A Person Deposits A Certain Amount Of Money In A Continuousl
A Person Deposits A Certain Amount Of Money In A Continuously Compo
Determine the initial deposit in a continuously compounded savings account given the future amount after a certain period, the amount deposited and the interest rate; find the future amount in a certain number of years given an initial deposit and interest rate; analyze a logistic growth model for population prediction; calculate remaining radioactive material after decay; estimate future population using exponential growth; apply Newton's law of cooling to estimate temperature change; evaluate stock value at a future time based on a given function; Compute bacteria growth over time with a continuous growth rate; and assess probabilities related to a normal distribution for test completion times.
Paper For Above instruction
The analysis of various exponential, logarithmic, logistic, and normal distribution models illustrates the broad application of mathematical principles in real-world scenarios ranging from finance and population dynamics to radioactive decay, thermodynamics, epidemiology, and performance analysis. These models serve as essential tools in predicting outcomes, making informed decisions, and understanding the behavior of complex systems.
1. Future value of an initial deposit with continuous compounding
Given the formula for continuous compounding, the future value \(A\) of an initial investment \(P\) after time \(t\) years at an annual interest rate \(r\) is expressed as:
\(A = Pe^{rt}\)
Using the information that after 15 years, \(A = 4552.24\), with an annual interest rate \(r = 0.0324\), the original amount deposited can be found by rearranging the formula:
\(P = \frac{A}{e^{rt}}\)
Calculating \(P\):
\[
P = \frac{4552.24}{e^{0.0324 \times 15}} \approx \frac{4552.24}{e^{0.486}} \approx \frac{4552.24}{1.626} \approx 2799
\]
Rounding to the nearest whole dollar, the original deposited amount was approximately \$2,799.
2. Future value of an initial deposit in 16 years
Given an initial deposit of \$3,550 in a continuously compounded account at 3.24%, the amount after 16 years is:
\(A = 3550 \times e^{0.0324 \times 16}\)
Calculating:
\[
A = 3550 \times e^{0.5184} \approx 3550 \times 1.678 \approx 5954.63
\]
Rounding to the nearest cent, the account will hold approximately \$5,954.63 after 16 years.
3. Population prediction using logistic function
The logistic function given is:
\[
P(t) = \frac{450}{1 + 9e^{-0.192t}}
\]
To find the population at \(t=26\) years, substitute into the formula:
\[
P(26) = \frac{450}{1 + 9e^{-0.192 \times 26}}
\]
Calculate exponent:
\[
-0.192 \times 26 = -4.992
\]
Calculate \(e^{-4.992} \approx 0.0068\):
\[
P(26) = \frac{450}{1 + 9 \times 0.0068} = \frac{450}{1 + 0.0612} = \frac{450}{1.0612} \approx 424
\]
Rounding to the nearest whole number, the population in 26 years would be approximately 424 individuals.
4. Radioactive decay after 17 years
The decay follows an exponential decay model:
\[
N(t) = N_0 e^{-\lambda t}
\]
where \(N_0=450\) grams, decay rate \(\lambda=0.032\) (since 3.2% per year):
\[
N(17) = 450 \times e^{-0.032 \times 17} \approx 450 \times e^{-0.544} \approx 450 \times 0.58 \approx 261
\]
Rounding to the nearest whole gram, approximately 261 grams remain after 17 years.
5. Future population of koala bears
Using exponential growth:
\[
N(t) = N_0 e^{rt}
\]
with \(N_0=58\) and \(r=0.123\), for \(t=21\) years (from 1999 to 2020):
\[
N(21) = 58 \times e^{0.123 \times 21} = 58 \times e^{2.583} \approx 58 \times 13.2 \approx 766
\]
Rounding to the nearest whole number, the population in 2020 will be approximately 766 koala bears.
6. Cooling calculation using Newton's law of cooling
Newton's law states:
\[
T(t) = T_{ambient} + (T_{initial} - T_{ambient}) e^{-kt}
\]
Given \(T_{initial} = 380°F\), \(T_{ambient} = 62°F\), \(k=0.114\), \(t=30\) minutes:
\[
T(30) = 62 + (380 - 62) e^{-0.114 \times 30} = 62 + 318 \times e^{-3.42} \approx 62 + 318 \times 0.033 \approx 62 + 10.5 \approx 72.5
\]
So, after half an hour, the ham's temperature is approximately 72.5°F.
7. Stock value estimation in 2009
Given:
\[
V(t) = e^{0.582 t}
\]
where \(t\) is years after 2000, in 2009, \(t=9\):
\[
V(9) = e^{0.582 \times 9} = e^{5.238} \approx 190.2
\]
Rounding to the nearest whole number, the stock value in 2009 is approximately \$190.
8. Bacterial growth after 35 minutes
Using exponential growth:
\[
N(t) = N_0 e^{rt}
\]
with \(N_0=2,500,000\), growth rate \(r=0.023\) per minute, \(t=35\):
\[
N(35) = 2,500,000 \times e^{0.023 \times 35} = 2,500,000 \times e^{0.805} \approx 2,500,000 \times 2.237 \approx 5,592,500
\]
Rounding, there will be approximately 5,592,500 bacteria after 35 minutes.
9. Probabilities related to normal distribution of test times
The test completion times follow a normal distribution with mean \(\mu=15\) minutes and standard deviation \(\sigma=2\) minutes.
- a. Probability of completing in less than 14 minutes: Calculate z-score:
\[
z = \frac{14 - 15}{2} = -0.5
\]
Using standard normal tables, \(P(Z
- b. Probability between 15 and 19 minutes: z-scores:
\[
z_{15} = 0,\quad z_{19} = \frac{19 - 15}{2} = 2
\]
Thus,
\[
P(15
\]
Approximately 47.72% of students take between 15 and 19 minutes.
- c. Completion time so only 10% take longer: Find z-score for upper 90%:
\[
z = 1.28
\]
Then,
\[
X = \mu + z \sigma = 15 + 1.28 \times 2 = 15 + 2.56 = 17.56
\]
Therefore, 90% of students finish in 17.56 minutes or less, and only 10% take longer than approximately 17.6 minutes.
- d. Probability that average of 5 tests is 14 minutes or more: For sample mean, the standard error:
\[
\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{5}} \approx 0.894
\]
Compute z-score:
\[
z = \frac{14 - 15}{0.894} \approx -1.12
\]
Using tables:
\[
P(\bar{X} \geq 14) = 1 - P(Z
\]
Approximately 86.86% probability that the mean of 5 tests will be 14 minutes or more.
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