A large container has the shape of a frustum of a cone with top radius 5 m, bottom radius 3m, and height 12m.
The container is being filled with water at the constant rate 4.9 m^3/min. At what rate is the level of water rising at the instant the water is 2 deep?
To determine the rate at which the water level is rising, we need to find the derivative of the volume of water with respect to time.
First, let's find the formula for the volume of the frustum of a cone. The formula for the volume of a frustum of a cone is given by:
V = (1/3)πh(R^2 + r^2 + Rr)
where V is the volume, h is the height of the frustum, R is the top radius, and r is the bottom radius.
In this problem, we are given the top radius (R) as 5m, the bottom radius (r) as 3m, and the height (h) as 12m.
Now, let's differentiate the volume formula with respect to time (t) to find the rate of change of volume with respect to time, which will give us the rate at which the water level is rising.
dV/dt = (1/3)π (2h)(R^2 + r^2 + Rr) dh/dt
where dV/dt is the rate of change of volume with respect to time, dh/dt is the rate of change of height with respect to time, and h is the current height of the water in the container.
In this problem, we are given that the rate of change of volume with respect to time (dV/dt) is 4.9 m^3/min.
Now, we need to find the rate at which the water level is rising, which is dh/dt when the height (h) is 2m.
Substituting the given values into the equation, we have:
4.9 = (1/3)π (2)(5^2 + 3^2 + 5*3) dh/dt
Simplifying the equation:
4.9 = 4π (25 + 9 + 15) dh/dt
4.9 = 4π (49) dh/dt
Now, let's solve for dh/dt:
dh/dt = 4.9 / (4π(49))
dh/dt ≈ 0.005m/min
Therefore, the rate at which the water level is rising at the instant when the water is 2m deep is approximately 0.005m/min.
Sorry Mr Steve
I submitted your answer and It was wrong answer
the right answer = 14.0 cm/min
Could You please tell me how to do this question because I have to do it again with different values.
When the water has depth y, the radius of the surface is
3 + (y/12)(5-3) = 3 + y/6
Its volume at that point is
v = 1/3 (R^2-r^2) h
= (1/3)((3+y/6)^2-3^2)y = y^3/108 + y^2/3
Now we are ready to begin. When y=2,
v = 8/108 + 4/3 = 38/27
dv/dt = (y^2/36 + 2y/3) dy/dt
4.9 = (4/36 + 8/3) dy/dt
dy/dt = 1.76 m/min
I guess you should have checked my math. The actual formula for the volume is
(π/3)((18+y)(3+y/6)^2-18*3^2)
= π(y^3/108 + y^2/2 + 9y)
because the water is in the bottom of the cone
dv/dt = π/36(y+18)^2 dy/dt
4.9 = π/36*20^2 dy/dt
dy/dt = .14 m/min or 14.0 cm/min