Answer:
Let's solve each part of the problem step by step:
(a) When the sand remaining in the upper cone has height y inches, its volume A in terms of y is:
First, let's find the volume of the upper cone with height y and radius 1 inch. The formula for the volume of a cone is given by:
V = (1/3) * π * r^2 * h
In this case, the radius (r) is 1 inch, and the height (h) is y inches. So, the volume of the upper cone (A) is:
A = (1/3) * π * (1^2) * y
A = (1/3) * π * y
(b) When the sand in the lower cone has reached a height of h inches, its volume B in terms of h is π - 27π(3 - h)^3:
Now, let's find the volume of the lower cone. The formula for the volume of a cone is the same:
V = (1/3) * π * r^2 * h
In this case, the radius (r) is 1 inch, and the height (h) is h inches. So, the volume of the lower cone is:
B = (1/3) * π * (1^2) * h
B = (1/3) * π * h
Now, let's subtract the volume of the empty space above the sand. The total height of the lower cone is 3 inches, and when the sand has reached a height of h inches, there is (3 - h) inches of empty space above the sand. So, the volume of the empty space is:
Volume of empty space = (1/3) * π * (1^2) * (3 - h)
Volume of empty space = (1/3) * π * (3 - h)
Now, subtract the volume of the empty space from the volume of the lower cone to find B:
B = (1/3) * π * h - (1/3) * π * (3 - h)
B = (1/3) * π * h - (1/3) * π * (3) + (1/3) * π * h
B = (1/3) * π * h - π + (1/3) * π * h
Now, simplify B:
B = 2/3 * π * h - π
(c) Assume the total volume of sand in the hourglass is 3π/4 cubic inches. At the instant when the sand in the lower cone is 1 inch high, the height of the sand in the lower cone is increasing at a rate of 401(245)^2 inches per second:
We are given that the total volume of sand in the hourglass is 3π/4 cubic inches. So, we can set up an equation for the volume of the sand in both the upper and lower cones at any given time t:
Volume of sand = A + B
Since A is the volume of the upper cone and B is the volume of the lower cone, we can use the expressions we derived in parts (a) and (b):
Volume of sand = (1/3) * π * y + (2/3) * π * h - π
Now, we need to find the values of y and h at the given instant. We are given that the sand in the upper cone is decreasing at a rate of 1/100 inches per second, so we can write:
dy/dt = -1/100
We are also given that the height of the sand in the lower cone is increasing at a rate of 401(245)^2 inches per second:
dh/dt = 401(245)^2
At the instant when the sand in the lower cone is 1 inch high, we can substitute h = 1 into the equation for the volume of sand:
Volume of sand = (1/3) * π * y + (2/3) * π * 1 - π
Now, we need to find y. To do that, we'll integrate the equation dy/dt = -1/100 with respect to t:
∫(dy/dt) dt = ∫(-1/100) dt
∫dy = -1/100 ∫dt
Integrating both sides:
y = (-1/100) * t + C
Now, we can find C by using the initial condition that at t = 0, y = 3 (since initially, the upper cone is full):
3 = (-1/100) * 0 + C
C = 3
So, the equation for y in terms of t is:
y = (-1/100) * t + 3
Now, we can substitute this into the equation for the volume of sand:
Volume of sand = (1/3) * π * [(-1/100) * t + 3] + (2/3) * π * 1 - π
Volume of sand = (π/300) * t + (2/3) * π - π
Now, we can find the rate of change of the volume of sand with respect to time (dV/dt) at the instant when the sand in the lower cone is 1 inch high:
dV/dt = (π/300) - 0 [since dh/dt = 401(245)^2, and at h = 1, dh/dt = 401(245)^2]
dV/dt = π/300 cubic inches per second
So, at that instant, the height of the sand in the lower cone is increasing at a rate of π/300 cubic inches per second.
Explanation: