Tank/Leak Rate Problem

[stevenrand]

Water is pumped into an underground tank at a constant rate of 8 gallons per minute. Water leaks out of the tank at the rate of sq.rt(t + 1) gallons per min. for 0 is less than or equal to 't' which is less than or equal to 120 minutes. At time t = 0 the tank contains 30 gallons of water. Sorry for not being able to "type math" .... usually I use the equation editor in Word, but it is not cooperating.

A) How many gallons of water leak out of the tank from time t = 0 to time t = 3? I said 3*sqrt(t + 1) ???
B) How many gallons of water are in the tank at time t =3? I said 30 + [24 - (3*sqrt(t + 1))] ???
C) Write an expression for A(t), the total number of gallons of water in the tank at time t I put A(t) = 30 + [8n - (n*sqrt(t + 1))] such that 'n' equals the number of minutes passed ???
D) This is where I think I am definitely wrong: At what time in the 0 - 120 min interval is the amount of water in the tank at a maximum? I mean I figured it would have the most at the 120 min/2 hour mark... yes?

 

[tutor_joel]

Water is pumped into an underground tank at a constant rate of 8 gallons per minute. Water leaks out of the tank at the rate of sq.rt(t + 1) gallons per min. for 0 is less than or equal to 't' which is less than or equal to 120 minutes. At time t = 0 the tank contains 30 gallons of water. Sorry for not being able to "type math" .... usually I use the equation editor in Word, but it is not cooperating.

A) How many gallons of water leak out of the tank from time t = 0 to time t = 3? I said 3*sqrt(t + 1) ???How did you get this? did you use calculus?
B) How many gallons of water are in the tank at time t =3? I said 30 + [24 - (3*sqrt(t + 1))] ???Did you use \(\displaystyle \int{\sqrt{t^2+1}dt}?\)
C) Write an expression for A(t), the total number of gallons of water in the tank at time t I put A(t) = 30 + [8n - (n*sqrt(t + 1))] such that 'n' equals the number of minutes passed ???
D) This is where I think I am definitely wrong: At what time in the 0 - 120 min interval is the amount of water in the tank at a maximum? I mean I figured it would have the most at the 120 min/2 hour mark... yes?

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ \frac{dI}{dt} \ = \ 8. \ \frac{dO}{dt} \ = \ (t+1)^{1/2}, \ t(0) \ = \ 30 \ gallons, \ and \ 0 \ \le \ t \ \le \ 120 \ minutes.\)

\(\displaystyle a) \ \int_{0}^{3}(t+1)^{1/2}dt \ = \ 4.\overline 6 \ gallons \ leaked \ out.\)

\(\displaystyle b) \ Note: \ I(t) \ = \ 8t \ and \ O(t) \ = \ \frac{2}{3}(t+1)^{3/2}-\frac{2}{3}.\)

\(\displaystyle I(3)+30-O(3) \ = \ 24 \ + \ 30 \ - \ \frac{14}{3} \ = \ 49.\overline 3 \ gallons.\)

\(\displaystyle c) \ A(t) \ = \ 8t+30-[\frac{2}{3}(t+1)^{3/2}-\frac{2}{3}]\)

\(\displaystyle d) \ A'(t) \ = \ 8-(t+1)^{1/2} \ = \ 0, \ t \ = \ 63, \ ergo, \ water \ is \ at \ its \ maximum \ when \ t \ = \ 63 \ minutes.\)

 

[stevenrand]

Wow. I really messed that up :/ I think I follow it a little now, but I will see on the next couple of questions. I may have to get a little more help