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.

Functions can be used to model real life scenarios

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

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