Answer
$V'(t) = \frac{500}{9}t-\frac{10,000}{3}$
The units of $V'(t)$ are $gallons/min$
Initially, the magnitude of the flow rate is $3333.3~gallons/min$
As water drains out of the tank, the magnitude of the flow rate gradually decreases to 0 during the 60 minutes when the tank is draining.
The magnitude of the flow rate is the greatest at $t =0$
The magnitude of the flow rate is the least at $t = 60$
Work Step by Step
$V(t) = 100,000(1-\frac{1}{60}~t)^2$
We can find $V'(t)$:
$V'(t) = \lim\limits_{h \to 0}\frac{V(t+h)-V(t)}{h}$
$V'(t) = \lim\limits_{h \to 0}\frac{100,000(1-\frac{t+h}{60}~)^2-100,000(1-\frac{1}{60}~t)^2}{h}$
$V'(t) = \lim\limits_{h \to 0}\frac{100,000(1-\frac{t+h}{30}+\frac{t^2+2th+h^2}{3600})-100,000(1-\frac{t}{30}+\frac{t^2}{3600})}{h}$
$V'(t) = \lim\limits_{h \to 0}\frac{100,000[(1-\frac{t+h}{30}+\frac{t^2+2th+h^2}{3600})-(1-\frac{t}{30}+\frac{t^2}{3600})]}{h}$
$V'(t) = \lim\limits_{h \to 0}\frac{100,000(-\frac{h}{30}+\frac{2th+h^2}{3600})}{h}$
$V'(t) = \lim\limits_{h \to 0}100,000(-\frac{1}{30}+\frac{2t+h}{3600})$
$V'(t) = 100,000(-\frac{1}{30}+\frac{t}{1800})$
$V'(t) = \frac{500}{9}t-\frac{10,000}{3}$
The units of $V'(t)$ are $gallons/min$
When $t=0$:
$V'(0) = \frac{500}{9}(0)-\frac{10,000}{3} = -\frac{10,000}{3}= -3333.3$
$V(0) = 100,000(1-\frac{0}{60})^2 = 100,000$
When $t=10$:
$V'(10) = \frac{500}{9}(10)-\frac{10,000}{3} = -\frac{10,000}{3}=-2777.8$
$V(10) = 100,000(1-\frac{10}{60})^2 = 69,444.4$
When $t=20$:
$V'(20) = \frac{500}{9}(20)-\frac{10,000}{3} = -\frac{10,000}{3}= -2222.2$
$V(20) = 100,000(1-\frac{20}{60})^2 = 44,444.4$
When $t=30$:
$V'(30) = \frac{500}{9}(30)-\frac{10,000}{3} = -\frac{10,000}{3} = -1666.7$
$V(30) = 100,000(1-\frac{30}{60})^2 = 25,000$
When $t=40$:
$V'(40) = \frac{500}{9}(40)-\frac{10,000}{3} = -\frac{10,000}{3} = -1111.1$
$V(40) = 100,000(1-\frac{40}{60})^2 = 11,111.1$
When $t=50$:
$V'(50) = \frac{500}{9}(50)-\frac{10,000}{3} = -\frac{10,000}{3} = -555.6$
$V(50) = 100,000(1-\frac{50}{60})^2 = 2777.8$
When $t=60$:
$V'(60) = \frac{500}{9}(60)-\frac{10,000}{3} = -\frac{10,000}{3} = 0$
$V(60) = 100,000(1-\frac{60}{60})^2 = 0$
Initially, the magnitude of the flow rate is $3333.3~gallons/min$
As water drains out of the tank, the magnitude of the flow rate gradually decreases to 0 during the 60 minutes when the tank is draining.
The magnitude of the flow rate is the greatest at $t =0$
The magnitude of the flow rate is the least at $t = 60$
