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Sagot :
Functions can be used to model real life scenarios
- The reasonable domain is [tex]\mathbf{[0,\infty)}[/tex].
- The average rate of change from t = 0 to 2 is 20 persons per week
- The instantaneous rate of change is [tex]\mathbf{V'(t) = -3t^2 + 2t + 12}[/tex].
- The slope of the tangent line at point (2,V(20) is 10
- The rate of infection at the maximum point is 8.79 people per week
The function is given as:
[tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
(a) Sketch V(t)
See attachment for the graph of [tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
(b) The reasonable domain
t represents the number of weeks.
This means that: t cannot be negative.
So, the reasonable domain is: [tex]\mathbf{[0,\infty)}[/tex]
(c) Average rate of change from t = 0 to 2
This is calculated as:
[tex]\mathbf{m = \frac{V(a) - V(b)}{a - b}}[/tex]
So, we have:
[tex]\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}[/tex]
[tex]\mathbf{m = \frac{V(2) - V(0)}{2}}[/tex]
Calculate V(2) and V(0)
[tex]\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}[/tex]
[tex]\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}[/tex]
So, we have:
[tex]\mathbf{m = \frac{20 - 0}{2}}[/tex]
[tex]\mathbf{m = \frac{20}{2}}[/tex]
[tex]\mathbf{m = 10}[/tex]
Hence, the average rate of change from t = 0 to 2 is 20
(d) The instantaneous rate of change using limits
[tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
The instantaneous rate of change is calculated as:
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}[/tex]
So, we have:
[tex]\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}[/tex]
[tex]\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}[/tex]
Expand
[tex]\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}[/tex][tex]\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}[/tex]
Subtract V(t) from both sides
[tex]\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}[/tex]
Substitute [tex]\mathbf{V(t) = -t^3 + t^2 + 12t}[/tex]
[tex]\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}[/tex]
Cancel out common terms
[tex]\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}[/tex]
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}[/tex] becomes
[tex]\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}[/tex]
[tex]\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}[/tex]
Limit h to 0
[tex]\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}[/tex]
[tex]\mathbf{V'(t) = -3t^2 + 2t + 12}[/tex]
(e) V(2) and V'(2)
Substitute 2 for t in V(t) and V'(t)
So, we have:
[tex]\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}[/tex]
[tex]\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}[/tex]
Interpretation
V(2) means that, 20 people were infected after 2 weeks of the virus spread
V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week
(f) Sketch the tangent line at (2,V(2))
See attachment for the tangent line
The slope of this line is:
[tex]\mathbf{m = \frac{V(2)}{2}}[/tex]
[tex]\mathbf{m = \frac{20}{2}}[/tex]
[tex]\mathbf{m = 10}[/tex]
The slope of the tangent line is 10
(g) Estimate V(2.1)
The value of 2.1 is
[tex]\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}[/tex]
[tex]\mathbf{V(2.1) = 20.35}[/tex]
(h) The maximum number of people infected at the same time
Using the graph, the maximum point on the graph is:
[tex]\mathbf{(t,V(t) = (2.361,20.745)}[/tex]
This means that:
The maximum number of people infected at the same time is approximately 21.
The rate of infection at this point is:
[tex]\mathbf{m = \frac{V(t)}{t}}[/tex]
[tex]\mathbf{m = \frac{20.745}{2.361}}[/tex]
[tex]\mathbf{m = 8.79}[/tex]
The rate of infection is 8.79 people per week
Read more about graphs and functions at:
https://brainly.com/question/18806107
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