Post by: tony3272 on June 24, 2011, 08:15:47 pm
Post by: Shark 774 on June 25, 2011, 02:37:06 pm
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.
Post by: b^3 on June 25, 2011, 03:36:43 pm
Post by: Shark 774 on June 25, 2011, 04:34:47 pm
couldn't you just usesince that is the arc length, sub it in and then find the volume?
We don't have a theta.
Post by: tony3272 on June 25, 2011, 04:44:29 pm
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 :-\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.
Post by: xZero on June 25, 2011, 05:20:26 pm
BTW is this question from essential text book?
Post by: gossamer on June 25, 2011, 05:43:54 pm
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]
Post by: tony3272 on June 25, 2011, 05:52:06 pm
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.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]
Post by: gossamer on June 25, 2011, 05:58:50 pm
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?
Post by: tony3272 on June 25, 2011, 06:06:06 pm
Post by: Shark 774 on June 26, 2011, 03:54:34 pm
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" ;)