Chapter 2.6, Problem 36E

a.

To determine

To find:The mass density of the material of an insect as a function of time.

Answer to Problem 36E

The density of the material of insect as the function of time is $\rho (t)=\frac{1+t^2}{1+2t}$ grams /cm³.

Explanation of Solution

Given: The given mass function $M\left(t\right)$ and the volume function $V\left(t\right)$ are: $M\left(t\right)=1+t^2$ and $V\left(t\right)=1+2t$ , where time t is measured in days, mass of insect M in grams, and volume of insect V in cm³.

Concept Used:

The mass M of an object is related to volume V of the object according to relation $M=V\times \rho$ , where $\rho$ is the density of the material of the object.

Calculation:

The mass and volume of the insect are functions time given as $M\left(t\right)=1+t^2$ , and $V\left(t\right)=1+2t$ .

Therefore, the density of the material the insect is made of is also time dependent given by

  $\rho (t)=\frac{M\left(t\right)}{V\left(t\right)}=\frac{1+t^2}{1+2t}$ .

Conclusion:

The density of the material of insect as the function of time is $\rho (t)=\frac{1+t^2}{1+2t}$ grams /cm³.

b.

To determine

To find:The derivative of the density $\rho (t)$ computed in part (a).

Answer to Problem 36E

The derivative of density $\rho (t)$ is $\frac{d\rho }{dt}=\frac{2\left(t^2+t-1\right)}{{\left(1+2t\right)}^2}$ .

Explanation of Solution

Given:

The expression for density is $\rho (t)=\frac{1+t^2}{1+2t}$

Formula used:

The derivative of function $y(x)=\frac{u\left(x\right)}{v\left(x\right)}$ is $\frac{dy}{dx}=\frac{v\cdot \frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ .

Calculation:

The density isthe function of t given by $\rho (t)=\frac{1+t^2}{1+2t}$.

Now, the derivative of $\rho (t)$ is

  $\frac{d\rho }{dt}=\frac{\left(1+2t\right)\cdot \frac{d}{dt}\left(1+t^2\right)-\left(1+t^2\right)\cdot \frac{d}{dt}\left(1+2t\right)}{{\left(1+2t\right)}^2}$

  $\frac{d\rho }{dt}=\frac{\left(1+2t\right)\left(2t\right)-\left(1+t^2\right)\cdot \left(2\right)}{{\left(1+2t\right)}^2}=\frac{\left(4t^2+2t\right)-2\left(1+t^2\right)}{{\left(1+2t\right)}^2}$

  $\Rightarrow \frac{d\rho }{dt}=\frac{2t^2+2t-2}{{\left(1+2t\right)}^2}$

Conclusion:

The derivative of density $\rho (t)$ is $\frac{d\rho }{dt}=\frac{2\left(t^2+t-1\right)}{{\left(1+2t\right)}^2}$ .

c.

To determine

To find: At what time the density $\rho (t)$ is increasing.

Answer to Problem 36E

The density will increase when $t>0.618\text{days}$ .

Explanation of Solution

Given:

The derivative density computed in part (a) as a function of time is $\frac{d\rho }{dt}=\frac{2\left(t^2+t-1\right)}{{\left(1+2t\right)}^2}$

Concept used:

A function $y=f\left(x\right)$ is an increasing when $f^{\prime }\left(x\right)>0$

Calculation:

The derivative of density is $\frac{d\rho }{dt}=\frac{2\left(t^2+t-1\right)}{{\left(1+t\right)}^2}$

The density will increase when $\frac{d\rho }{dt}>0$

  $\Rightarrow \frac{2\left(t^2+t-1\right)}{{\left(1+2t\right)}^2}>0$

  $\Rightarrow 2\left(t^2+t-1\right)>0$$\because {\left(1+2t\right)}^2$ , always a positive quantity.

  $\Rightarrow t^2+t-1>0$

  $\Rightarrow t^2+t>1$

  $\Rightarrow t^2+2t\cdot \frac{1}{2}+\frac{1}{4}>1+\frac{1}{4}=\frac{5}{4}$

  $\Rightarrow {\left(t+\frac{1}{2}\right)}^2>\frac{5}{4}$

  $\Rightarrow \left(t+\frac{1}{2}\right)>\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

  $\Rightarrow t>\frac{\sqrt{5}}{2}-\frac{1}{2}=\frac{\sqrt{5}-1}{2}=0.618$ .

Conclusion:

The density will increase when $t>0.618\text{days}$ .

d.

To determine

To graph:The density given in part (a) for first 5 days.

Explanation of Solution

Given:

The density function computed in part (a)

Concept used:

We use graphing calculator for the graph.

Graph:

The graph of the function $\rho (t)=\frac{1+t^2}{1+2t}$ is shown in figure.

  

Interpretation:

We see clearly that the function $\rho (t)$ is increasing when $t>0.618$ that the density start increasing after 0.618 days.