Math Problem Statement

An environmental study of a certain community suggests that the average daily level of carbon monoxide in the air will be C(p) = 0.5p + 1 parts per million when the population is p thousand. It is estimated that t years from now the population of the community will be P(t) = 10 + 0.1t 2 thousand.

a. Express the level of carbon monoxide in the air as a function of time. b. When will the carbon monoxide level reach 6.8 parts per million?

Solution

To solve this problem, we need to express the carbon monoxide level as a function of time and then determine when it reaches a specific level. Here’s the step-by-step solution:

Part (a): Express the Level of Carbon Monoxide as a Function of Time

Given:

  • The carbon monoxide level as a function of population is C(p)=0.5p+1C(p) = 0.5p + 1 parts per million.
  • The population as a function of time is P(t)=10+0.1t2P(t) = 10 + 0.1t^2 thousand.

We want to find C(t)C(t), the level of carbon monoxide as a function of time.

  1. Substitute P(t)P(t) into C(p)C(p): C(t)=0.5P(t)+1C(t) = 0.5P(t) + 1

  2. Substitute the expression for P(t)P(t): C(t)=0.5(10+0.1t2)+1C(t) = 0.5(10 + 0.1t^2) + 1

  3. Simplify the expression: C(t)=0.5×10+0.5×0.1t2+1C(t) = 0.5 \times 10 + 0.5 \times 0.1t^2 + 1 C(t)=5+0.05t2+1C(t) = 5 + 0.05t^2 + 1 C(t)=6+0.05t2C(t) = 6 + 0.05t^2

So, the level of carbon monoxide as a function of time is:

C(t)=6+0.05t2\boxed{C(t) = 6 + 0.05t^2}

Part (b): When Will the Carbon Monoxide Level Reach 6.8 Parts per Million?

To find when the carbon monoxide level reaches 6.8 parts per million, we need to solve for tt when C(t)=6.8C(t) = 6.8.

  1. Set C(t)=6.8C(t) = 6.8: 6+0.05t2=6.86 + 0.05t^2 = 6.8

  2. Subtract 6 from both sides: 0.05t2=0.80.05t^2 = 0.8

  3. Divide both sides by 0.05: t2=0.80.05t^2 = \frac{0.8}{0.05} t2=16t^2 = 16

  4. Take the square root of both sides: t=16t = \sqrt{16} t=4t = 4

So, the carbon monoxide level will reach 6.8 parts per million in 4 years.

Final Answer:

  • (a) The level of carbon monoxide as a function of time is C(t)=6+0.05t2\boxed{C(t) = 6 + 0.05t^2}.
  • (b) The carbon monoxide level will reach 6.8 parts per million in 4\boxed{4} years.

Would you like more details or have any other questions?

Related Questions:

  1. What would be the carbon monoxide level after 10 years?
  2. How does the rate of increase of carbon monoxide change with time?
  3. At what population level does the carbon monoxide level reach 10 parts per million?
  4. If the rate of population increase changed, how would that affect the carbon monoxide level over time?
  5. What is the carbon monoxide level when the population is 15 thousand?

Tip: Understanding the relationship between variables helps in constructing and interpreting mathematical models effectively.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Functions
Quadratic Equations

Formulas

Function of time from population
Carbon monoxide function from population

Theorems

-

Suitable Grade Level

Advanced High School