https://file.aiquickdraw.com/mathbot/file/1730486496541gu1634ux.jpg

Let's go through each part of this problem.

The given formula for the average level of carbon dioxide in the air is:

\[
q(t) = 0.05t^2 + 0.1t + 3.4
\]

where \( q(t) \) represents the CO\(_2\) level in parts per million, and \( t \) is the time in years from now.

### Part (a)
To find the rate at which the CO\(_2\) level is changing with respect to time one year from now, we need to calculate the derivative \( q'(t) \) and evaluate it at \( t = 1 \).

1. Differentiate \( q(t) \) with respect to \( t \):
   \[
   q'(t) = \frac{d}{dt} (0.05t^2 + 0.1t + 3.4)
   \]
   \[
   q'(t) = 2 \cdot 0.05 \cdot t + 0.1 = 0.1t + 0.1
   \]

2. Substitute \( t = 1 \) into \( q'(t) \):
   \[
   q'(1) = 0.1 \cdot 1 + 0.1 = 0.1 + 0.1 = 0.2
   \]

So, the rate at which the CO\(_2\) level is changing one year from now is \( 0.2 \) parts per million per year.

### Part (b)
To find how much the CO\(_2\) level changes in the first year, we need to evaluate \( q(1) - q(0) \).

1. Calculate \( q(1) \):
   \[
   q(1) = 0.05 \cdot (1)^2 + 0.1 \cdot 1 + 3.4 = 0.05 + 0.1 + 3.4 = 3.55
   \]

2. Calculate \( q(0) \):
   \[
   q(0) = 0.05 \cdot (0)^2 + 0.1 \cdot 0 + 3.4 = 3.4
   \]

3. Find the change in CO\(_2\) level:
   \[
   q(1) - q(0) = 3.55 - 3.4 = 0.15
   \]

Thus, the CO\(_2\) level increases by \( 0.15 \) parts per million in the first year.

### Part (c)
To determine how much the CO\(_2\) level changes over the next (second) year, we need to calculate \( q(2) - q(1) \).

1. Calculate \( q(2) \):
   \[
   q(2) = 0.05 \cdot (2)^2 + 0.1 \cdot 2 + 3.4 = 0.05 \cdot 4 + 0.2 + 3.4 = 0.2 + 0.2 + 3.4 = 3.8
   \]

2. Find the change in CO\(_2\) level:
   \[
   q(2) - q(1) = 3.8 - 3.55 = 0.25
   \]

Therefore, the CO\(_2\) level increases by \( 0.25 \) parts per million over the second year.

### Summary of Answers
- **(a)** The rate of change in CO\(_2\) level one year from now is \( 0.2 \) parts per million per year.
- **(b)** The CO\(_2\) level increases by \( 0.15 \) parts per million in the first year.
- **(c)** The CO\(_2\) level increases by \( 0.25 \) parts per million over the second year.

Would you like more details on any part of the solution, or do you have any questions?

---

Here are 5 related questions that could expand your understanding:
1. How would the rate of change vary if the function \( q(t) \) had a different coefficient for \( t^2 \)?
2. How can we interpret the second derivative \( q''(t) \) in terms of CO\(_2\) concentration change over time?
3. What would happen to the CO\(_2\) levels if the initial concentration (\( 3.4 \)) were different?
4. How could we predict the CO\(_2\) level for a future year, say \( t = 5 \)?
5. How does the constant term \( 0.1 \) in the derivative affect the rate of change over time?

**Tip:** Differentiating a quadratic function \( ax^2 + bx + c \) always results in a linear function \( 2ax + b \), which can make it easier to analyze rate of change patterns.

An environmental study of a certain suburban community suggests that t years from now, the average level of carbon dioxide in the air can be modeled by the formula q(t) = 0.05t^2 + 0.1t + 3.4 parts per million. Part (a): At what rate will the CO2 level be changing with respect to time one year from now? Part (b): By how much will the CO2 level change in the first year? Part (c): By how much will the CO2 level change over the next (second) year?

Explore how to calculate the rate of change and changes in CO2 levels over time using calculus and quadratic functions. This problem provides a model for predicting CO2 levels in a suburban area, demonstrating the application of differentiation and function analysis. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation