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 cm3/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 given by V = (1/3)πr^2h, where r is the radius and h is the height.
At the start of pouring water, the height of the water is 0 cm.
After two minutes, the volume of water poured into the tank is 9π x 2 = 18π cm3.
Therefore, the height of the water after two minutes is h = (18π/πr^2) = (18/r^2) cm.
The rate of change of height of water two minutes after the start of pouring water is:
Rate of change = (18/r^2) - 0 = 18/r^2 cm/min
Since the radius of the base is 5 cm, the rate of change of height of water two minutes after the start of pouring water is:
Rate of change = 18/5^2 = 7.2 cm/min
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To calculate the rate of change of the height of water in the tank, we need to find the rate at which the volume of water is increasing and then divide it by the surface area of the water.
First, let's find the volume of the cone-shaped water tank. The formula for the volume of a cone is:
V = (1/3) * π * r^2 * h
where V is the volume, π is a mathematical constant (approximately equal to 3.14159), r is the radius of the base, and h is the height of the cone.
In this case, the base radius (r) of the tank is 5 cm and the height (h) is given as 10 cm. Let's calculate the initial volume (V₀) when pouring water starts:
V₀ = (1/3) * π * (5 cm)^2 * 10 cm
≈ 261.79 cm³
Next, we'll find the rate at which the volume of water is increasing. It is given as 9π cm³/min.
Let's calculate the volume (V) of water in the tank after two minutes:
V = V₀ + (rate of increase of volume) * (time)
= 261.79 cm³ + (9π cm³/min) * (2 min)
= 261.79 cm³ + 18π cm³
≈ 261.79 cm³ + 56.55 cm³
≈ 318.34 cm³
Finally, let's find the rate of change of the height of water (dh/dt) using the relation between volume and height of the cone:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (1/3) * π * (2r) * h * (dh/dt)
We know that dV/dt = 9π cm³/min and we need to find dh/dt. Substituting the known values, we can solve for dh/dt:
9π = (1/3) * π * (2 * 5 cm) * h * (dh/dt)
dh/dt = (9π * 3) / (π * 2 * 5 cm)
= (27/10) cm/min
≈ 2.7 cm/min
Therefore, the rate of change of the height of water two minutes after the start of pouring water is approximately 2.7 cm/min.