A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 15 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.
Answers
Step-by-step explanation:
Given A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 15 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.
- So A (o) = 200 litres
- A (5) = 60 litres
- So A(15) = ?
- Now A(t) = 200 e^kt
- 60 = 200 e^k(5)
- 0.3 = e^5k
- Now log (0.3) = log e^5k
- = 5k
- So k = log (0.3) / 5
- Now A(t) = 200 e^(log 0.3 / 5)t
- A(15) = 200 e^(log 0.3 / 5)(15)
- = 200 e^3 log(0.3)
- = 200 e^ log (0.3)^3
- = 200 (0.3)^3
- = 5.4 litres
Water remaining after 15 mins = 5.4 litres
Step-by-step explanation:
Given: Initial amount of water = Q = 200 L
At t = 5 mins, 30% of water escapes = 200 - (30/100 * 200) = 140 L
To find: Volume of water in the tank after 15 minutes.
Solution:
W(0) = 200 and W(5) = 140.
W(t) = 200 e^kt
60 = 200 e^k(5)
0.3 = e^5k
Now log (0.3) = log e^5k
= 5k
So k = log (0.3) / 5
Now W(t) = 200 e^(log 0.3 / 5)t
W(15) = 200 e^(log 0.3 / 5)(15)
= 200 e^3 log(0.3)
= 200 e^ log (0.3)^3
= 200 (0.3)^3
= 5.4 L