Business Applications of Differentiation (Continuation)
Example 8:
The gross domestic product (GDP) of a certain country was
[tex]$N(t)=t^2+5 t+106$[/tex]
billion Ghana cedi [tex]$t$[/tex] years after 1998.
(a) At what rate was the GDP changing with respect to time in 2008?
(b) At what percentage rate was the GDP changing with respect to time in 2008?
Example 9:
A medical research team determines that [tex]$t$[/tex] days after an epidemic begins,
[tex]$N(t)=10 t^3+5 t+\sqrt{t}$[/tex]
People will be infected, for [tex]$0 \leq t \leq 20$[/tex].
At what rate is the infected population increasing on the ninth day?
Answer (2)
For Example 8(a), find the rate of change of GDP by differentiating N ( t ) and evaluating at t = 10 , resulting in 25 billion Ghana cedi per year.
For Example 8(b), calculate the percentage rate of change by dividing N ′ ( 10 ) by N ( 10 ) and multiplying by 100, yielding approximately 9.765625% .
For Example 9, determine the rate of increase of infected population by differentiating N ( t ) and evaluating at t = 9 , which is approximately 2435.1667 people per day.
Explanation
Problem Analysis Let's analyze the given problems and find the required rates of change using differentiation. We have three examples to solve.
Solving Example 8(a) Example 8(a):
The gross domestic product (GDP) of a country is given by the function:
N ( t ) = t 2 + 5 t + 106
billion Ghana cedi, where t is the number of years after 1998. We want to find the rate at which the GDP was changing in 2008.
First, we need to find the value of t for the year 2008:
t = 2008 − 1998 = 10
Next, we find the derivative of N ( t ) with respect to t :
$N'(t) =
d t d ( t 2 + 5 t + 106 ) = 2 t + 5
Now, we evaluate N ′ ( t ) at t = 10 :
N ′ ( 10 ) = 2 ( 10 ) + 5 = 20 + 5 = 25
So, the GDP was changing at a rate of 25 billion Ghana cedi per year in 2008.
Solving Example 8(b) Example 8(b):
We want to find the percentage rate at which the GDP was changing in 2008. We already found that N ′ ( 10 ) = 25 . Now we need to find the GDP in 2008, which is N ( 10 ) :
N ( 10 ) = ( 10 ) 2 + 5 ( 10 ) + 106 = 100 + 50 + 106 = 256
Now, we can find the percentage rate of change:
Percentage rate = N ( 10 ) N ′ ( 10 ) × 100 = 256 25 × 100 ≈ 9.765625%
So, the GDP was changing at approximately 9.77% in 2008.
Solving Example 9 Example 9:
The number of infected people t days after an epidemic begins is given by:
N ( t ) = 10 t 3 + 5 t + t
We want to find the rate at which the infected population is increasing on the ninth day, i.e., at t = 9 .
First, we find the derivative of N ( t ) with respect to t :
$N'(t) =
d t d ( 10 t 3 + 5 t + t ) = 30 t 2 + 5 + 2 t 1
Now, we evaluate N ′ ( t ) at t = 9 :
N ′ ( 9 ) = 30 ( 9 ) 2 + 5 + 2 9 1 = 30 ( 81 ) + 5 + 2 ( 3 ) 1 = 2430 + 5 + 6 1 = 2435 + 6 1 ≈ 2435.1667
So, the infected population is increasing at a rate of approximately 2435.17 people per day on the ninth day.
Final Answers Final Answers:
(a) The GDP was changing at a rate of 25 billion Ghana cedi per year in 2008.
(b) The GDP was changing at a percentage rate of approximately 9.765625% in 2008.
(c) The infected population is increasing at a rate of approximately 2435.1667 people per day on the ninth day.
Examples
Differentiation is a powerful tool used across various fields. In economics, it helps determine rates of change in economic indicators like GDP, providing insights into economic growth. In epidemiology, it models the spread of infectious diseases, aiding in public health planning and response. Understanding these rates of change allows for informed decision-making, whether it's adjusting economic policies or implementing measures to control disease outbreaks. By analyzing derivatives, economists and healthcare professionals can better predict future trends and optimize strategies.
In 2008, the GDP was changing at a rate of 25 billion Ghana cedi per year and approximately 9.77% per year. On the ninth day of the epidemic, the infected population was increasing at a rate of approximately 2435.17 people per day.
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