APC.2.8.5 related rates

In summary, the conversation discusses the rate at which water is being pumped into a tank, given by the function $r(t) = 30(1-e^{-0.16t})$ where t is the number of minutes since the pump was turned on. The question asks how much water is in the tank after 20 minutes, given that there were initially 800 gallons of water. Using the integral of the function, the estimated amount of water in the tank after 20 minutes is approximately 1220 gallons.
  • #1

karush

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Water is pumped into a tank at a rate of $r(t) = 30(1-e^{e-0.16t})$ gallons per minute,
where t is the number of minutes since the pump was turned on.
If the tank contained 800 gallons of water when the pump was turned on,
how much water, to the nearest gallon, is in the tank after 20 minutes?
\begin{array}{ll}
a. &380 \textit{ gal}\\
b. &420\textit{ gal}\\
c. &829\textit{ gal}\\
d. &1220\textit{ gal}\\
e. &1376\textit{ gal}
\end{array}
so starting take the integral
why is e in the d/dt
$$\displaystyle800-\int_0^{20} 30(1- e^{-0.16t})\ dt $$
 
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  • #2
water is pumped into the tank at a rate of $r(t) = 30(1-e^{-0.16t})$

$\displaystyle 800 + \int_0^{20} r(t) \, dt \approx 1220 \, gal$

fyi, this is a calculator active question
 
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  • #3
ok
into not out of
 
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