Volume of a Cone by Parker Emmerson (1) H1 ê 3L height * area of base = H1 ê 3L 4 p r 2 q- r 2 q 2 2 p p r - r q 2 p ^2 Theorem 1 When a sector of angle q is removed from a circle of radius r and the resulting shape is folded into a cone, then the base of the cone has radius r 1 given by r 1 = r - Hr qL 2 p ; and height h, given by h = r 2 - r 1 2 =r Sin[b] Proof. The circumference of the initial circle is 2 p r and the wedge removed has an arc length r q. Therefore, the remaining circumference is of length r (2 p - q), and after the fold, this is the circumference of the base of the cone. Establishing the circumference of the base of the cone, from the equation, q r = 2 p r - 2 p r 1 , we calculate that its radius r 1 is 2 p r-r q 2 p , which simplifies to r - r q 2 p . Thus, we have proved the first part of the theorem. To find the height of the cone, h, we apply the Pythagorean theorem to a right triangle formed between the apex of the cone, the center of the base, and a point on the circumference of the base. This gives h = r 2 - r 1 2 = r Sin[b], where b is the angle formed by the slant of the cone and the base of the cone. The initial radius is always equal to the slant of the cone, and the height of the cone is always orthogonal to the center of the base of the cone. Lemma 1 The height of the cone can be caluclated in terms of r and q. Proof. q r = 2 p r - 2 p r 1 h = r 2 - r 1 2 q r = 2 p r - 2 p Hr^2 -h ^2L Printed by Mathematica for Students
Transcript
Volume of a Coneby Parker Emmerson
(1)H1 ê 3L height * area of base = H1 ê 3L4 p r2 q - r2 q2
2 pp r -
r q
2 p^2
Theorem 1 When a sector of angle q is removed from a circle of radius r and the resulting shape is folded into a cone, then
the base of the cone has radius r 1 given by r 1= r - Hr qL2 p
; and height h, given by h = r2 - r12 =r Sin[b]
Proof. The circumference of the initial circle is 2 p r and the wedge removed has an arc length r q. Therefore, the remaining circumference is of length r (2 p - q), and after the fold, this is the circumference of the base of the cone.
Establishing the circumference of the base of the cone, from the equation, q r = 2 p r - 2 p r 1, we calculate that its radius r 1 is 2 p r-r q
2 p, which simplifies to r - r q
2 p. Thus, we have proved the first part of the theorem.
To find the height of the cone, h, we apply the Pythagorean theorem to a right triangle formed between the apex of the cone, the
center of the base, and a point on the circumference of the base. This gives h = r2 - r12 = r Sin[b], where b is the angle formed by the slant of the cone and the base of the cone. The initial radius is always equal to the slant of the cone, and the height of the cone is always orthogonal to the center of the base of the cone.
Lemma 1 The height of the cone can be caluclated in terms of r and q.
8r, -1, 1<, 8q, -2 p, 2 p<, 8b, -p ê 2, p ê 2<, 8AxesLabel Ø 8q, r, b<<F
Out[30]=
In[31]:= Plot3D@H1 ê 3L p b^2 * h, 8b, -1, 1<, 8h, -1, 1<D
Out[31]=
Lemma 2 The angle q can be calculated in terms of r and h.
Proof
SolveBh ==4 p r2 q-r2 q2
2 p, qF
::q Ø2 p r2- r4-r2 h2
r2>, :q Ø
2 p r2+ r4-r2 h2
r2>>
Lemma 3 The initial radius is a function of q and h.
Volume of a Cone by Parker Emmerson.nb 3
Printed by Mathematica for Students
SolveB4 p r2 q - r2 q2
2 pã h, rF
::r Ø -2 p h
4 p q - q2>, :r Ø
2 p h
4 p q - q2>>
Lemma 4 The height of the cone can be calculated in terms of only r and q, thus b is a function of q alone.
Proof. Since we have shown that q r = 2 p r - 2 p r 1 and r1 Ø r2 - h2 , we can substitute the expression for r1, calculated from the Pythagorean theorem in terms of the height of the cone and the initial radius of the circle, into the expression for q r in terms of the change in circumference of the initial circle to the circle, which is the base of the cone. q r = 2 p r - 2 p
Hr^2 - h^2L , thus, h =4 p r2 q-r2 q2
2 p = (r Sin[b]). From 2 p h
4 p q-q2= r, we note that r = 2 p r Sin@bD
4 p q-q2, so
q = 2 p ± p2 - p2 Sin@bD2 =2 p r2+ r4-r2 h2
r2, because 1 = 2 p Sin@bD
4 p q-q2
Lemma 5 The height of the cone can be calculated in terms of only r and q, thus q is a function of b alone.
b = ArcSinBH4 p - qL q
2 pF
The elapse of one unit of time, t, can be expressed by a constant function of the angle q. The simplest expression is t = q2 p
; q = k
t , where k is 2p, because one unit of time is equal to one revolution of q through a circle.
Proof. q r = 2 p r - 2 p (r - r t) yields t Ø q
2 p.
V = H1 ê 3L
4 p r22 p r2+ r4-r2 h2
r2- r2
2 p r2+ r4-r2 h2
r2
2
2 pp r -
r2 p r2+ r4-r2 h2
r2
2 p^2
4 Volume of a Cone by Parker Emmerson.nb
Printed by Mathematica for Students
In[32]:= Plot3DBH1 ê 3L
4 p r22 p r2+ r4-r2 h2
r2- r2
2 p r2+ r4-r2 h2
r2
2
2 pp r -
r2 p r2+ r4-r2 h2
r2
2 p^2,
8r, -1, 1<, 8h, -1, 1<, 8AxesLabel Ø 8h, r<<F
Out[32]=
V = H1 ê 3L
4 p r2 2 p + p2 - p2 Sin@bD2 - r2 2 p + p2 - p2 Sin@bD22
2 p
p r -
r 2 p + p2 - p2 Sin@bD2
2 p^2
Volume of a Cone by Parker Emmerson.nb 5
Printed by Mathematica for Students
In[33]:= RevolutionPlot3DBH1 ê 3L
4 p r2 2 p + p2 - p2 Sin@bD2 - r2 2 p + p2 - p2 Sin@bD22
