Chapter 5.10, Problem 55E

a.

To determine

To make: a rough sketch of graph of F .

Answer to Problem 55E

Explanation of Solution

Given information: A manufacture of lightbulbs wants to produce bulbs that last about

700 hours but, of course, some bulbs burns out faster than others. Let F (t) be the fraction of the company’s bulbs that burn out before t hours, so F (t) always lies between 0 1nd 1.

Calculation:

The graph of F (t) is the fraction of bulbs that will burn out at time t , so it will approach 1 as $t\to \infty$ , with 1 representing 100%.

The graph of F (t) is shown below.

  

b.

To determine

To find: the meaning of the derivative r (t) =F’ (t).

Answer to Problem 55E

r (t) =F’ (t) is the rate at which the fraction F (t) of burnt-out bulbs increases as t increases . This could be interpreted as a fractional burnout rate.

Explanation of Solution

Given information: A manufacture of lightbulbs wants to produce bulbs that last about

700 hours but, of course, some bulbs burns out faster than others. Let F (t) be the fraction of the company’s bulbs that burn out before t hours, so F (t) always lies between 0 1nd 1.

Calculation:

r (t) =F’ (t) is the rate at which the fraction F (t) of burnt-out bulbs increases as t increases . This could be interpreted as a fractional burnout rate.

c.

To determine

To find: the value of ${\displaystyle \underset{0}{\overset{\infty }{\int }}r(t)dt}$ .

Answer to Problem 55E

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}r(t)dt}$ =1, where 1 represents 100%.

Explanation of Solution

Given information: A manufacture of lightbulbs wants to produce bulbs that last about

700 hours but, of course, some bulbs burns out faster than others. Let F(t) be the fraction of the company’s bulbs that burn out before t hours, so F(t) always lies between 0 1nd 1.

Calculation:

Recall that if r(t) = F(t) then ${\displaystyle \underset{0}{\overset{\infty }{\int }}r(t)dt}$ represents the net change of F(t) starts at 0 bulbs out , to eventually all bulbs burnt out ( where 1 represents 100%) ,then the net change should be 1 as $t\to \infty$ .

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}r(t)dt}$ =1.