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5. Water is being pumped into two tanks at variable rates. The volume of water in Tank A is modeled by [tex]V_A(t)=-0.067 t^3 + t^2[/tex], and the volume of water in Tank B is modeled by [tex]V_B(t)=-0.083 t^3 + 1.125 t^2 + 2.5 t[/tex], where [tex]t[/tex] is the time in minutes since the tanks started to be filled. The volume is measured in gallons.

A. Write an expression to model the total amount of water in tanks A and B after [tex]t[/tex] minutes. How much water is in both tanks after 3 minutes?

B. Write an expression to model how much more water Tank B has than Tank A after [tex]t[/tex] minutes. How much more water is in Tank B after 4 minutes?


Sagot :

Certainly! Let's break down the given problem step-by-step.

### A. Total Amount of Water in Both Tanks

#### Step 1: Model the Total Volume
The volume functions for tanks A and B are given as:
[tex]\[ V_A(t) = -0.067t^3 + t^2 \][/tex]
[tex]\[ V_B(t) = -0.083t^3 + 1.125t^2 + 2.5t \][/tex]

The total volume of water in both tanks after [tex]\( t \)[/tex] minutes can be found by summing these two functions:
[tex]\[ V_{total}(t) = V_A(t) + V_B(t) \][/tex]
[tex]\[ V_{total}(t) = (-0.067t^3 + t^2) + (-0.083t^3 + 1.125t^2 + 2.5t) \][/tex]
Combine like terms:
[tex]\[ V_{total}(t) = (-0.067t^3 - 0.083t^3) + (t^2 + 1.125t^2) + 2.5t \][/tex]
[tex]\[ V_{total}(t) = -0.15t^3 + 2.125t^2 + 2.5t \][/tex]

So, the expression modeling the total volume of water in both tanks after [tex]\( t \)[/tex] minutes is:
[tex]\[ V_{total}(t) = -0.15t^3 + 2.125t^2 + 2.5t \][/tex]

#### Step 2: Calculate the Total Volume After 3 Minutes
To find the total volume of water in both tanks after 3 minutes, substitute [tex]\( t = 3 \)[/tex] into the total volume expression:
[tex]\[ V_{total}(3) = -0.15(3)^3 + 2.125(3)^2 + 2.5(3) \][/tex]
Evaluating this gives us:
[tex]\[ V_{total}(3) = -0.15(27) + 2.125(9) + 2.5(3) \][/tex]
[tex]\[ V_{total}(3) = -4.05 + 19.125 + 7.5 \][/tex]
[tex]\[ V_{total}(3) = 22.575 \][/tex]

So, the total volume of water in both tanks after 3 minutes is 22.575 gallons.

### B. Difference in Volume Between Tanks B and A

#### Step 1: Model the Difference in Volume
The difference in volume between Tank B and Tank A after [tex]\( t \)[/tex] minutes is:
[tex]\[ \Delta V(t) = V_B(t) - V_A(t) \][/tex]
[tex]\[ \Delta V(t) = (-0.083t^3 + 1.125t^2 + 2.5t) - (-0.067t^3 + t^2) \][/tex]
Combine like terms:
[tex]\[ \Delta V(t) = (-0.083t^3 + 0.067t^3) + (1.125t^2 - t^2) + 2.5t \][/tex]
[tex]\[ \Delta V(t) = -0.016t^3 + 0.125t^2 + 2.5t \][/tex]

So, the expression modeling the difference in volume between Tank B and Tank A after [tex]\( t \)[/tex] minutes is:
[tex]\[ \Delta V(t) = -0.016t^3 + 0.125t^2 + 2.5t \][/tex]

#### Step 2: Calculate the Difference in Volume After 4 Minutes
To find the difference in volume between Tank B and Tank A after 4 minutes, substitute [tex]\( t = 4 \)[/tex] into the difference volume expression:
[tex]\[ \Delta V(4) = -0.016(4)^3 + 0.125(4)^2 + 2.5(4) \][/tex]
Evaluating this gives us:
[tex]\[ \Delta V(4) = -0.016(64) + 0.125(16) + 2.5(4) \][/tex]
[tex]\[ \Delta V(4) = -1.024 + 2 + 10 \][/tex]
[tex]\[ \Delta V(4) = 10.976 \][/tex]

So, after 4 minutes, Tank B has 10.976 gallons more water than Tank A.

Therefore, the final answers are:
The total amount of water in both tanks after 3 minutes is 22.575 gallons.
Tank B has 10.976 gallons more water than Tank A after 4 minutes.