Answer the following questions concerning the given word problem. All answers in Change of Base Form and round 4 decimals.
Suppose a researcher initially as 2 mg of Carbon-14. After 20 years there are 1.9 mg .
a. Find . where is the amount of Carbon-14 the researcher has after tyears.
b. Interpret " " from Part a.
c. How much Carbon-14 does the researcher have after 10,000 years?
d. When will the researcher have 1 mg of Carbon-14?
Use the editor to formot your answer

Ask by Goodwin Fernandez. in the United States

Nov 14,2024

Question

Answer the following questions concerning the given word problem. All answers in Change of Base Form and round 4 decimals.
Suppose a researcher initially as 2 mg of Carbon-14. After 20 years there are 1.9 mg .
a. Find . where is the amount of Carbon-14 the researcher has after tyears.
b. Interpret " " from Part a.
c. How much Carbon-14 does the researcher have after 10,000 years?
d. When will the researcher have 1 mg of Carbon-14?
Use the editor to formot your answer

Ask by Goodwin Fernandez. in the United States

Nov 14,2024

UpStudy AI · Accepted Answer

To solve the problem, we will use the exponential decay model for Carbon-14, which can be expressed as:

$$
P(t) = A e^{rt}
$$

where:
- $$ P(t) $$ is the amount of Carbon-14 at time $$ t $$,
- $$ A $$ is the initial amount of Carbon-14,
- $$ r $$ is the decay constant,
- $$ t $$ is the time in years.

### Part a: Find $$ P(t) = A e^{rt} $$

1. **Initial amount**: $$ A = 2 $$ mg (the initial amount of Carbon-14).
2. **Amount after 20 years**: $$ P(20) = 1.9 $$ mg.

Using the formula:

$$
P(20) = 2 e^{20r}
$$

Setting this equal to the amount after 20 years:

$$
1.9 = 2 e^{20r}
$$

Now, we can solve for $$ e^{20r} $$:

$$
e^{20r} = \frac{1.9}{2} = 0.95
$$

Taking the natural logarithm of both sides:

$$
20r = \ln(0.95)
$$

Now, solving for $$ r $$:

$$
r = \frac{\ln(0.95)}{20}
$$

Calculating $$ r $$:

$$
r \approx \frac{-0.051293}{20} \approx -0.0025647
$$

Thus, the function $$ P(t) $$ becomes:

$$
P(t) = 2 e^{-0.0025647t}
$$

### Part b: Interpret " $$ r $$ $$ from Part a

The value of $$ r $$ represents the decay constant of Carbon-14. Specifically, it indicates the rate at which the amount of Carbon-14 decreases over time. A negative value signifies that the substance is decaying, and the magnitude of $$ r $$ indicates how quickly this decay occurs. In this case, $$ r \approx -0.0026 $$ means that the amount of Carbon-14 decreases by approximately 0.25647% per year.

### Part c: How much Carbon-14 does the researcher have after 10,000 years?

Using the formula $$ P(t) = 2 e^{-0.0025647t} $$:

$$
P(10000) = 2 e^{-0.0025647 \times 10000}
$$

Calculating the exponent:

$$
-0.0025647 \times 10000 \approx -25.647
$$

Now, calculating $$ P(10000) $$:

$$
P(10000) = 2 e^{-25.647} \approx 2 \times 0.0000001 \approx 0.0000002 \text{ mg}
$$

Thus, after rounding to four decimal places, the amount of Carbon-14 after 10,000 years is approximately:

$$
P(10000) \approx 0.0000 \text{ mg}
$$

### Part d: When will the researcher have 1 mg of Carbon-14?

We need to find $$ t $$ such that $$ P(t) = 1 $$:

$$
1 = 2 e^{-0.0025647t}
$$

Dividing both sides by 2:

$$
0.5 = e^{-0.0025647t}
$$

Taking the natural logarithm of both sides:

$$
\ln(0.5) = -0.0025647t
$$

Solving for $$ t $$:

$$
t = \frac{\ln(0.5)}{-0.0025647}
$$

Calculating $$ t $$:

$$
t \approx \frac{-0.693147}{-0.0025647} \approx 270.4
$$

Thus, the researcher will have 1 mg of Carbon-14 after approximately:

$$
t \approx 270.4 \text{ years}
$$

### Summary of Answers

- **Part a**: $$ P(t) = 2 e^{-0.0025647t} $$
- **Part b**: $$ r \approx -0.0026 $$ (decay constant)
- **Part c**: $$ P(10000) \approx 0.0000 $$ mg
- **Part d**: $$ t \approx 270.4 $$ years
a. $$ P(t) = 2 e^{-0.0025647t} $$ b. $$ r \approx -0.0026 $$ (decay constant) c. $$ P(10000) \approx 0.0000 $$ mg d. $$ t \approx 270.4 $$ years

Upstudy AI Solution

Answer

Solution

Extra Insights

a. Find

b. Interpret " " from Part a

c. How much Carbon-14 does the researcher have after 10,000 years?

d. When will the researcher have 1 mg of Carbon-14?

Related Questions

Latest Calculus Questions