V = (1/3) * pi * (x/2pi)^2 * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = (1/3) * pi * x^2/4pi^2 * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = (1/3) * x^2/4pi * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = [x^3 * sqrt(129600 -pi*theta^2)] / [24*pi^2*theta]

[tony3272]

Hey guys. Just wondering, if you had a circle and you cut out a sector of arc length x, and radius r in order to make a cone, how would you express the volume just in terms of x? The equation i've got at the moment still has an r in it and it's not helping my cause at all 

I'm not sure that there's a way to express just in terms of x - the volume would also be dependent upon the angle of the sector.

Anyway: (Sorry, it's a bit messy and probably hard to read because I haven't used latex)

x = (2 * pi * r * theta) / 360, where theta is the angle of the sector in degrees
Rearrage:
r = (360 * x) / (2 * pi * theta)
  = (180 * x) / (pi * theta)

x = 2 * pi * r1 as well, where r1 is the radius of the base circle of the cone
r1 = x / (2 * pi)

If you look at the cross-section of the cone, the height of the cone can be found using pythag:
r1^2+h^2=r^2
h=sqrt(r^2-r1^2)
that is,
(x / (2 * pi))^2 + h^2 = ((180*x) / (pi*theta))^2
h = sqrt( ((180 * x) / (pi * theta))^2 - (x / (2*pi))^2 )
h = sqrt( (32400x^2 / pi^2 * theta^2) - ( x^2 / 4 pi^2 ) )
Make the thing inside the sqrt one fraction:
h = sqrt( (129600 x^2 - x^2*pi*theta^2) / (4*pi^2*theta^2) )
h = sqrt( (x^2(129600 -pi*theta^2)) / (4*pi^2*theta^2) ) --> factor x^2 out of the numerator
h = (x * sqrt(129600 -pi*theta^2) ) / 2*pi*theta

Volume of cone = 1/3 * pi*r^2*h

The r used for the formula is what we've called r1 (radius of the base circle of the cone) = x/2pi

So, V = (1/3) * pi * (x/2pi)^2 * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = (1/3) * pi * x^2/4pi^2 * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = (1/3) * x^2/4pi * (x * sqrt(129600 -pi*theta^2) )/(2*pi*theta)
        = [x^3 * sqrt(129600 -pi*theta^2)] / [24*pi^2*theta]

Yeah that's what i've done to get my volume, but i decided not to use that value for r as it's a lot more messy and you still have the variable theta,so when i differentiate it it wont change anything.

[gossamer]

Yeah that's what i've done to get my volume, but i decided not to use that value for r as it's a lot more messy and you still have the variable theta,so when i differentiate it it wont change anything.

Yeah, the other r would make the equation slightly nicer.
Why are you needing to differentiate it?

[tony3272]

Part of an analysis task question, find the max volume, draw and describe the graph, state domain and range etc. etc...

[Shark 774]

you can, the equations required are 'x=r(theta), circumference = 2*pi*r and a relationship between the circle part of the cone and the old circle'. You dont need to know theta, just leave it as a variable. I suggest you to get a piece of paper and do some experiment with it.

BTW is this question from essential text book?

Technically you can't, coz he said "just in terms of x"