Chapter 2.6, Problem 25E

(a)

To determine

The total mass as a function of time

Answer to Problem 25E

  $T\left(t\right)==160\times {10}^6+80\times {10}^3{t}^2-1\times {10}^{6}t-0.5\times {10}^3{t}^3\text{kg}$

Explanation of Solution

Given:

The population size is

  $P\left(t\right)=2.0\times {10}^6+1000t^2$

The weight per person is

  $W\left(t\right)=80-0.5t$

Calculation:

The total mass is the product of the number of individuals and the mass of each individual.

Hence, the total mass of the function will be

  $\begin{array}{c}T\left(t\right)=P\left(t\right)W\left(t\right)\\ =\left(2.0\times {10}^6+1000t^2\right)\left(80-0.5t\right)\\ =\left(160\times {10}^6+{10}^3{t}^2\right)80-0.5t\\ =160\times {10}^6+80\times {10}^3{t}^2-1\times {10}^{6}t-0.5\times {10}^3{t}^3\text{kg}\end{array}$

(b)

To determine

The value of derivative

Answer to Problem 25E

  $T\text{'}\left(t\right)=-1\times {10}^6+1.6\times {10}^{5}t-1.5\times {10}^3{t}^2$

Explanation of Solution

Given:

The population size is

  $P\left(t\right)=2.0\times {10}^6+1000t^2$

The weight per person is

  $W\left(t\right)=80-0.5t$

Calculation:

The derivative of total mass of the function will be

  $\begin{array}{l}T\left(t\right)=160\times {10}^6+80\times {10}^3{t}^2-1\times {10}^{6}t-0.5\times {10}^3{t}^3\text{kg}\\ T\text{'}\left(t\right)=\frac{d}{dt}\left(160\times {10}^6+80\times {10}^3{t}^2-1\times {10}^{6}t-0.5\times {10}^3{t}^3\text{kg}\right)\\ T\text{'}\left(t\right)=80\times {10}^3\left(2t\right)-1\times {10}^6-0.5\times {10}^3\left(3t^2\right)\\ T\text{'}\left(t\right)=-1\times {10}^6+1.6\times {10}^{5}t-1.5\times {10}^3{t}^2\end{array}$

(c)

To determine

The value of population, the mass of each individual and the total mass

Answer to Problem 25E

  $\begin{array}{l}P\left(6.67\right)=2.04\times {10}^6\\ W\left(6.67\right)=76.67\text{kg}\\ T\left(6.67\right)=1.56\times {10}^8\text{kg}\\ \\ P\left(100\right)=1.2\times {10}^7\\ W\left(100\right)=30\text{kg}\\ T\left(100\right)=3.6\times {10}^8\text{kg}\end{array}$

Explanation of Solution

Given:

The population size is

  $P\left(t\right)=2.0\times {10}^6+1000t^2$

The weight per person is

  $W\left(t\right)=80-0.5t$

Calculation:

The derivative is equal to zero, therefore,

  $\begin{array}{l}T\text{'}\left(t\right)=0\\ T\text{'}\left(t\right)=-1\times {10}^6+1.6\times {10}^{5}t-1.5\times {10}^3{t}^2=0\\ t=6.67\text{or }t=100\end{array}$

The population at t = 6.67 will be

  $\begin{array}{l}P\left(t\right)=2.0\times {10}^6+1000t^2\\ P\left(6.67\right)=2.0\times {10}^6+1000{\left(6.67\right)}^2\\ P\left(6.67\right)=2.0\times {10}^6+44.49\times {10}^3\\ P\left(6.67\right)=2.0\times {10}^6+0.04\times {10}^6\\ P\left(6.67\right)=2.04\times {10}^6\end{array}$

The weight per person will be

  $\begin{array}{l}W\left(t\right)=80-0.5t\\ W\left(6.67\right)=80-0.5\left(6.67\right)\\ W\left(6.67\right)=76.67\text{kg}\end{array}$

The total weight will be

  $\begin{array}{l}T\left(6.67\right)=P\left(6.67\right)W\left(6.67\right)\\ T\left(6.67\right)=\left(2.04\times {10}^6\right)\left(76.67\right)\\ T\left(6.67\right)=1.56\times {10}^8\text{kg}\end{array}$

The population at t = 100 will be

  $\begin{array}{l}P\left(t\right)=2.0\times {10}^6+1000t^2\\ P\left(100\right)=2.0\times {10}^6+1000{\left(100\right)}^2\\ P\left(100\right)=2.0\times {10}^6+{10}^7\\ P\left(100\right)=1.2\times {10}^7\end{array}$

The weight per person will be

  $\begin{array}{l}W\left(t\right)=80-0.5t\\ W\left(100\right)=80-0.5\left(100\right)\\ W\left(100\right)=30\text{kg}\end{array}$

The total weight will be

  $\begin{array}{l}T\left(100\right)=P\left(100\right)W\left(100\right)\\ T\left(100\right)=\left(1.2\times {10}^7\right)\left(30\right)\\ T\left(100\right)=3.6\times {10}^8\text{kg}\end{array}$

(d)

To determine

To sketch: The total mass over the next 100 years.

Explanation of Solution

Given:

The population size is

  $P\left(t\right)=2.0\times {10}^6+1000t^2$

The weight per person is

  $W\left(t\right)=80-0.5t$

Calculation:

The graph for total mass can be drawn as