a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9 pi cm 3/min. calculate the rate of change of height of water two minutes after the start of pouring water.
The volume of the water tank is (1/3)πr2h = (1/3)π(5)2(10) = 250π cm3.
The rate of change of height of water two minutes after the start of pouring water is (18π cm3)/(2 min) = 9π cm3/min.
a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9 pi cm 3/min. calculate the rate of change of height of water 2 minutes after the start of pouring water.
a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9pi cm^3/min. calculate the rate of change of height of water 2 minutes after the start of pouring water.
a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9pi cm3/min. calculate the rate of change of height of water two minutes after the start of pouring water.
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To calculate the rate of change of the height of water two minutes after the start of pouring water, we can use related rates.
First, let's establish some notations:
- r is the radius of the circular cross-section of the cone.
- h is the height of water in the cone.
We are given that the base radius of the cone is 5 cm and the height is 10 cm.
We are pouring water into the tank at a rate of 9π cm^3/min. This implies that the volume of water in the cone is changing at a rate of 9π cm^3/min.
The volume V of a cone is given by the formula V = (1/3)πr^2h.
We can differentiate both sides of this equation with respect to time (t) using the chain rule:
dV/dt = (1/3)(d/dt)(πr^2h)
The rate of change of volume with respect to time (dV/dt) is given as 9π cm^3/min, but we are interested in finding the rate of change of the height of water (dh/dt) at t = 2 minutes. So we need to solve for dh/dt.
Substituting the given values (r = 5 cm, h = 10 cm, dV/dt = 9π cm^3/min) into the above equation, we get:
9π = (1/3)(π)(5^2)(10)(dh/dt)
Canceling π from both sides and solving for dh/dt:
9 = (1/3)(5^2)(dh/dt)
dh/dt = (9 * 3) / (5^2)
dh/dt = 27/25 cm/min
Therefore, the rate of change of the height of water two minutes after the start of pouring water is 27/25 cm/min.