Question

The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,

Answer 1

Functions can be used to model real life scenarios

• The reasonable domain is $\mathbf{[0,\infty)}$.
• The average rate of change from t = 0 to 2 is 20 persons per week
• The instantaneous rate of change is $\mathbf{V'(t) = -3t^2 + 2t + 12}$.
• 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:

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

(a) Sketch V(t)

See attachment for the graph of $\mathbf{V(t) = -t^3 + t^2 + 12t}$

(b) The reasonable domain

t represents the number of weeks.

This means that: t cannot be negative.

So, the reasonable domain is: $\mathbf{[0,\infty)}$

(c) Average rate of change from t = 0 to 2

This is calculated as:

$\mathbf{m = \frac{V(a) - V(b)}{a - b}}$

So, we have:

$\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}$

$\mathbf{m = \frac{V(2) - V(0)}{2}}$

Calculate V(2) and V(0)

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}$

So, we have:

$\mathbf{m = \frac{20 - 0}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

Hence, the average rate of change from t = 0 to 2 is 20

(d) The instantaneous rate of change using limits

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

The instantaneous rate of change is calculated as:

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$

So, we have:

$\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}$

$\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}$

Expand

$\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}$$\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}$

Subtract V(t) from both sides

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}$

Substitute $\mathbf{V(t) = -t^3 + t^2 + 12t}$

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}$

Cancel out common terms

$\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}$

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$ becomes

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}$

$\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}$

Limit h to 0

$\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}$

$\mathbf{V'(t) = -3t^2 + 2t + 12}$

(e) V(2) and V'(2)

Substitute 2 for t in V(t) and V'(t)

So, we have:

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}$

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:

$\mathbf{m = \frac{V(2)}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

The slope of the tangent line is 10

(g) Estimate V(2.1)

The value of 2.1 is

$\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}$

$\mathbf{V(2.1) = 20.35}$

(h) The maximum number of people infected at the same time

Using the graph, the maximum point on the graph is:

$\mathbf{(t,V(t) = (2.361,20.745)}$

This means that:

The maximum number of people infected at the same time is approximately 21.

The rate of infection at this point is:

$\mathbf{m = \frac{V(t)}{t}}$

$\mathbf{m = \frac{20.745}{2.361}}$

$\mathbf{m = 8.79}$

The rate of infection is 8.79 people per week

Read more about graphs and functions at:

brainly.com/question/18806107