Answer
(a) $V'(t) = -218.75$
(b) $V'(t) = -187.5$
(c) $V'(t) = -125$
(d) $V'(t) = 0$
Water flows out the fastest at $t = 0$. The flow of water gradually decreases until $t = 40~min$ when the flow is 0, because the tank is then empty.
Work Step by Step
$V(t) = 5000(1-\frac{1}{40}t)^2$
$V'(t) = 10,000~(1-\frac{1}{40}t)~(-\frac{1}{40})$
$V'(t) = 250~(\frac{1}{40}t-1)$
(a) We can find the draining rate after $5~min$:
$V'(t) = 250~(\frac{1}{40}t-1)$
$V'(t) = 250~[(\frac{1}{40})(5)-1]$
$V'(t) = -218.75$
(b) We can find the draining rate after $10~min$:
$V'(t) = 250~(\frac{1}{40}t-1)$
$V'(t) = 250~[(\frac{1}{40})(10)-1]$
$V'(t) = -187.5$
(c) We can find the draining rate after $20~min$:
$V'(t) = 250~(\frac{1}{40}t-1)$
$V'(t) = 250~[(\frac{1}{40})(20)-1]$
$V'(t) = -125$
(d) We can find the draining rate after $40~min$:
$V'(t) = 250~(\frac{1}{40}t-1)$
$V'(t) = 250~[(\frac{1}{40})(40)-1]$
$V'(t) = 0$
Water flows out the fastest at $t = 0$. The flow of water gradually decreases until $t = 40~min$ when the flow is 0, because the tank is then empty.
