# [SOLVED]2.8.292 AP Calculus Exam .... related rates

#### karush

##### Well-known member
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
$$\displaystyle800-\int_0^{20} 30(1- e^{-0.16t})\ dt$$

#### skeeter

##### Well-known member
MHB Math Helper
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

Last edited:

ok
into not out of