Two full 48-gallon tanks begin draining at t = 0. Tank Alpha's volume is changing
at a constant rate of -16/5 gal/min. The rate at which Tank Bravo's volume is
changing is given by r(t)=-1/3t-1 gal/min.
a) How much water is in each tank after 5 minutes?
b) Which tub drains �rst? How do you know?
Cannot parse -1/3t-1
Is that (-1/3)(t-1) or -1/(3t)-1 or -1/(3t-1)?
none of those choices makes sense at t=0.
negative one third t minus 1
ok. If you mean (-1/3 t) - 1 then we have for the two tanks:
Va = 48 - ∫16/5 dt = 48 - 16/5 t
Vb = 48 - ∫1/3 t + 1 dt = 48 - t - 1/6 t^2
So,
Va(5) = 48 - 16 = 32
Vb(5) = 48 - 5 - 25/6 = 38 5/6
Now just solve for when Va and Vb = 0 to find which drains first.
If I still didn't get it right, and I suspect I might not have, since "negative one third t minus 1" is -1 at t=0, then just make the fix and reevaluate the integrals.
Tyler, if yu do not know how to use parentheses try to type it with lines like
1
--------
(3 t - 1)
or something.
To find how much water is in each tank after 5 minutes, we need to calculate the volume of water that each tank has lost in that time.
a) To find the volume of water in Tank Alpha after 5 minutes, we can use the formula:
Volume of Alpha = Initial volume of Alpha - Rate of change of Alpha x Time
where Initial volume of Alpha is 48 gallons, and the Rate of change of Alpha is -16/5 gallons per minute.
Volume of Alpha = 48 - (-16/5) x 5
= 48 + 16 x 5/5
= 48 + 16
= 64 gallons
Therefore, after 5 minutes, Tank Alpha has 64 gallons of water.
To find the volume of water in Tank Bravo after 5 minutes, we need to integrate the rate function r(t) with respect to time:
Volume of Bravo = ∫ (rate of Bravo) dt
The rate function for Bravo is r(t) = -1/3t - 1 gallons per minute.
Integrating r(t) gives us:
Volume of Bravo = ∫ (-1/3t - 1) dt
= -1/3 ∫ t dt - ∫ 1 dt
= -1/3 (t^2/2) - t + C
To find the value of C, we need to know the initial volume of Bravo at t = 0. Assuming it is also 48 gallons, we can substitute the given values:
48 = -1/3 (0^2/2) - 0 + C
48 = 0 + 0 + C
C = 48
Substituting C back into the equation for volume, we get:
Volume of Bravo = -1/3 (t^2/2) - t + 48
Plugging in t = 5, we can find the volume of Bravo after 5 minutes:
Volume of Bravo = -1/3 (5^2/2) - 5 + 48
= -1/3 (25/2) - 5 + 48
= -25/6 - 5 + 288/6
= -25/6 + 288/6 - 30/6
= 233/6
≈ 38.83 gallons
Therefore, after 5 minutes, Tank Bravo has approximately 38.83 gallons of water.
b) To determine which tank drains first, we need to compare their rates of volume change. The rate at which a tank is draining is given by the derivative of its volume function.
For Tank Alpha, the rate of change is constant at -16/5 gallons per minute.
For Tank Bravo, the rate of change is given by r(t) = -1/3t - 1 gallons per minute.
Since the rate of Tank Bravo is dependent on time and becomes more negative as time progresses, while the rate of Tank Alpha stays constant, we can conclude that Tank Bravo drains first.
We can also compare the volume functions at different times to see when each tank reaches 0 gallons, but in this case, it is sufficient to compare the rates of change.