A right circular cone is circumscribed about a sphere of radius a .If h is the distance from the centre of the sphere to the vertex of the cone. Show that the volume of the cone is 1/3pi a^2(a+h)^2/(h-a). Find the angle of a cone when the volume is minimum
Find the point on the line
x/a+ y/b=1
that is closest to the origin.
Be sure to enter it as an ordered pair.
To find the volume of the cone, we need to make use of the concepts of similar triangles and the formula for the volume of a cone.
Let's start by drawing a diagram to better understand the given situation:
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a | \
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In the diagram, the cone is drawn such that it is tangent to the sphere at the base of the cone. The radius of the sphere is denoted as 'a', the distance from the center of the sphere to the vertex of the cone is denoted as 'h', and the slant height of the cone is denoted as 'l'.
Now, let's consider the right triangle formed by the height 'h', the slant height 'l', and the radius of the sphere 'a'. By the Pythagorean Theorem, we have:
h^2 + a^2 = l^2 ---(1)
Next, we can express the volume of the cone using the formula:
V = (1/3) * π * r^2 * h, where r is the radius of the base of the cone.
In this case, the radius of the base of the cone is 'a'. Thus, we can rewrite the volume formula as:
V = (1/3) * π * a^2 * h ---(2)
Now, we need to eliminate 'l' from equations (1) and (2). Solving equation (1) for 'l', we have:
l^2 = h^2 + a^2
l = √(h^2 + a^2)
Substituting this value of 'l' into equation (2), we get:
V = (1/3) * π * a^2 * h
= (1/3) * π * a^2 * √(h^2 + a^2)
Simplifying further, we can express the volume as:
V = (1/3) * π * a^2 * (h^2 + a^2)^(1/2) ---(3)
Finally, to find the angle of the cone when the volume is at its minimum, we can take the derivative of equation (3) with respect to 'h' and set it equal to zero:
dV/dh = 0
Differentiating equation (3) with respect to 'h', we have:
dV/dh = (1/3) * π * a^2 * (1/2) * (h^2 + a^2)^(-1/2) * 2h
= (2/3) * π * a^2 * h / √(h^2 + a^2)
Setting this derivative equal to zero:
(2/3) * π * a^2 * h / √(h^2 + a^2) = 0
Since the term (2/3) * π * a^2 is always positive and non-zero, we can ignore it. Hence, we have:
h / √(h^2 + a^2) = 0
Multiplying both sides by √(h^2 + a^2), we get:
h = 0
Therefore, the volume of the cone is minimum when the height 'h' is equal to zero, which means the cone degenerates into a flat disk.
Note: Since there is no cone with zero height, we can assume that the volume is minimum when the height is approaching zero.