CALCULATION OF MATHS

QUESTION

Assignment 2
Due: 4
1

th
May 2012 (by 5 pm)

Marks will be given for working, clarity of explanation and mathematical correctness.
Solutions without working may not receive any marks.

It is very strongly recommended that you start this assignment early.

NUMBER OF QUESTIONS: 12

You should leave all answers in terms of   and  . That is, do not give your answers as a
decimal value unless the question specifically asks for a decimal value.

QUESTION 1

Find:   (You DO NOT have to simplify the solution)
(a)

(b)

(

)
(c)

(

(d)

(

(   )
)

)

QUESTION 2
For each of the following functions:
i.
(

)

ii.
(

)

2

(a) Determine the equations of the asymptotes, giving reasons for your answers.
(b) Find   ( ) and use it to sketch a sign diagram and determine the nature of any
stationary points.
(c) Find the axes intercepts. i.e. where it cuts the   and   axes.
(d) On graph paper, accurately plot the graph of  ( ), showing all asymptotes and
axes intercepts (that is, show all the information found in a, b, and c above).

QUESTION 3
A particle P moves in a straight line. The displacement of the particle from its starting point
(origin) is given by:

(

)

(

)
metres (m)

where   is the time in seconds (s),    .

(a) Find the velocity  ( ) and acceleration  ( ) functions.
(b) Find the initial position, initial velocity and initial acceleration of P. Give your answers
to 2 decimal places and specify the units for position, velocity and acceleration.
(c) Discuss the acceleration as    .
(d) Sketch the graph of the acceleration function. Show all important points.
(e) Find the mathematical relationship between the displacement and the acceleration of P.
(Hint: Can acceleration be represented in terms of displacement or vice versa?)
QUESTION 4
An object ‘O’ oscillates (that is, moves back and forth) in a straight line along the horizontal
axis. The position of O is given by:

(

)
cm

where   is the time in seconds (s), 0     .

Leave all answers in terms of   unless otherwise specified in the question.

(a) Find the velocity  ( ) and acceleration  ( ) functions.
(b) Find the initial position, initial velocity and initial acceleration of P. Give your answers
to 2 decimal places and specify units for position, velocity and acceleration.
(c) On graph paper (if not using a computer), accurately plot the graphs for  ( ),  ( ) and
( ). Show the time on the horizontal axis in terms of  . (plots must be accurate and
correctly scaled). For this part (c), the use of a computer to plot the graph is permitted.
It is up to you whether you use graph paper or plot using a computer. For either
method, all working must be shown.
(d) Find the amplitude of oscillation.
(e) Find the period of oscillation.
(f) Find the points in time (during the time given above), when the magnitude of the velocity
is a maximum.
(g) Find the points in time when the acceleration is a maximum.
(h) Find the position at which O has the maximum velocity. In 20 words or less, describe the
relationship between the position of the particle and its maximum velocity.

3

QUESTION 5
(a) Find the EXACT area bound by the two functions
(

)
and  ( ).
4

(

)

(

)

(b) Write the definite integral for the area bound by the   axis and the function  ( ),
Calculate this area to 2 decimal places.

QUESTION 6
Consider two quadratic functions,  ( ) and  ( ) with the properties given in the table below.
Property
( )  ( )
Roots (-2,0) and (5,0) (-6,0) and (2,0)
intercept
(0,-10) (0,12)

(a) Determine the equations of the functions
(

)
and  ( ).
(b) On graph paper, accurately plot the two functions on the SAME graph, showing all
intercepts and the turning points.
(c) Find the area bound by the function  ( ) and the   axis.
(d) Find the area bound by the function  ( ) and the   axis.
(e) Find the area bound by the two functions  ( ) and  ( ) to 3 decimal places.

QUESTION 7
A manufacturer of open top cylindrical tin cans that hold 0.3 litre (L) of water has asked you
to minimise the cost of metal used by designing a can with dimensions that will make the
total surface area A must be as small as possible. Find the values of r and h which makes A as
small as possible.

QUESTION 8
If
(

)
(  ), show that

(

)

(

)
.
QUESTION 9
(a) Find the equation of the normal to the curve        at the point where    .
(b) For the equation       ,       , find the co-ordinates (   ) where the
gradient of the tangent to the curve is zero. Leave your answers in terms of  . (Hint:
there are two sets of co-ordinates)

h cm
5

r cm
Top is open. No
metal used here
Base is closed.
Metal used.
For questions 10, 11 and 12, you will need to refer extensively to your text book on the topics
of stationary points, and inflections and shape (Chapter 18). The aim of this question is to
covered the majority of the topic in class. You now need to apply your knowledge of
calculus, functions and sign diagrams to explain and answer this question.

If you choose, you may do this question in collaboration with another student in your class. If
you do choose to collaborate, then please put the name of your partner next to the question on
your solution paper. Maximum group size is two students. All group members must submit a
copy of the solution with their assignment.

QUESTION 10
Explain the following:
(a) Point of inflection.
(b) Stationary inflection.
(c) Non-stationary inflection.
(d) How do you determine that a function  ( ) has a point of inflection at a particular
point    ? Explain using calculus and sign diagrams.
(e) Explain why the function
(

)

6

has no points of inflection.

QUESTION 11

For the function
(

)
above, complete the table below by identifying the property of the
points A, B and C by identifying whether they are + (positive), or – (negative) or zero. Two
cells have already been filled out.

Function A B C
( )   +
( )   0
( )

Sketch the likely shape for   ( ) and    ( ).

7

QUESTION 12

For the derivative function   ( ) shown in the graph, sketch the likely shape for  ( ) and
( ). (Note: You are not required to find a polynomial function from the curve as it is not
possible to do it accurately enough)

8

SOLUTION

1. (a)

(b)

(c)

(d)

1. (a)

(i)

The equation of horizontal asymptote is

The equation of vertical asymptote is

Therefore, the equation of vertical asymptote is

(ii)

The equation of horizontal asymptote is

The equation of vertical asymptote is

Therefore, the equation of vertical asymptote is

(b) (i)

Now, for all x,

Therefore, the sign diagram will be

Therefore, for all x,

(ii)

Now, for all x,

Therefore, the sign diagram wiil be

Therefore, for all x,

Therefore, there are no stationery points.

(c) (i) for x-intercept,

Therefore, x-intercept is -2/3.

for y-intercept,

Therefore, y-intercept is -1.

(ii) for x-intercept,

Therefore, there is no x-intercept.

for y-intercept,

Therefore, y-intercept is 1.

(d) the graph of f(x) is shown below.

1. Given,

(a)  Then, the velocity,

The acceleration,

(b) The initial position is at

Therefore, the initial position is

The initial velocity is

The initial acceleration is

(c)  As

Therefore, the acceleration will approach zero.

(d) We construct a table for acceleration.

 t a(t) 0 3.68 2 1.35 4 0.5 6 0.18

We can sketch the graph of acceleration using these values.

(e)  We have,

And,

Therefore,

1. Given,

Where,

(a)  The velocity is

The acceleration is,

(b) The initial position is at time t=0

The initial velocity is

The initial acceleration is

(c)  We construct table of values.

 t s(t) v(t) a(t) 0 6.3 0 -4 π/6 5.8 -1.73 -2 π/4 5.3 -2 0 π/3 4.8 -1.73 2 π/2 4.3 0 4

We sketch the graph of s(t), v(t), and a(t) as shown below.

(d) We compare

with

We get

Therefore, the amplitude of oscillation

(e)  The period of oscillation is

(f)  When the velocity is maximum, then

Now,

At

Therefore, the velocity is minimum.

At

Therefore, the velocity is maximum at

(g) When the acceleration is maximum, then

Now,

At

Therefore, the acceleration is maximum.

At

Therefore, the acceleration is maximum at

(h) The velocity is maximum at

Therefore, the position will be

Therefore, when the velocity is maximum, the displacement is zero, therefore, the particle will oscillate about its equilibrium position.

1. Given,

We find the points where the two curves intersect.

Either,

Or,

Therefore, the sketch of the region is shown below.

Therefore, the exact area bounded by the two functions

(b) The curve f(x) intersects the x-axis, where,

Therefore, the area bound by the x-axis and the function f(x)

1. (a) Since the roots of f(x) are

The equation of f(x) is

Since the roots of g(x) are

The equation of f(x) is

Now,

Therefore,

For

For

At

Therefore, the turning point is

Therefore,

For

For

Therefore, the turning point is

We can sketch of the curves as shown below.

(c) The area bound by f(x) and the x-axis is,

(f)  The area bound by g(x) and the x-axis is,

(g) The curves f(x) and g(x) intersect when,

Therefore, The area bounded by two functions f(x) and g(x)

1. Given, the capacity of can

We suppose that volume of the can be V.

Therefore,

Now, surface area when can is open at the top, is

Therefore,

For maxima or minima,

Now,

Therefore, when

Then,

Therefore, A is minimum when,

Therefore,

Therefore,  For A to be as small as possible,

8. given,

Therefore,

9. Given,

Therefore,

At the point where,

We have,

Therefore, the slope of the normal is

At

We have,

Therefore, the point is

The equation of normal is

(b) The gradient of tangent to the curve

Is

If the gradient is zero, then

At

At

Therefore, the points are

10. (a) A point of inflection occurs at a point where,

And where there is a change in concavity of the curve at that point.

(b) A stationery point of inflection is a point where,

But,

(c) A non-stationery point of inflection is a point where,

But,

(d) To find the point of inflection, we find

Then, if

And,

Then it is a point of inflection. But, if

And,

Then it is not a point of inflection.

From sign diagram, we can say that the point of inflection is a point where,

Changes sign.

(d) Given,

Therefore,

Therefore,

Therefore, at

Therefore, at

Therefore, at

Therefore, since,

Therefore, x=0 is not a point of inflection.

11.

Since the graph of the function crosses negative y-axis at A, therefore, at A,

At B, the graph is in fourth quadrant, therefore, at B,

Since, A is a stationery point and the graph is minimum, therefore, at A,

Since, B is a non-stationery point of inflection and the graph is increasing, therefore,

Since C is a stationery point of inflection, therefore,

Therefore,

Therefore, the graph of

can be drawn as shown below.

12. From the graph, we can see that the graph has either local maximum or local minimum at

For,

So. f(x) is decreasing.

For,

Therefore, f(x) is decreasing.

For,

Therefore, f(x) is decreasing.

For,

Therefore, the function is increasing.

Therefore, x=3 is a point of local minimum.

Therefore, x=0 is a saddle point.

Therefore, graph of f(x) will look like as shown below.

 3

For drawing the graph of

We consider

As a function

And we have to draw the graph of

We see that

Has a local maximum at

And a local minimum at

For,

Slope of g(x) is positive.

Therefore,

For

The slope is negative. Therefore,

For,

The slope is positive. Therefore,

Therefore, the graph of

Will look like as shown below.

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