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l l
✿ Spring ✿
Answer:
285,753 cm³/min
Step-by-step explanation:
The rate of change of volume is the product of the water's surface area and its rate of change of depth.
At a depth of 2 m, the water has filled 1/3 of the 6 m depth of the tank. So, the radius at the water's surface will be 1/3 of the tank's radius of 2 m. The water's surface area is ...
A = πr² = π(2/3 m)² = 4π/9 m²
The rate of change of depth is 0.2 m/min, so the volume of water is increasing at the rate ...
dV/dt = (0.20 m/min)(4π/9 m²) = 8π/90 m³/min ≈ 279253 cm³/min
This change in volume is the difference between the rate at which water is being pumped in and the rate at which it is leaking out:
2.8×10⁵ cm³/min = (input rate) - 6500 cm³/min
Adding 6500 cm³/min to the equation, we get ...
input rate ≈ 285,753 cm³/min
The given data is
t, h: 0 2 4 6 8 10
r(t), L/h: 8.6 7.9 6.8 6.4 5.7 5.3
The lower and upper estimates for the total amount that leaked may be computed as the Left and Right Riemann sums.
The shape of the graph of r versus will determine which of the two sums yields an upper or lower sum.
The plot of the graph is shown below.
The Left Riemann sum is
Sl = 2*(8.6+7.9+6.8+6.4+5.7) = 70.8 L
The Right Riemann sum is
Sr = 2*(7.9+6.8+6.4+5.7+5.3) = 64.2 L
Answer:
The lower estimate for oil leakage is 64.2 L
The upper estimate for oil leakage is 70.8 L
<h3>Rounding it to the nearest thousand:</h3>
11,000
<h2>Answer:</h2>
11,000