Welcome to our community

Be a part of something great, join today!

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

karush

Well-known member
Jan 31, 2012
2,685
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
Mar 1, 2012
658
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:

karush

Well-known member
Jan 31, 2012
2,685
ok
into not out of