Answer to Question #277534 in Calculus for chav

Question

A swimming pool is to be drained for cleaning. The quantity of Q of water (in gallons) in the pool at any given time is a function of the amount of time t (in minutes) which has elapsed since the drain was opened. the function is: Q(t)= 10(20-t)³

find Q’(t)=

what is the rate of change of the quantity of water in the pool exactly 10 minutes after the drain is opened? (labeled units)

what is the average rate of change of the quantity of water during the first 10 minutes after the drain is open?

how long will it take to drain the pool completely? (labeled units)

Answer 1

$Q’(t)=-30(20-t)^2$

$Q’(10)=-30(20-10)^2=-3000$ gallons/min

average rate of change:

$\frac{\Delta Q}{\Delta t}(10)=\frac{Q(10)-Q(0)}{10}$

$Q(10)=10000$ gallons

$Q(0)=80000$ gallons

$\frac{\Delta Q}{\Delta t}(10)=\frac{10000-80000}{10}=-7000$ gallons/min

drain the pool completely:

$Q(t)= 10(20-t)^3=0$

$t=20$ min