Mathematical FoundationsTutorial problems

Juan Carlos Ponce Campuzano

j.ponce@uq.edu.au

October 15, 2015

1

2

Contents

1 Week 4 5

2 Week 5 10

3 Week 7 15

4 Week 8 17

5 Week 9 22

6 Week 11 28

3

Important:

This notes contain problems and exercises from the tutorial sessions of the courseMathematical Foundations and are constantly updated. If you find a typo or afactual error, by all means let me know: j.ponce@uq.edu.au

4

1 Week 4

Problem 1.1. Find the domain and range of the following functions:

a) f (z) =6z 3 b) f (y) =

127+ |y|

Solution.

Part a). For the domain of f (z) = 6z3 we need to keep in mind that:

First, the denominator of a fraction cannot be zero. Thus z 3 6= 0. Thismeans that

z 6= 3, and so z 6= 9.

Second, we can only take the square root of positive numbers or zero. That is,z 0.

Therefore, if z is in the domain of the function, it must satisfy that z 6= 9 and 0 z.That is, the domain of f (z) is

D( f ) = [0, 9) (9,).

On the other hand, for the range, we need to consider z [0, 9) (9,). So we havetwo cases.

Case 1. If z [0, 9), then0 z < 9

0 z < 33 z 3 < 3 33 z 3 < 0

1z 3

13 < 0

6z 3

63 > 0

f (z) =6z 3 2 > 0.

Therefore, for z [0, 9), we have that f (z) [2,).In this step we used the property: If a, b R such that a b < 0, then 1

b 1

a< 0.

5

Case 2. If z (9,), thenz > 9z > 3

z 3 > 3 3z 3 > 01z 3 > 0

f (z) =6z 3 < 0.

Hence, for z (9, 0), we have that f (z) (, 0).

Consequently, we conclude that the range of the function f (z) is

R( f ) = (, 0) [2,).

Figure 1: Domain (green) and range (blue) representation of f (z) = 6z3 . To use this applet click Here.

Part b). In order to find the domain of f (y) =12

7+ |y| , we need to keep in mind that:

The denominator of a fraction cannot be zero. Thus 7 + |y| 6= 0, or |y| 6= 7.This means that y can be any real number.

6

Therefore, the domain of f (y) is D( f ) = (,) = R.

On the other hand, in order to find the range we need to consider y (,). Thismeans that

0 |y|0 < 7 7+ |y|

0 4

3x > 1x > 1

3.

Case 2. If 5+ 3x < 0, then

(5+ 3x) > 45 3x > 4

9 > 3x3 = 9

3> x.

Therefore, |5+ 3x| > 4 provided that x > 1/3 or x < 3. In other words,

x (,3) (1

3,)

.

9

2 Week 5

Problem 2.1. Solve the inequalities:

a)x2 5xx2 9 0. b) |x 3| < |x+ 4| 3.

Solution.

Part a) Notice that

x2 5xx2 9 0

x(x 5)(x 3)(x+ 3) 0.

This means that we need to analyse two cases.

Case 1. Consider x(x 5) 0 and (x 3)(x+ 3) > 0. Notice that each part splitsinto two cases. That is

x(x 5) 0Case i Case ii

x 0 and x 5 0 or x 0 and x 5 0x 0 and x 5 x 0 and x 5

Combining both cases we have

x 0 or x 5.

On the other hand

(x 3)(x+ 3) > 0Case i Case ii

x 3 > 0 and x+ 3 > 0 or x 3 < 0 and x+ 3 < 0x > 3 and x > 3 x < 3 and x < 3

Which means thatx > 3 or x < 3.

10

Thus

x 0 or x 5 and x > 3 or x < 3

Therefore, we get the following

x < 3 or x 5.In other words

x (,3) [5,).

Figure 5: Representation of the condition x < 3 or x 5 in the real line.

Case 2. Now consider x(x 5) 0 and (x 3)(x+ 3) < 0. Again, notice that eachpart splits into two cases. That is

x(x 5) 0Case i Case ii

x 0 and x 5 0 or x 0 and x 5 0x 0 and x 5 x 0 and x 5

The Case i must be rejected (Why?). So from Case ii we have that

0 x 5.

On the other hand

(x 3)(x+ 3) < 0Case i Case ii

x 3 > 0 and x+ 3 < 0 or x 3 < 0 and x+ 3 > 0x > 3 and x < 3 x < 3 and x > 3

11

The Case i must be rejected (Why?). Hence, from Case ii, we have that

3 < x < 3.

Thus

0 x 5 and 3 < x < 3

Therefore, we obtain0 x < 3.

In other wordsx [0, 3).

Figure 6: Representation of the condition 0 x 5 and 3 < x < 3 in the real line.

Finally, from Case 1 and Case 2, we conclude that

x < 3 or 0 x < 3 or x 5.In other words

x (,3) [0, 3) [5,).

12

Part b) One way to solve the inequality |x 3| < |x + 4| 3 is considering thefollowing results:

Theorem 2.1. If x, y R such that |x| < y, then y < x < y.Theorem 2.2. If x, y R such that |x| > y, then x > y or x < y.

By Theorem 2.1, we have to analyse two cases.

Case 1. Consider (|x+ 4| 3) < x 3. Notice that|x+ 4| < x 6 |x+ 4| > 6 x.

Now, by Theorem 2.2, we have to consider two more cases. That is

|x+ 4| > 6 xCase i Case ii

x+ 4 > 6 x x+ 4 < (6 x)2x > 2 or x+ 4 < 6+ xx > 1 4 < 6

The Case ii must be rejected (Why?). Hence, from Case i, we have x > 1, orx (1,).

Case 2. Consider now x 3 < |x+ 4| 3. Observe thatx 3 < |x+ 4| 3 x < |x+ 4|.

Now, by Theorem 2.2, we have to consider two more cases. That is

|x+ 4| > xCase i Case ii

x+ 4 > x x+ 4 < x4 > 0 or 2x < 2

x < 213

Observe that, from Case i, x can take any value, that is, x (,). On theother hand, from Case ii, we get x < 2, or x (,2).Hence

x (,) or x (,2),which is equivalent to

x (,) (,2) = (,).

Finally, combining the results obtained in Case 1 and Case 2, we have that

x (1,) and x (,).In other words

x (1,) (,) = (1,).

14

3 Week 7

Problem 3.1. Find y for the following functions:

a) y =2xx+ 1

b) y = sin3 x ln (x)Solution. a) First, let us change the radical for a fractional exponent. That is

y =2xx+ 1

=2x

(x+ 1)1/2.

Then

y =(x+ 1)1/2 2 2x

(1

2(x+1)1/2

)((x+ 1)1/2)2

=2(x+ 1)1/2 x

(x+1)1/2

x+ 1

=

2(x+1)1/2(x+1)1/2x(x+1)

12

x+ 1

=

2(x+1)x(x+1)1/2

x+ 1

=

x+2(x+1)1/2

x+ 1

=x+ 2

(x+ 1)3/2

Or if you prefer

y =x+ 2(x+ 1

)3 .Now for part b) we have

y = sin3 x (

1x 1

2x

)+ ln

(x) 3 sin2 x cos x

=sin3 x

2x+ 3 ln

(x) sin2 x cos x

= sin2 x(

sin x2x

+ 3 cos x ln (x)) .

15

Problem 3.2. Find the derivative of

y3 + 3y = 6x2 2.Then find the equation of the tangent line to the graph of

y3 + 3y = 6x2 2,at the point A = (0, 2).

Solution. First, let us find the derivative. In this case, we need to use implicit differ-entiation. Thus we have

ddx(y3 + 3y) = d

dx(6x2 2)

3y2dydx

+ 3dydx

= 12x

dydx(3 3y2) = 12x

dydx

=12x

3 3y2 =4x

1 y2 .

Evaluating in the point A = (0, 2) we have

dydx

A=(0,2)

=4(0)

1 (2)2 = 0.

This means that the slope of the tangent to the curve at the point A = (0, 2) is 0.Hence, the equation of the tangent has the form

y = b, or f (x) = b

with b a real number.

Because the point A = (0, 2) belongs to both the curve and the tangent, we find thatthe value of b must be 2. In other words, the equation of the tangent to the curve is

y = 2.

Exercise 3.1. Find the equation of the tangent to the curve

x3 15x2 + 48x = 27y2 64.at the point A = (11,2).

Note: The curve defined in the previous expression is known as the Tschirnhausen cubic.

Do you want to check your answer? Click Here.

16

4 Week 8

Guidelines to solve optimisation problems:

Read the problem carefully. If appropriate, draw a sketch or diagram of the problem to be solved. Define variables to be used. You can used these variables to label your diagram. Write down all the equations which are related to the problem. Clearly denote

the equation which you are asked to maximise or minimise.

Remember, the equation to maximise or minimise must be in terms of just onevariable.

Finally, find the maxima or minima.

Problem 4.1. A company produces juice containers which have the shape of a boxwith squares on the base and the top, and a capacity of 2 litres.

Figure 7: Box with a constant capacity of 2 litres.

a) Find a formula for the surface area, A, of the container in terms of the apothema of the base. The apothem is the segment from the centre of the square to themidpoint of one of its sides.

b) Draw a rough sketch of A versus a.

c) Find the apothem of the container which will minimise the cost of the materialrequired to make the box.

d) Find the corresponding height and side of the square.

17

Solution. If a represents the apothem and h the height of the box, then we have thefollowing:

Surface area: A = 4 (2a) h+ 2(2a)2Volume of box: 2 = (2a)2 h

From the Volume of box, we obtain: h =2

4a2=

12a2

.

Thus, substituting the value of h in the equation of the Surface area, we obtain:

A = 4(2a)(

12a2

)+ 2(2a)2 =

4a+ 8a2.

This is the Surface area in terms of the apothem a. Notice that the function A(a)is defined for all a 6= 0, which means that a > 0 or a < 0. However, we are justinterested in the positive values of a, see Figure 8. To play with the dynamic repre-sentation of the box click Here.

Figure 8: Graph of A(a) = 4a + 8a2 on the interval (0, 2).

Now let us find the minimum of the function A(a). In this case we have that:

A(a) = 16a 4a2

18

If A(a) = 0, then

16a 4a2

= 0

16a =4a2

16a3 = 4

a3 =416

=14=

122

a = 3

122

=1

3

22 0.6299

Remark: Notice that we are using the laws of exponents:

nam = am/n, an am = an+m, an/am = anm and (an)m = anm.

The second derivative of A(a) is:

A(a) = 16+8a3

.

Substituting the value a =1

3

22, we obtain

A(

13

22

)= 16+

8(1

3

22

)3 = 16+ 8122

= 16+ 32 = 48

which is a positive value. Hence, the function of the surface area has a minimum

when a =1

3

22.

If S denotes the side of the square, then

S = 2a = 2(

13

22

)=

3

2 1.2599 . . .

Finally, the height of the box is:

h =1

2a2=

1

2(

1322)2 = 32 1.2599 . . .19

To check these values, let us calculate the volume of the box which is

Volume of box = S2 h,so we get

Volume of box =(

3

2)2 ( 32) = 22/3 21/3 = 22/3+1/3 = 2.

Exercise 4.1. Find two nonnegative numbers whose sum is 9 and so that the productof one number and the square of the other number is a maximum.

Answer: Here

Exercise 4.2. A container in the shape of a right cylinder with no top has surface area3pi m2. What height h and base radius r will maximise the volume of the cylinder?

To play with a dynamic representation of the cylinder click: Here

Answer: Here

20

Problem 4.2. Finddydx

, for 2x2 + 3y2 4y = 2.

Solution. Here we use implicit differentiation. Thus we have

ddx(2x2 + 3y2 4y) = d

dx(2)

ddx(2x2)+

ddx(3y2) d

dx(4y) = 0

4x+ 6ydydx 4dy

dx= 0

dydx

(6y 4) = 4xdydx

=4x

6y 4dydx

=2x

2 3y .

Remark: Remember that implicit differentiation is nothing more than aspecial case of the well-known chain rule for derivatives.

For example, to differentiate the expression 2x2, with respect of x, is simply

ddx(2x2)= 4x

But to differentiate with respect of x, an expression like 3x2y3, you need toconsider that the variable y has an implicit variable, which in this case is x.That is

ddx(3x2y3

)= 3x2

(3y2

dydx

)+ y3(6x) = 3xy2

(3x

dydx

+ 2y)

.

Notice also that we used the product rule.

Exercise 4.3. Finddydx

, for 3xy 5x2 + 3y2 = 0.

Answer: Here

For more problems about implicit differentiation, click Here

21

5 Week 9

Problem 5.1. Determine the following integrals, if possible. If not, state why.

a) 11

x3 dx.

b) 12

x 1x2 1 dx.

Solution.

a) The function f (x) = x3 is well-defined and continuous over the interval [1, 1].Hence, we can apply the Fundamental Theorem of Calculus (FTC) with F(x) = x4/4as an antiderivative of f (x). Thus, we have

11

x3 dx =x4

4

11

=(1)4

4 (1)

4

4

=14 1

4= 0.

Figure 9: Geometrical representation of the integral 11 x

3 dx.

22

b) The function

f (x) =x 1x2 1

is not defined for x = 1 or x = 1. Notice also that 1, 1 [2, 1]. Because f (x) isnot continuous on the interval [2, 1], we can not apply the FTC.

Notice that this function behaves similarly to the function 1/x, because

x 1x2 1 =

x 1(x 1)(x+ 1) =

1x+ 1

Hence, we can not calculate the definite integral: 12

x 1x2 1 dx

Figure 10: Graph of f (x) =x 1x2 1 .

Problem 5.2. Calculate the following indefinite integrals

a) (

2x3 2x) (x4 2x2)3 dx.b) 6e2x 9e3x

e2x e3x + 7 dx.

c)

x cos x dx.

23

Solution. a) Let u = x4 2x2. Thendudx

= 4x3 4x = du2= (2x3 2x) dx

Thus

(2x3 2x) (x4 2x2)3 dx = ( x4 2x2

This is u

)3 This is du/2 (

2x3 2x) dx=

u3

du2

=12

u3 du

=12 u

4

4+ C =

18u4 + C

=18(x4 2x2)4 + C.

b) If u = e2x e3x + 7, thendudx

= 2e2x 3e3x = du = (2e2x 3e3x)dx.

Multiplying by 3 we have3du = (6e2x 9e3x)dx.

Thus, 6e2x 9e3xe2x e3x + 7 dx =

3duu

= 3 1

udu = 3 ln |u|+ C = 3 ln e2x e3x + 7+ C.

c) For this part, we need to use the method of integration by parts. Recall the formulau dv = uv

v du+ C.

We write

u = x, dv = cos x dx,

du = dx, v =

cos x dx = sin x.

Hence x cos x dx = x sin x

sin x dx+ C = x sin x+ cos x+ C.

24

Problem 5.3. Use integration by parts to deduce the formulasin2 x dx = sin x cos x+

cos2 x dx.

Use the identities

sin2 + cos2 = 1 and sin(+ ) = sin cos + cos sin

to deduce the formula sin2 x dx =

12x 1

4sin 2x.

Finally, use integration by parts and the previous results to prove thatsinn x dx = 1

ncos x sinn1 x+

n 1n

sinn2 x dx.

Solution.

First, we write

u = sin x, dv = sin x dx,

du = cos x dx, v =

sin x dx = cos x.

Thussin2 x dx = sin x cos x

( cos2 x) dx = sin x cos x+ cos2 x dx.Since cos2 x = 1 sin2 x, we have

sin2 x dx = sin x cos x+(1 sin2 x) dx = sin x cos x+

dx

sin2 x dx.

and then we obtainsin2 x dx = sin x cos x+ x

sin2 x dx.

Notice that we have the same integral

sin2 x dx in both sides with oposite sign.Hence

sin2 x dx+

sin2 x dx = sin x cos x+ x

2

sin2 x dx = x sin x cos xsin2 x dx =

12x 1

2sin x cos x.

25

Since sin x cos x = 12 sin 2x, we havesin2 x dx =

12x 1

2

(12

sin 2x)=

12x 1

4sin 2x

which is the required result. To be more precise we writesin2 x dx =

12x 1

4sin 2x+ C.

Now let us prove the general formula. First we write

u = sinn1 x, dv = sin x dx,

du = (n 1) sinn2 x cos x dx, v = sin x dx = cos x.Thus, using the formula of integration by parts, we have

sinn x dx = cos x sinn1 x+ (n 1)

sinn2 x cos2 dx.

But cos2 x = 1 sin2 x, thensinn x dx = cos x sinn1 x+ (n 1)

sinn2 x(1 sin2 x)dx

= cos x sinn1 x+ (n 1)

sinn2 x dx (n 1)

sin2 x dx.

By grouping similar terms we obtainsinn x dx+ (n 1)

sinn x dx = cos x sinn1 x+ (n 1)

sinn2 x dx

n

sinn x dx = cos x sinn1 x+ (n 1)

sinn2 x dxsinn x dx = 1

ncos x sinn1 x+

n 1n

sinn2 x dx.

This proves the general formula. Finally, to be more precise we writesinn x dx = 1

ncos x sinn1 x+

n 1n

sinn2 x dx+ C.

26

Exercise 5.1. Determine the following integrals, if possible. If not, state why.

a) 15

(4x3+ 2x4 2x2 1

)dx Ans. a) Here

b) pi/2pi/2

sin (2x)cos (2x)

dx Ans. b) Here

c) 2xex2

ex2dx Ans. c) Here

d) 7x+ 8

x2 + x 2 dx Ans. d) Here

e)

sin5 x dx Ans. e) Here

27

6 Week 11

Problem 6.1. A river flows due West at a speed of 2.5 metres per second and has aconstant width of 1 km. You want to cross the river from point A (South) to a pointB (North) directly opposite with a motor boat that can manage to a speed of 5 metresper second.

a) If you head out pointing your boat at an angle of 90 degrees to the bank. Howlong does it take to get from point A to point B?

b) After crossing the river you realised that it took longer than expected. In whatdirection should you point you motor boat in order to reduce the time to crossthe river? How long will it take you to get from point A to point B? Is it a bettertime?

Remark: As you already noticed, if you head out at 90, the boat doesnot reach point A directly. Instead, it reaches a point C, and then it needsto cover a second distance from C to B (see Figure 11). This problem isan example of relative velocity, which means that we have to consider aparticular frame of reference. For this case we have that

Vb represents the velocity of the boat with respect to the water,

Vc represents the velocity of the water with respect to the Earth (cur-rent), and

VR = Vb+Vc represents the resultant velocity of the boat with respectto the Earth.

The magnitude of VR represents the relative speed of the boat with respectto the Earth.

For more information about relative motion click Here.

If you want to play with the applet shown in the tutorial click Here.

28

Solution. With the information given in the problem, it is easy to calculate the com-ponent form of the resultant velocity VR, which is

VR = 2.5 i+ 5 j with magnitude ||VR|| = 5

52

and direction R = 116.57.

Notice also that the resultant displacement DR has the same direction as VR (seeFigure 11).

Figure 11: Geometrical representation of velocity and displacement vectors. The angle R represents thedirection of VR and DR.

Considering the triangle ABC in Figure 11, observe that

tan (116.57 90) = ||D2||||D1|| .Thus

||D2|| = ||D1|| tan (116.57 90) = 1000 tan(26.57) 500 m.

Hence, the magnitude of DR is

||DR|| =||D1||2 + ||D2||2 =

10002 + 5002 = 500 m.

Now, using the formula of average speed v = d/t, the time it takes to get from pointA to C is

t1 =||DR||||VR|| = 200 s,

and from point C to B is

t2 =||D2||||Vc|| = 200 s.

Therefore, the total time is t1 + t2 = 400 seconds or 6.6 minutes approximately. Thissolves part a).

29

In order to solve part b), we need to consider a new position of the velocity anddisplacement vectors as it is shown in Figure 12. Notice that the vectors VR (resultantvelocity) and DR (resultant displacement) have the same direction 90 up to North.

Figure 12: Left-hand diagram: The angle R represents the direction of VR and DR. Right-hand diagram:The b represents the direction of Vb and D1.

Figure 13: Triangle ABD.

To calculate the direction of the boat (that is, thedirection of the vectors D1 and Vb), consider thetriangle ABD in Figure 13. Thus we have

cos b =2.55

=12

.

Hence b = 60 or pi/3 radians. This is the direc-tion you should point the boat.

Knowing this angle, it is easy to calculate thecomponent form of the resultant velocity VR,which is

VR =5

32

i+ 0 j.

Therefore, the time it takes to get from point A topoint B is

t =||DR||||VR|| =

10005

32

230.94 seconds,

or approximately 3.84 minutes, which is, ofcourse, a better time.

30

Problem 6.2. A 3 kg brick is sitting on a inclined plane at pi/6 radians to the hori-zontal. The only forces acting on it are weight, normal reaction from the plane andfriction. Determine the magnitude of each force.

Figure 14: Representation of forces: An object rests on an inclined plane that makes an angle with thehorizontal.

Solution. First, we have that W = m g, where m = 3 kg and g = 9.8 m/s2. ThusW = 3 (9.8 N) = 29.4 N

ThusF = W sin

(pi6

)= (29.4 N) 1

2= 14.7 N

and

N = W cos(pi

6

)= (29.4 N)

3

2= 14.7

3 N

Exercise 6.1. A 12 kg brick is sitting on a inclined plane at pi/3 radians to the hori-zontal. The only forces acting on it are weight, normal reaction from the plane andfriction. Determine the magnitude of each force. Click here to check your answer

31

Problem 6.3. Consider a painting that hangs on a wall. The painting is in a stateof equilibrium, and thus all the forces acting upon it must be balanced. Thus theleftward pull of cable A must balance the rightward pull of cable B and the sum ofthe upward pull of cable A and cable B must balance the weight of the sign (seeFigure 15). Suppose the tension in both of the cables is 5 N and that the anglethat each cable makes with the horizontal is 30 degrees. What is the weight of thepainting?

Figure 15: Diagram for representing the vertical component of the tension.

Solution. Considering the triangle on the right hand in Figure 15, we have that

sin 30 =Fy

5 N.

Thus Fy = 2.5 N. Since each cable pulls upwards with a force of 2.5 N, the totalupward pull of the painting is 5 N. Therefore, the force of gravity (or weight) is 5 N,down. The painting weighs 5 N.

Exercise 6.2. Consider the painting hanging on a wall as shown in Figure 16. Thepainting has a mass of 500 g and the angle between the two cables is 100 degrees.Determine the tension on the cable. Click here to check your answer

Figure 16: Eschers paiting.

32

33

References

[1] Hirst, K. E. (2006). Calculus of one variable. Springer-Verlag London Limited.

[2] Kline, M. (1972) Mathematical Thought From Ancient to Modern Times, Oxford Uni-versity Press.

[3] Polya, G. (2004). How to solve it? Expanded Princeton Science Library Edition.Princeton University Press. New Jersey (First published 1945).

[4] University of Queensland. (2015). MATH1050: Mathematical foundations (Coursenotes) University of Queensland

34

Week 4Week 5Week 7Week 8Week 9Week 11