a vessel comtaining water has the shape of an inverted circular cone of base radius 5 feet and height 10 feet.the water flow from the apex of the cone at a constant rate of 3 cubic feet per minute.how fast is the water lavel rising when the level is 4 feet(rate of change)
Quickly because I have to go baby sit for grand kids:
find the radius of the cone when depth = 4 feet, call it r
then
the surface area times the change in height dh = small volume change dv
so
dv = pi r^2 dh
dv/dt = (pi r^2) dh/dt
so
dh/dt = dv/dt /(pi r^2)
and dv/dt = 3 ft^2/min
To find out how fast the water level is rising, we need to relate the height of the water level in the cone to the rate at which water flows out of the cone.
Let's call the height of the water level in the cone "h" feet. We need to find dh/dt, the rate of change of h with respect to time.
The volume of a cone can be calculated using the formula V = (1/3) * π * r^2 * h, where V is the volume, π is approximately 3.14159, r is the radius of the base, and h is the height.
Since the water is flowing out at a constant rate of 3 cubic feet per minute, the rate of change of volume with respect to time is dV/dt = -3 ft^3/min (negative because the volume is decreasing).
We can differentiate the volume function with respect to time t to find the expression for dV/dt:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Given that the base radius r = 5 feet and dh/dt = ? feet/min, and we want to find out dh/dt when h = 4 feet, we can plug these values into the equation and solve for dh/dt.
Plugging in the values, we get:
-3 = (1/3) * π * (2 * 5 * 0 + 5^2 * dh/dt)
Simplifying the equation:
-3 = (1/3) * π * (25 * dh/dt)
-3 = (25/3) * π * dh/dt
Now, we can solve for dh/dt by isolating it:
dh/dt = (-3 * 3) / (25π)
dh/dt ≈ -0.36 feet/minute
Therefore, when the water level is 4 feet, the water level is decreasing at a rate of approximately 0.36 feet per minute.