Math Problem Statement

here is an example

A tank has 50 gallons of water in it at time t = 0 minutes. Water begins to be pumped into the tank at time t = 0. A different pipe is draining water from the tank starting at t = 0. Water is being pumped into a tank at a rate modeled by the function K(t) = -9(0.87)^t + 15, where K(t) is measured in gallons per minute. The total amount of water, in gallons, that has been removed from the tank t minutes after the pump was turned on can be modeled by the function Z(t) = 22(1.26)^t - 22. What is the total amount of water in the tank at time t = 5?

Solution

It seems like you've uploaded an image with a mathematical problem related to water being pumped into and drained from a tank. I'll walk you through the solution based on the provided information.

The problem provides the following functions:

  • The rate of water being pumped into the tank is modeled by K(t)=9(0.87)t+15K(t) = -9(0.87)^t + 15, where K(t)K(t) is in gallons per minute and tt is the time in minutes.
  • The total amount of water removed from the tank is modeled by Z(t)=22(1.26)t22Z(t) = 22(1.26)^t - 22, where Z(t)Z(t) gives the total water removed in gallons.

The goal is to determine the total amount of water in the tank at t=5t = 5 minutes.

Step-by-step solution:

  1. Initial Condition:

    • The tank starts with 50 gallons of water at t=0t = 0.
    • Thus, the initial value is 50 gallons.
  2. Total water in:

    • The function K(t)K(t) represents the rate at which water is pumped in. To find the total amount of water pumped into the tank from t=0t = 0 to t=5t = 5, we integrate K(t)K(t): total water in=05K(t)dt=05(9(0.87)t+15)dt\text{total water in} = \int_0^5 K(t) \, dt = \int_0^5 \left(-9(0.87)^t + 15 \right) \, dt After calculating this integral, the result is: total water in=42.584772gallons.\text{total water in} = 42.584772 \, \text{gallons}.
  3. Total water out:

    • The function Z(t)Z(t) gives the total amount of water removed from the tank. To find the amount of water removed at t=5t = 5, we evaluate Z(5)Z(5): Z(5)=22(1.26)522=47.867533gallons.Z(5) = 22(1.26)^5 - 22 = 47.867533 \, \text{gallons}.
  4. Total amount of water in the tank at t=5t = 5:

    • The total amount of water in the tank is given by: Total water=(initial value)+(total water in)(total water out)\text{Total water} = (\text{initial value}) + (\text{total water in}) - (\text{total water out}) Substituting the values we have: Total water=50+42.58477247.867533=44.717gallons.\text{Total water} = 50 + 42.584772 - 47.867533 = 44.717 \, \text{gallons}.

So, at t=5t = 5, the total amount of water in the tank is approximately 44.717 gallons.

Let me know if you'd like more details or have any questions about this!

Related questions:

  1. How do you calculate the total water in if the rate is not constant?
  2. What happens to the amount of water in the tank if the rate of pumping changes?
  3. How can you model similar problems where the water is drained at a different rate?
  4. How does the initial amount of water affect the solution in such problems?
  5. How can we solve for the total water when the time is given as a variable instead of a constant?

Tip: When calculating integrals involving exponential functions like this, break them into parts you can easily handle, and use a calculator or software tool to assist with the integration.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Rate of change
Exponential functions

Formulas

K(t) = -9(0.87)^t + 15
Z(t) = 22(1.26)^t - 22
Total water = (initial value) + (total water in) - (total water out)

Theorems

Integral of rate functions
Evaluation of exponential functions

Suitable Grade Level

Grades 11-12