Show that the height of the right circular cone of minimum volume which circumscribed a sphere of radius r is 4r . Also show semi verticle angle of the cone is sin^-1(1/3)

Question

Answer

Solution:

Finding the Height of the Cone

Let the radius of the sphere be r and the height of the cone be h.

The cone will be circumscribed around the sphere in such a way that the base of the cone will be tangent to the sphere.

Let A be the center of the sphere and B be the point of tangency between the sphere and the base of the cone.

Let C be the apex of the cone.

Draw a perpendicular from B to the center of the sphere as shown below:

Since BC is perpendicular to AB, we have

AB^2 + BC^2 = AC^2

But AB = r and AC = h, so

r^2 + BC^2 = h^2

Now, draw a line from A to C as shown below:

Since AC is perpendicular to the base of the cone, we have

tan(theta) = r/h

where theta is the semi-vertical angle of the cone.

Solving the two equations above for BC^2, we get

BC^2 = h^2 - r^2

Substituting this into the equation for the volume of a cone, we get

V = (1/3)pi*r^2*h = (1/3)pi*r^2*(h^2-r^2)^(1/2)

To find the minimum volume, we need to minimize V with respect to h.

Taking the derivative of V with respect to h and setting it equal to zero, we get

(1/3)pi*r^2*(h^2-r^2)^(-1/2)*2h = 0

Solving for h, we get

h = 2^(1/2)*r

Therefore, the height of the cone is 4r.

Finding the Semi-Vertical Angle of the Cone

Using the equation above, we can solve for the semi-vertical angle of the cone:

tan(theta) = r/h = r/(2^(1/2)*r) = 1/(2^(1/2))

Therefore,

theta = sin^(-1)(1/3)

Answer

8

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