calculus: Sphere fitted in a regular cone, volume of cone=?

2008-06-15 5:13 pm

Suppose a sphere with radius R is fitted into a regular cone, such that the sphere touches the sides of the cone, and also the base of the cone, with a height of h. (Draw your own diagram.)

When the cone is placed steadily on the table on its base, (with the sphere inside it), the angle of elevation, measured from the centre of the base of the cone surface, to the centre of gravity in the sphere inside the cone is x degrees.
(a) show that volume of cone V = 1/3 (pi R^3 tan 2 x cotan^3 x
(b) show that mini volume of cone occurs at cos 2x = 1/3
(c) find the mini volume of the cone.

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(The angle of elevation x for the sphere's CG, is measured from the centre of gravity lying in this vertical line, the centre of the disk, and the other end of the flat disk.

回答 (1)

[匿名]

2008-06-17 8:30 am

✔ 最佳答案

The following is the answer.

(a) Step1: Find the raduis of the base of the cone r.
tanx=R/r
r=R/tanx=Rcotx

Step2: Find the height of the cone h.
tan(90-2x)=r/h
cot2x=Rcotx/h
h=Rcotx/cot2x=Rtan2xcotx

Step3: Find the volume of the cone V.
V=1/3(base area of the cone)(height)
V=1/3(pi r^2)h
V=1/3[pi (Rcotx)^2](Rtan2xcotx)
V=1/3(pi)R^3tan2xcot^3(x)

(b)To find the mini volume, find dV/dx=0.
dV/dx=1/3(pi)R^3[2sec^2(2x)cot^3(x)-3cot^2(x)csc^2(x)tan2x]

V attains its minimum value when dV/dx=0, i.e.
2sec^2(2x)cot^3(x)-3cot^2(x)csc^2(x)tan2x=0
2sec^2(2x)cot^3(x)=3cot^2(x)csc^2(x)tan2x
2cos^3(x)/[cos^2(2x)sin^3(x)]=3sin2xcos^2(x)/[sin^4(x)cos2x]
2cosx/cos2x=3sin2x/sinx
2cosx/cos2x=6sinxcosx/sinx
1/cos2x=3
cos2x=1/3

(c) when cos2x=1/3,
2x=70.5
x=35.3

The mini volume of the cone
= 1/3(pi)R^3tan(70.5)cot^3(35.3)
= 8/3(pi)R^3

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