A right circular cone of base radius r cm and height h cm fits exactly into a sphere of internal radius 12cm.
a)Express r in terms of h.
b)If the volume of the cone is Vcm cube, show that V=8¶h^2 - 1/3¶h^3
c)Find the value of r for which V has a turning value, and determine whether the value of V is a maximum of a minimum. Hence find this volume.
(¶ is Pi, i typed it out, looks weird)
P.S. I know how to determine whether a value is a max of min value, but i'm not sure about other parts to this question! I'd love to know how to do this question!
To solve this problem, let's break it down into smaller steps:
a) Expressing r in terms of h:
We know that the sphere's radius is 12 cm. Since the cone fits exactly into the sphere, its height will be twice the sphere's radius, which is 24 cm.
In a right circular cone, the height, radius, and slant height form a right triangle. Using the Pythagorean theorem, we can relate the height, radius, and slant height of the cone:
(r^2) + (h^2) = (24^2)
Simplifying this equation, we get:
r^2 = (24^2) - h^2
r = sqrt((24^2) - h^2)
Therefore, r is expressed in terms of h as r = sqrt((24^2) - h^2).
b) Showing that V = 8πh^2 - (1/3)πh^3:
The volume of a cone is given by the formula V = (1/3)πr^2h.
Substituting the expression for r from part (a) into the formula for V, we get:
V = (1/3)π(sqrt((24^2) - h^2))^2h
Simplifying this equation, we get:
V = (1/3)π((24^2) - h^2)h
V = (1/3)π((24^2)h - h^3)
V = (1/3)π(576h - h^3)
V = (1/3)πh(576 - h^2)
V = (1/3)πh(576 - h^2)
Expanding further, we get:
V = (1/3)π(576h - h^3)
V = (1/3)π(576h) - (1/3)π(h^3)
V = 192πh - (1/3)πh^3
Therefore, we have shown that V = 8πh^2 - (1/3)πh^3.
c) Finding the value of r for which V has a turning value, and determining if V is a maximum or minimum:
To find the turning points of a function, we need to find where its derivative equals zero. So, let's find the derivative of V with respect to h.
dV/dh = 8π(2h) - (1/3)π(3h^2)
dV/dh = 16πh - πh^2
Setting dV/dh equal to zero and solving for h, we get:
16πh - πh^2 = 0
πh(16 - h) = 0
This equation has two solutions: h = 0 and h = 16.
To determine whether these values correspond to a maximum or minimum, we can use the second derivative test. Let's find the second derivative of V with respect to h:
d^2V/dh^2 = -2πh
At h = 0, the second derivative is negative (-2π(0)), which indicates a maximum value for V.
At h = 16, the second derivative is positive (-2π(16)), which indicates a minimum value for V.
Therefore, the value of h for which V has a turning point is h = 16, and the volume V is a minimum at this point.
To find the volume at this turning point, we substitute h = 16 into the equation we derived in part b:
V = 8π(16^2) - (1/3)π(16^3)
V = 128π(256 - 1024/3)
V = 128π(768/3)
V = 32768π
Thus, the volume V at the turning point is 32768π cubic cm.