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An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 105e−0.01t liters per minute. How much oil leaks out during the first hour? (Round your answer to the nearest liter.)

Answer :

Answer:

4738 L

Explanation:

[tex]r(t)=105e^{-0.01t}[/tex]

Let V be the volume leaks out in 1 hour or 60 minutes

So, [tex]V=\int_{0}^{60}105e^{-0.01t}dt[/tex]

[tex]V=\frac{105}{-0.01}\times \left ( e^{-0.01t} \right )_{0}^{60}[/tex]

[tex]V=\frac{105}{-0.01}\times \left (\left ( e^{-0.01\times 60} \right )-e^{-0}}  \right )[/tex]

V = - 10500 (0.5488 - 1) = 4737.6 L

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