Applications of Calculus

Rates.jpgIn previous topics, calculus was used to find the gradient of curves and the area under curves.

Integration and differentiation have many other uses.

Rates of Change

If a car travels 100 km in 2 hours then the car has an average speed of 50 km/h. Speed here is a measure of the average rate of change of distance against time.

Differentiation gives a measure of a rate of change at a particular instant. It gives an instantaneous rate of change.

Example 1
Example 2

Find the rate of change of y with respect to x for the function y = x2 − 3x − 4 at the point x = 5.

Differentiating ,Y11_Applications_of_Calculus_01.gif = 2x − 3

When x = 5

Y11_Applications_of_Calculus_02.gif = 2 x 5 − 3 = 7

The rate of change, at the point x = 5, of y with respect to x is 7.

Find the rate of change of volume with respect to time at the instant when t = 5 given that 
V = 1.6t2 + 5 
where V is the volume of oil leaking from a tanker in m3 and t is the time in minutes.

Differentiating , V ' = 3.2t

When t = 5

V '(5) = 3.2 x 5 = 16

The rate of change, after 5 minutes, of volume with respect to time is 16 m3/min.

 

Maximum and Minimum Problems

In previous topic, calculus was used to find the coordinates of maximum and minimum points on graphs such as parabolas and cubic functions. These techniques are also useful for solving real-life type problems where values need to be maximised or minimised.

Example
Solution

A farmer has 200 metres of fencing. He wants to enclose a rectangular paddock.

  
a. Show that if the length of the paddock is x metres then the width will be (100 − x) metres.

Draw a sketch

Y11_Applications_of_Calculus_03.gif

Let the width = w

2w = 200 − 2x
w = 100 − x
b. Write down an expression for the area, A, of the paddock.

Area = length x width

A = x(100 − x)

A = 100x − x2

c. Find the maximum possible area for the paddock.

 

Differentiate A
A' = 100 − 2x

To find maximum or minimum value put A' = 0

0 = 100 − 2x
2x = 100
x = 50

When x = 50

w = 100 − 50
w = 50

So maximum area = x.w = 50 x 50 = 2500 m2

(Because the graph of A = 100x − x2 is an inverted parabola we know that the stationary point at x =50 is a maximum.)

 

Distance, Velocity and Acceleration Problems

The study of problems involving distance, velocity and acceleration is called kinematics.
Solving problems of this type involves the use of differentiation and integration because both velocity and acceleration are rates of change.

The table summarises the terminology used in kinematics.

Term
Symbol
Typical units
Distance (displacement)
s
metres (m)
Time
t
seconds (s)
Velocity (speed)
v
metres/second (m/s)
Acceleration
a
metres/second/second (m/s2)

Velocity is the rate of change of distance with respect to time.

v = s' = Y11_Applications_of_Calculus_04.gif

Acceleration is the rate of change of velocity with respect to time

a = v ' = Y11_Applications_of_Calculus_05.gif

Because of these relationships, integrating a function for acceleration will give a velocity function and integrating a velocity function will give the distance or displacement.

To summarise:

Y11_Applications_of_Calculus_06.gif

When solving problems it is important to be aware of the following:

Distance
s = 0 Object is at the starting point.
s is positive Object is above or to right of starting point.
s is negative Object is below or to left of starting point.
Velocity
v = 0 Object is stationary.
v is positive Object is moving to right or upwards.
v is negative Object is moving to left or downwards.
Acceleration
a = 0 Object is moving at constant speed or velocity.
a is positive Object is speeding up.
a is negative Object is slowing down.

 

Example 1
Solution
A particle is s metres from a fixed point given by the formula s = t2 − 4t − 5 
where t is the time in seconds.

 

  
a. Find the velocity after 5 seconds.

s = t2 − 4t − 5

Differentiate, s' = v = 2t − 4

After 5 seconds, t = 5

v = 2 x 5 − 4 = 6 m/s
b. Find the acceleration after 5 seconds.

a = v ' = 2

The acceleration after 5 seconds is 2 m/s2
Example 2
Solution

An object is moving in a straight line with a velocity given by the formula v = 3t2 − 2t .
The object is 3 m from a fixed point O on the line, after 5 seconds.
How far from O will the object be after 10 seconds?

v = 3t2 − 2t

By integrating, s = t3 − t2 + c

To find c, substitute s = 3 when t = 5

3 = 53 − 52 + c
c = -97

Thus s = t3 − t2 -97

When t = 10

s = 103 − 102 − 97
s = 803

The object will be 803 m from O after 10 seconds.