Applications-differentiation1

2010-11-22 8:14 am

圖片參考：http://imgcld.yimg.com/8/n/HA00265776/o/701011220155013873413840.jpg

The figure shows a sphere of radius R inscribed in a right circular cone of volume V. The line joining the centre of the sphere and any point on the edge of the base of the cone makes an angle θ with the base.

(a) Show that V = 1/3 兀 (R^3) tan2θ cot^3 θ

(b) Show that the volume of the cone is at the minimum when cos2θ = 1/3

(c) Find the minimum volume of the cone in terms of R.

回答 (1)

蛙人

2010-11-22 8:43 am

✔ 最佳答案

a. Note that the slant edge of the cone is at a angle 2θ to the horizontal plane.
base radius of the cone r = Rcotθ
Height of the cone h = rtan2θ = Rcotθtan2θ
So V = (1/3)(pi)(Rcotθ)^2Rcotθtan2θ
= (1/3)(pi)R^3tan2θcot^3θ

b. dV/dθ = (1/3)(pi)(R^3)[2(sec2θ)^2cot^3θ - 3cot^2θcsc^2θtan2θ]
dV/dθ = 0 <=> 2(sec2θ)^2cot^3θ - 3cot^2θcsc^2θtan2θ = 0
2(sec2θ)^2cotθ - 3csc^2θtan2θ = 0 (since cotθ =/= 0)
2(1/cos2θ)^2cotθ - 3sin2θ/cos2θsin^2θ = 0
2cosθ/cos2θsinθ - 6sinθcosθ/sin^2θ = 0
1/cos2θ - 3 = 0 (since cosθ =/= 0)
cos2θ = 1/3
Let cos2a = 1/3, where 0 < a < 90.
When 0 < θ < a, dV/dθ < 0
When a < θ < 90, dV/dθ > 0
So when V attains minimum, cos2θ = 1/3

c. cos2θ = 1/3
2cos^2θ - 1 = 1/3
cosθ = root(2/3)
So sinθ = root(1/3)
Hence cotθ = cosθ/sinθ = root2
Also, sin2θ = root(1 - 1/9) = 2root2/3
Hence tan2θ = sin2θ/cos2θ = 2root2
Thus minimum value of V
= (piR^3/3)(2root2)(root2)^3
= 8piR^3/3

參考: me

👍 0 👎 0