Thomas' Calculus
Tenth Edition
Section 2.2- The Derivative as a Rate of Change
5.
At time , the position of a body moving along the s-axis is        m.
a) Find the body's acceleration each time the velocity is zero.
The velocity  is the derivative of 
      
                  
when    sec and    sec .
The acceleration  is the derivative of  .
    
When              m/sec 
When             m/sec 
b) Find the body's speed each time the acceleration is zero.
       when  when    sec .
                    m/sec .
The speed when    sec is      m/sec .
c) Find the total distance traveled by the body from    to    
To find the total distance traveled, 1st look at a graph of 
10
10
5
3t
2
12 t
9
0
1
2
t
2
5 5
0
 is positive from      which means that the body is
moving to the right. Then at   1 the velocity is 0 which means
the body stops and will turn around because the velocity is negative
from      . When the velocity is negative the body is moving
to the left.
So from 0  t  1, the body goes from
   m to              m
which is a distance of  m .
From      the body goes from
   m to             2 m 
which is a distance of    or 2 m .
Therefore the total distance traveled is      m .
13.
The figure below shows the velocity       m /sec) of a moving body
along a coordinate line.
v (m / sec)
3
v = f (t)
t (sec)
1
3
6
8
10
-3
a) When does the body reverse direction?
The body reverses direction when  changes from positive to negative.
When it does this, its velocity is 0. This occurs when   2 and 7 seconds.
b) When (approximately) is the body moving at a constant speed?
The speed is constant when the graph is horizontal. This occurs
between 3 and 6 seconds.
c) Graph the body's speed for      
To get the speed, we don't have to change the graph when   
However, when    we want the opposite positive number.
To do this, just rotate the graph 180° about the t-axis. Only do this
when   
v (m / sec)
Speed
3
t (sec)
1
3
6
8
10
d) Graph the acceleration where defined.
  
Note that acceleration is not defined when there are sharp corner points
on the graph of . Therefore  is not defined when      and 8.
To get  from 0 to 1 seconds, get the slope of the line segment from  
to  . This slope is 3. The value of  from 1 to 3 seconds is the slope
of the line from   to 3   which is    Similarly the value of 
from 3 to 6 seconds is 0, the value of  from 6 to  seconds is , and from
8 to 10 seconds,  is    Note to get   we get the slope of the

line from   to   which is
   
  
dv/dt (m / sec)
Acceleration
3
t (sec)
1
3
6
8
10
-3
21.
Suppose that the revenue from selling  washing machines is
    

 dollars

a) Find the marginal revenue when 100 machines are produced.
Marginal revenue is the derivative of       
          



  
 
 
 2 dollars
 
 

b) Use the function   to estimate the increase in revenue that will result from
increasing production from 100 machines a week to 101 machines a week.
The increase in revenue will be approximately    2 dollars.
c) Find the limit of   as     How would you interpret this number?
lim 

 


This means that as production of machines increases, our additional
revenue will be getting smaller and smaller, and the amount will be
getting closer and closer to 0.
25.
The volume     of a spherical balloon changes with the radius.
a) At what rate (ft /ft) does the volume change with respect to the radius
when    ft?
The rate of change of the volume with respect to the radius is

 

When   , the rate is 4   ft /ft .
b) By approximately how much does the volume increase when the radius
changes from 2 to 2.2 ft?
From part a, we know the rate of change of  with respect to  is
16 when    This means that when  changes from 2 to 3,
 will change by approximately 16 ft .
Now if  changes from 2 to 2.2, a change of only 0.2, then  will
change by approximately   16  3.2 ft .