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The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,
where V (t) is the number of people (in hundreds) with the virus and t is the number of weeks since the first case was observed.
(a) Sketch V (t).
(b) What is a reasonable domain of t for this problem?
(c) Find the average rate of infection from t = 0 to t = 2.
(d) Find the instantaneous rate of infection as a function of t using the limit definition of the derivative.
(e) Find V (2) and V ‘ (2). Write a sentence interpreting V (2) and V ‘ (2) in terms of the number of infected people. Make sure to include units.
(f) Sketch the tangent line to the graph you drew in a. at the point (2, V (2)). State the slope of the tangent line.
(g) Use V (2) and V ‘ (2) to estimate the value of V (2.1).
(h) Find the maximum number of people infected at the same time and when the maximum occurs. Determine the rate of infection at this time.

Sagot :

MrRoyal

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

View image MrRoyal
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