V = (1/3)πh^2 (18-h)
= 6πh^2 - (1/3)πh^3
dV/dt = 12πh dh/dt - πh^2 dh/dt
4 = dh/dt(12πh - πh^2)
when h = 2
4 = dh/dt(24π - 4π)
dh/dt = 4/(20π) = 1/(5π) cm/s
= 6πh^2 - (1/3)πh^3
dV/dt = 12πh dh/dt - πh^2 dh/dt
4 = dh/dt(12πh - πh^2)
when h = 2
4 = dh/dt(24π - 4π)
dh/dt = 4/(20π) = 1/(5π) cm/s
Given that the volume of the water in the bowl is given by the equation V = (1/3)πh^2(18 - h), where V is the volume and h is the depth of the water.
Taking the derivative of V with respect to time:
dV/dt = (1/3)π * 2h * dh/dt + (1/3)πh^2 * (-dh/dt)
Where dh/dt represents the rate at which the depth is changing (or the rate at which the water level is rising).
We are given that dh/dt = 4 cm^3/s, and we want to find dh/dt when h = 2 cm.
Substituting the given values into the derivative equation:
dV/dt = (1/3)π * 2(2) * 4 + (1/3)π * (2^2) * (-4)
= (8/3)π + (4/3)π
= (12/3)π
= 4π cm^3/s
Therefore, when the depth is 2 cm, the rate at which the water level is rising is 4π cm^3/s.
Given:
Radius of the hemispherical bowl = 6 cm
Volume of the water = 1/3πh^2(18-h)
Rate at which water is poured into the bowl = 4 cm^3/s
1. Differentiate the volume formula with respect to time:
dV/dt = d/dt(1/3πh^2(18-h))
2. Simplify the derivative:
dV/dt = 1/3π * (2h * dh/dt) * (18 - h) + 1/3π * h^2 * (-1 * dh/dt)
3. Substitute the given values:
dh/dt = 4 cm^3/s
h = 2 cm
4. Calculate dV/dt when h = 2 cm and dh/dt = 4 cm^3/s:
dV/dt = 1/3π * (2(2) * 4) * (18 - 2) + 1/3π * (2^2) * (-1 * 4)
5. Simplify the expression:
dV/dt = 1/3π * (16) * (16) + 1/3π * (4) * (-4)
6. Calculate the final answer:
dV/dt = (256/3)π + (-16/3)π
dV/dt = (240/3)π
dV/dt = 80π cm^3/s
Therefore, the rate at which the water level is rising when the depth is 2 cm is 80π cm^3/s.