Question

A 1000 gallon tank, initially full of water, develops a leak at the bottom. Given that 300 gallons of water leak out in the first 30 minutes, find the amount of water, A(t), left in the tank t minutes after the leak develops if the water drains off at a rate proportional to the amount of water present.
a. A(t)= 1000(1/3)^1/10t
b. A(t)= 1000(1/2)^-1/10t
c. A(t)= 1000(1001/2)^t
d. A(t)= 1000e ^-1/10t
e. A(t)= 1000(1/2)^1/10t
f. None of these

Answer 1

Answer:

option F

Step-by-step explanation:

given,

capacity of tank = 1000 gallon

volume of water leak = 300 gallon

time = 30 min

water is leaking at the rate

formula used

$A = B e^{-kt}$

At t = 0   A = 1000 gallons

$1000= B e^{-k\times 0}$

B = 1000

$A = 1000 e^{-kt}$

at t = 30        A  = 700

$700= 1000 e^{-k\times 30}$

$e^{-k\times 30}=0.7$

taking ln both side

$-k\times 30=ln(0.7)$

$k=-\dfrac{ln(0.7)}{30}$

now,

$A = 1000 e^{\dfrac{ln(0.7)}{30}t}$

$A = 1000\times 0.7 e^{\dfrac{t}{30}}$

$A = 700 e^{\dfrac{t}{30}}$

hence, the correct answer is option F