CHAPTER 17
Applications of
differential calculus
A Kinematics
B Rates of change
C Optimization
Displacement
Displacement - the moving of something from its place or
position.
s(t) is the displacement function which
represents the separation between
the origin and point P with respect
to time.
On the horizontal axis through O:
if s(t) > 0, P is located to the right of O
if s(t) = 0, P is located at O
if S(t) < 0, P is located to the left of O.
Motion Graph
A motion graph is a horizontal graph showing the location of
the object in motion with respect to time.
s(t) = t2 + 2t – 3
t=0
s(0) = -3 cm
t=1
s(1) = 0 cm
t=2
s(2) = 5 cm
t=3
s(3) = 12 cm
t=4
s(4) = 21 cm
Velocity
The average velocity of an object moving in a straight line in
the time interval from t = t1 to t = t2 is the ratio of the change in
displacement to the time taken.
If s(t) is the displacement function then average velocity is
st 2   st1 
t 2  t1
If s(t) is the displacement function of an object moving in a
straight line, then v(t) = s’(t) is the instantaneous velocity or
velocity function of the object at time t.
Acceleration
If an object moves in a straight line with velocity function v(t)
then its average acceleration for the time interval from t = t1
to t = t2 is the ratio of the change in velocity to the time taken.
average acceleration 
v(t 2 )  v(t1 )
t 2  t1
If a particle moves in a straight line with velocity function v(t),
then the instantaneous acceleration at time t is a(t) = v’(t).
Units
Units of velocity is m/s or ms-1
Units of acceleration is m/s2 or ms-2
Example 1
Exercise 17A.1 #1
a.
s t 2   3  33  2  9  9  2  16
2
s t1   1  31  2  1  3  2  2
2
s t 2   s t1  16  2 14
average velocity 


 7 m/s
t 2  t1
3 1
2
b.
s t 2   1  h   31  h   2
2
 1  2h  h 2  3  3h  2
 h 2  5h  2
s t1   1  31  2  1  3  2  2
2
s t 2   s t1  h 2  5h  2  2 h 2  5h
average velocity 


 h5 m/s
t 2  t1
1 h 1
h
c.
lim h  5  5
h 0
The instantaneous velocity at t = 1 is 5 m/s
d.
s t 2   t  h   3t  h   2
2
 t 2  2th  h 2  3t  3h  2
s t1   t   3t   2  t 2  3t  2
2

s t 2   s t1  t 2  2th  h 2  3t  3h  2  t 2  3t  2
average velocity 

t 2  t1
t  ht
2th  h 2  3h

 2t  h  3 m / s
h
lim 2t  h  3  2t  3
h 0
The instantaneous velocity of object with respect to time

Velocity and Acceleration Function
If a particle P moves in a straight line and its position is given
by the displacement function s(t), t > 0, then:
1st derivative
2nd derivative
Sign Interpretation
Suppose a particle P moves in a straight line with
displacement function s(t) relative to an origin O.
Its velocity function is v(t) and its acceleration function is a(t).
We can use sign diagrams to interpret:
a. where the particle is located relative to O
b. the direction of motion and where a change of
direction occurs
c. when the particle’s velocity is increasing or
decreasing.
Sign of s(t)
Location of the object with respect to the original location.
Sign of v(t)
The direction of the velocity of the object.
Sign of a(t)
Increasing or decreasing velocity. (getting faster or slowing down)
Frame of reference of the “0”
Speed
Velocity of a vector quantity which means it has both
magnitude and direction. Velocity may have positive or
negative values depending on its direction.
Speed is the magnitude of the velocity which means it has no
direction and can never be negative.
Speed
To determine when the speed S(t) of an object P with
displacement s(t) is increasing or decreasing, we need to
employ a sign test.
Example 2
B. Rate of Change
Examples of other real world rates of change
temperature varies continuously
the height of a tree varies as it grows
the prices of stocks and shares vary with each day’s
trading.
Example 3
EXERCISE 17B #1
Example 4
Exercise 17B. #5
Example 5
Area = ½ bc sin A
(chapter 9 part A)
C. Optimization
There are many problems for which we need to find the maximum or
minimum value of a function.
The solution is often referred to as the optimum solution and the process is
called optimization.
We can find optimum solutions in several ways:
using technology to graph the function and search for the maximum
or minimum value
using analytical methods such as the formula x =-b/2a
for the vertex of a parabola
using differential calculus to locate the turning points of a
function.
These last two methods are useful especially when exact solutions are
required.
Warning!!
The maximum or minimum value does not always occur when
the first derivative is zero. It is essential to also examine the
values of the function at the endpoint(s) of the interval under
consideration for global maxima and minima.
OPTIMIZATION PROBLEM SOLVING METHOD
Example 6
A rectangular cake dish is made by cutting out squares
from the corners of a 25 cm by 40 cm rectangle of tinplate, and then folding the metal to form the container.
What size squares must be cut out to produce the cake
dish of maximum volume?
Step 1
Step 2
Find the volume of the dish.
Step 3
Determine the x-value of the stationary points.
Step 4
Draw a sign diagram to determine the maximum.
Step 5
Summarize the answer in a sentence.
The maximum volume is obtained when x = 5, which is when 5
cm squares are cut from the corners.
Note: If they ask for the maximum volume, then simply
substitute 5 in for the x-value and solve for the volume.
Example 7
A 4 liter container must have a square base, vertical sides, and
an open top. Find the most economical shape which minimizes
the surface area of material needed.
Step 1
Step 2
Step 3
Step 4
Step 5
The most economical shape has a square base
20 cm x 20 cm, and height 10 cm.
Example 8
Exercise 17C #2