When the annual rate of inflation averages 6% over the next 10 years, the approximate cost C of goods or services during any year in that decade is given below, where t is the time in years and P is the present cost.
C(t) = P(1.06)t
Find the rates of change of C with respect to t when
t = 1
and
t = 5.
(Round your coefficients to three decimal places.)
At
t = 1
	$$
Incorrect: Your answer is incorrect.
At
t = 5
  	$$

To find the rate of change of $ C(t) $ with respect to $ t $, we need to compute the derivative of $ C(t) $ with respect to $ t $. The function given is:

\[
C(t) = P(1.06)^t
\]

Let's differentiate $ C(t) $ with respect to $ t $.

### Step 1: Differentiate $ C(t) $ with respect to $ t $
The derivative of $ C(t) $ with respect to $ t $ is:

\[
\frac{dC(t)}{dt} = \frac{d}{dt} \left[P(1.06)^t\right]
\]

Since $ P $ is a constant, it can be factored out:

\[
\frac{dC(t)}{dt} = P \cdot \frac{d}{dt} \left[(1.06)^t\right]
\]

Now, recall that the derivative of $ a^t $ (where $ a $ is a constant) with respect to $ t $ is given by:

\[
\frac{d}{dt} \left[a^t\right] = a^t \ln(a)
\]

So:

\[
\frac{dC(t)}{dt} = P \cdot (1.06)^t \cdot \ln(1.06)
\]

### Step 2: Calculate the rate of change at $ t = 1 $ and $ t = 5 $

1. **At $ t = 1 $:**

\[
\frac{dC(1)}{dt} = P \cdot (1.06)^1 \cdot \ln(1.06)
\]

Now, calculate the value:

\[
\frac{dC(1)}{dt} \approx P \cdot 1.06 \cdot 0.05827 \approx P \cdot 0.0617662
\]

So, the rate of change at $ t = 1 $ is approximately:

\[
\frac{dC(1)}{dt} \approx 0.062P
\]

2. **At $ t = 5 $:**

\[
\frac{dC(5)}{dt} = P \cdot (1.06)^5 \cdot \ln(1.06)
\]

Now, calculate the value:

\[
(1.06)^5 \approx 1.338225
\]
\[
\frac{dC(5)}{dt} \approx P \cdot 1.338225 \cdot 0.05827 \approx P \cdot 0.077972
\]

So, the rate of change at $ t = 5 $ is approximately:

\[
\frac{dC(5)}{dt} \approx 0.078P
\]

### Final Answers:
- At $ t = 1 $, the rate of change is approximately $ \boxed{0.062P} $.
- At $ t = 5 $, the rate of change is approximately $ \boxed{0.078P} $.

Would you like more details on any part of this solution? Here are five questions to expand on this topic:

1. How would the rate of change be different if the inflation rate were higher?
2. What would the function look like if the inflation rate were compounded monthly instead of annually?
3. How do we interpret the rate of change in the context of this problem?
4. How would you find the exact value of $ C(t) $ at a particular time $ t $?
5. What would happen if inflation decreased after a few years?

**Tip:** Understanding the derivative as a rate of change is crucial for interpreting real-world scenarios like inflation.

When the annual rate of inflation averages 6% over the next 10 years, the approximate cost C of goods or services during any year in that decade is given below, where t is the time in years and P is the present cost. C(t) = P(1.06)^t Find the rates of change of C with respect to t when t = 1 and t = 5. (Round your coefficients to three decimal places.)

Explore how to calculate the rate of change for the exponential function C(t) = P(1.06)^t, which models inflation over a decade. Learn how to find the derivative with respect to time (t) and solve for specific years. This problem involves key calculus concepts, including exponential functions and derivatives. Suitable for Grades 11-12.

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