1. A solid is constructed so that it has a circular base of radius r centimeters and every plane section perpendicular to a certain diameter of the base is a square, with a side of the square being a chord of the circle.
a. Find the volume of the solid.
b. If the solid describe expands so that the radius of the base increases at a constant rate of ½ centimeters per minute, how fast is the volume changing when the radius is 4 centimeters?
TEARS
To find the volume of the solid, we first need to understand its shape. Let's analyze the problem step by step.
1. Circular Base:
The solid has a circular base with a radius of r centimeters. The base is a 2-dimensional circle.
2. Plane Sections:
Every plane section perpendicular to a certain diameter of the base is a square. This means that if we cut the solid with a plane that is perpendicular to a diameter of the circular base, the resulting shape is a square.
3. Chord of the Circle:
The side length of the square is equal to the length of a chord of the circle. A chord is a line segment that connects two points on a circle's circumference.
Now, let's proceed to find the volume of the solid.
a. Finding the Volume:
To find the volume of the solid, we need to integrate the area of each square section along the height of the solid.
Since the plane sections are squares, the area of each section is equal to the side length squared. Therefore, the area of each plane section is (2r)^2 = 4r^2.
Since the solid is constructed by stacking these square sections, the height of the solid will be equal to the diameter of the base (2r), as each stacking level corresponds to a perpendicular plane section.
Therefore, the volume of the solid can be found by integrating the area of each square section along the height of the solid, which gives us:
Volume = ∫ [0, 2r] 4r^2 dh
This simplifies to:
Volume = 4r^2 * (2r - 0)
Volume = 8r^3
Therefore, the volume of the solid is 8r^3 cubic centimeters.
b. Finding the Rate of Change:
To find how fast the volume is changing when the radius is 4 centimeters, we need to take the derivative of the volume formula with respect to time and substitute in r = 4.
dV/dt = d(8r^3)/dt
dV/dt = 24r^2 (dr/dt)
Since we know that dr/dt (the rate of change of the radius) is 1/2 cm/min, we can substitute r = 4 and dr/dt = 1/2 into the expression:
dV/dt = 24(4)^2 * (1/2)
dV/dt = 192 cm^3/min
Therefore, when the radius is 4 centimeters, the volume of the solid is changing at a rate of 192 cubic centimeters per minute.
a. To find the volume of the solid, we need to find the area of the circular base and multiply it by the height.
The area of the circular base is given by the formula A = πr^2, where r is the radius of the circle.
Since every plane section perpendicular to a certain diameter of the base is a square, the side of the square is the diameter of the circle. Therefore, the side length of the square is 2r.
Now, we can find the height of the solid. The height is equal to the diagonal of the square, which can be found using the Pythagorean theorem:
h = √(s^2 + s^2) = √(2r^2) = r√2
Now we can find the volume:
V = A * h = πr^2 * r√2 = πr^3√2
b. To find out how fast the volume is changing, we can use the chain rule of differentiation.
Let V be the volume and r be the radius. We are given that dr/dt = 1/2 cm/min.
Differentiating both sides of the volume equation with respect to time t:
dV/dt = d/dt (πr^3√2)
Using the chain rule:
dV/dt = (3πr^2√2 * dr/dt)
Substituting r = 4 cm and dr/dt = 1/2 cm/min:
dV/dt = (3π(4^2)√2 * 1/2) = 12π√2 cm^3/min
Therefore, the volume is changing at a rate of 12π√2 cubic centimeters per minute when the radius is 4 centimeters.