Queen's The University of Belfast FACULTY OF ENGINEERING AND PHYSICAL SCIENCES 105FOE101 Mathematics 1 and 105CIV216 Additional Mathematics 2 Professor W. Sha ([e-mail]), David Keir Building Room 1002 Overview This course seeks to cover engineering mathematics topics that are deemed to be essential to a professional civil engineer. This is an objective that is shared between this half-module and a second half-module taken in the second semester, however, with the emphasis in this half-module being put on basic algebra, functions and calculus. Learning Outcomes Successful completion of this module will lead to the following learning outcomes: Knowledge and Understanding of Polynomials Other mathematic functions Complex numbers Non-linear equation Differentiation and its applications Integration and its applications Intellectual abilities You will be able to: Manipulate, factor and solve functions and equations containing elementary mathematical functions. Use complex numbers to extend mathematically models. Solve systems of non-linear equations. Differentiate and integrate a range of functions. Apply differentiation and integration to the mathematical modelling of engineering applications. Practical skills You will be able to: Use mathematics to model engineering problems. Think logically. Transferable skills The course will enhance the following skills: Solve engineering problems through mathematical analysis. Solve non-routine problems. Learn independently. Syllabus Polynomials, other mathematic functions, complex numbers, non-linear equation, differentiation and its applications, integration and its applications Background Reading Text Modern Engineering Mathematics, Glyn James, Addison Wesley Engineering Mathematics, K.A. Stroud, Palgrave Mathematics for Engineers, A. Croft, R. Davidson, Pearson/Prentice Hall A Guide to Microsoft Excel, Bernard V. Liengme, Arnold Foundation Maths, Anthony Croft and Robert Davison, Prentice Hall Higher Engineering Mathematics, John Bird, Newnes Provisional lecture timetable 1
Lecture 1 2 3 4-5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20-21 22
Topic Polynomials:- The basic shape, factorisation of polynomials in linear and irreducible quadratic factors. Multiplication and division with polynomials, completing the square and partial fractions. The laws of indices, the function y = a x and y = e x , the log function and its rules. Trigonometric functions:- Definition, formulae, inverse functions. A sin ( t ) form. Hyperbolic functions:- Definition, formulae, inverse functions. Non-linear functions:- Graphical solution, bisection method, Newton's method. Complex numbers:- Definition of j, plotting on an Argand Diagram, modulus and argument. Complex numbers:- Addition, multiplication and division. Graphical approach to differentiation, the definition of a derivative, the derivatives of simple functions. Product rule and chain rule, implicit differentiation, logarithmic differentiation. Derivatives in mathematical modelling, approximation, Taylor's series. Optimisation. Minimisation. Integration as the anti-derivative, table of anti-derivatives. Definite integrals. Improper integrals. Techniques of integration:- By change of variable. Integration by parts. Integration by partial fractions. Numerical integration:- Trapezium rule, Simpson's rule. Finding areas. Finding moments. Finding second moments of areas. Parallel axis theorem.
Assessment This will be based on a written examination (2 hours) worth 80% and two class tests worth 10% each. Tutorials Tutorials are held once a week. Maths Help sessions at 2:30 on Wednesdays, Ashby 4.5.
2
Polynomials
f(x)=x^3-3x^2-x+9 Increment x -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 -2.8E-16 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4 f(x) -9 -4.752 -1.176 1.776 4.152 6 7.368 8.304 8.856 9.072 9 8.688 8.184 7.536 6.792 6 5.208 4.464 3.816 3.312 3 2.928 3.144 3.696 4.632 6 7.848 10.224 13.176 16.752 21 25.968 31.704 0.2 Start -2
f(x) =x^3 - 3x^2 - x +9
35 30 25 20 15 f(x) 10 5 0 -4 -2 -5 0 -10 -15 x 2 4 6
Factorisation A factor of a polynomial is an exact divisor of that polynomial. Thus if we have a linear factor (x a) then Polynomial function = (x - a)*other factors If (x - a) is a factor of a polynomial then putting x = a makes the polynomial have a value of zero. Putting x = a will solve the equation f(x) and is therefore a root. Rational Functions Rational functions have the general form p(x) q(x) where p(x) and q(x) are polynomials. If the degree of p is less than the degree of q, f(x) is said to be a strictly proper rational function. If p = q the f(x) is a proper rational function and if p > q then f(x) is an improper rational. An improper or proper rational function can always be expressed as a polynomial plus a strictly proper rational function. f (x) = 3
Partial Fractions Any strictly proper rational function can be expressed as a sum of simple functions whose denominators are linear or irreducible quadratic functions. x2 +1 1 1 4x + 7 = + ( 1 + x ) ( 1 x ) ( 2 + 2x + x 2 ) 1 + x 5 ( 1 x ) 5 ( 2 + 2x + x 2 ) These simpler functions are called the partial fractions. The method of forming these partial fractions is as follows for a function p(x) f ( x) = q(x) 1. If the degree of p is greater than or equal to the degree of q, divide q into p to obtain s(x) f ( x ) = r(x) + q(x) where the degree of s is less than the degree of q. 2. Factorise q(x) fully into real linear and irreducible quadratic factors, collecting together all like factors. Each linear factor ax + b in q(x) will give rise to a fraction of the type Each repeated linear factor ( ax + b )
n
3.
A ; ax + b will give rise to n fractions of the type
A1 A2 A3 An + + + ... 2 3 n; ax + b ( ax + b ) ( ax + b ) ( ax + b ) Each irreducible quadratic factor ax 2 + bx + c in q(x) will give rise to a fraction of the type Ax + B ; ax 2 + bx + c
n
Each irreducible quadratic factor ( ax 2 + bx + c ) in q(x) will give rise to n fractions of the type A1x + B1 A 2 x + B2 A n x + Bn + + ... 2 n 2 ax + bx + c ( ax 2 + bx + c ) ( ax 2 + bx + c ) .
4. 5.
Multiply both sides of the equations by q(x) to obtain an identity involving polynomials from which the multiplying constants may be found. To find these coefficients, two strategies are used. Strategy 1 Choose special values of x that make finding the values of the unknowns easy. Strategy 2 Compare the coefficients of like powers of x on both sides of the identity, starting with the highest and lowest powers usually makes it easier. Check the answer either by choosing a test value for x or by putting the partial fractions over a common denominator.
6.
4
Exponential curves
(1/3)^x (1/2)^x 10 9 8 7 6 5 4 3 2 1 0 0 x 3^x 2^x
f(x)
C
-4
-2
2
4
6
The exponential function e^x
10 e^-x f(x) 4 2 0 -6 -4 -2 0 x Gradient = e^x0 e^x0 x0 2 4 6 8 6 e^x
Natural logarithm ln x
2 0 -2 ln x -4 -6 -8 x 0 1 2 3 4 5
5
Trigonometric Functions sine and cosine
1.5 1 0.5 f(x) -5 0 -0.5 -1 -1.5 x 0 5 10 15 sin x cosx
tan x
15 10 5 tan x -15 -10 -5 0 -5 -10 -15 x 0 5 10 15
6
x = a sin t, y = a cos t
2.5 2 1.5 1 0.5 y -3 -2 -1 0 -0.5 0 -1 -1.5 -2 -2.5 x 1 2 3
Asin(wt+g) 2.5 2 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5
f(t) -2
2
4
6
8
10
12
t
7
Trigonometric Formulae sin = cos 2 cos = sin 2 cosec = 1 sin 1 sec = cos 1 cot = tan
cos 2 + sin 2 = 1 1 + tan 2 = sec 2 cot 2 + 1 = cos ec 2 sin ( A + B ) = sin A cos B + cos A sin B sin ( A B ) = sin A cos B cos A sin B cos ( A + B ) = cos A cos B sin A sin B cos ( A B ) = cos A cos B + sin A sin B tan A + tan B 1 tan A tan B tan A tan B tan ( A B ) = 1 + tan A tan B tan ( A + B ) = 2sin A cos B = sin ( A + B ) + sin ( A B )
2 cos A cos B = cos ( A + B ) + cos ( A B ) 2sin A sin B = cos ( A B ) cos ( A + B ) 2 cos A sin B = sin ( A + B ) sin ( A B )
sin 2 = 2sin cos cos 2 = cos 2 sin 2 = 1 2sin 2 = 2 cos 2 1 2 tan tan 2 = 1 tan 2
8
Inverse Functions sin x versus x
15
10
5
0 -1.5 -1 -0.5 -5 0 0.5 1 1.5
-10
-15
Arc sin x
2 1.5 1 0.5 0 -1.5 -1 -0.5 -0.5 0 -1 -1.5 -2 0.5 1 1.5
9
cos x versus x
15
10
5
0 -1.5 -1 -0.5 -5 0 0.5 1 1.5
-10
-15
Arc cos x
3.5 3 2.5 2 1.5 1 0.5 0 -1.5 -1 -0.5 -0.5 0 0.5 1 1.5
10
tan x versus x
15
10
5
0 -15 -10 -5 -5 0 5 10 15
-10
-15
Arc tan x
2 1.5 1 0.5 0 -15 -10 -5 -0.5 0 -1 -1.5 -2 5 10 15
11
Hyperbolic Functions cosh x and sinh x
15 10 5 f(x) -3 -2 -1 0 -5 -10 -15 x 0 1 2 3 cosh x sinh x
Hyperbolic functions x = cosh t, y = sinh t
2 1.5 1 0.5 y 0 0 -0.5 -1 -1.5 -2 x 0.5 1 1.5 2
12
Non-linear Functions and Solving Non-linear Equations A non-linear function is basically a function containing something more complicated than just constants and the variable by itself. The function f(x) = ax + b is linear and when plotted a straight line is obtained. If a non-linear function is plotted a curve is obtained. The functions x 2, ex, sin x and cos x are all non-linear functions and any function which contains them in any form will likewise be non-linear. An equation is obtained when two functions are put equal to each other. An equation is equivalent to applying some condition. Example
2sin x = x
Solving an equation requires finding a value or values for the variable, which make both sides of the equation equal. Plots of 2sin x and x
4 3 2 1 f(x) 0 -4 -3 -2 -1 -1 -2 -3 -4 x 0 1 2 2 sin x 3 4 x
The roots are given by the intersection of the curves y = sin x and y = x. There are three roots to this problem, note that it is not known how many roots, if any, there are to a non-linear equation.
I 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 x 0.1 0.199667 0.396686 0.772727 1.396181 1.969587 1.843063 1.926328 1.874923 1.908218 1.887223 1.900707 1.892143 1.897622 1.894132 1.896361 1.89494 1.895847 1.895269 1.895638 2*sin x 0.199667 0.396686 0.772727 1.396181 1.969587 1.843063 1.926328 1.874923 1.908218 1.887223 1.900707 1.892143 1.897622 1.894132 1.896361 1.89494 1.895847 1.895269 1.895638 1.895402
The equation can be expressed as f(x) = 2 sin x x = 0. 13
f(x) = 2 sin x - x
3 2 1 f(x) -4 -3 -2 -1 0 -1 -2 -3 x 0 1 2 3 4
The roots are now defined where the curve cuts the x-axis Bisection Method 1. Start with two values of x, x1 and x2 which enclose the root. 2. Let xm = (x1 + x2)/2 3. If f(xm)f(x1) < 0 then set x2 = xm otherwise set x1 = xm 4. Repeat steps 2 and 3 until |f(xm)| < tolerance or |x1 - x2| < tolerance.
Bisection Method I 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x1 1 1 1.5 1.75 1.875 1.875 1.875 1.890625 1.890625 1.894531 1.894531 1.894531 1.89502 1.895264 1.895386 1.895447 1.895477 1.895493 1.895493 1.895493 1.895493 fx=2sin x -x xm 2 1.5 1.75 1.875 1.9375 1.90625 1.890625 1.898438 1.894531 1.896484 1.895508 1.89502 1.895264 1.895386 1.895447 1.895477 1.895493 1.8955 1.895496 1.895494 1.895494 f(xm) -0.18141 0.49499 0.217972 0.033172 -0.07047 -0.01773 0.007954 -0.00483 0.001577 -0.00162 -2.2E-05 0.000777 0.000378 0.000178 7.78E-05 2.78E-05 2.81E-06 -9.7E-06 -3.4E-06 -3.2E-07 1.24E-06 f(x1) 0.682942 0.682942 0.49499 0.217972 0.033172 0.033172 0.033172 0.007954 0.007954 0.001577 0.001577 0.001577 0.000777 0.000378 0.000178 7.78E-05 2.78E-05 2.81E-06 2.81E-06 2.81E-06 2.81E-06 f(xm)f(x1) -0.12389 0.338049 0.107894 0.00723 -0.00234 -0.00059 0.000264 -3.8E-05 1.25E-05 -2.6E-06 -3.5E-08 1.23E-06 2.94E-07 6.71E-08 1.38E-08 2.16E-09 7.8E-11 -2.7E-11 -9.7E-12 -8.9E-13 3.49E-12 abs(f(xm)) abs(x1-x2) 0.181405 2 0.49499 1 0.217972 0.5 0.033172 0.25 0.070471 0.125 0.017728 0.0625 0.007954 0.03125 0.004829 0.015625 0.001577 0.007813 0.001623 0.003906 2.22E-05 0.001953 0.000777 0.000977 0.000378 0.000488 0.000178 0.000244 7.78E-05 0.000122 2.78E-05 6.1E-05 2.81E-06 3.05E-05 9.69E-06 1.53E-05 3.44E-06 7.63E-06 3.18E-07 3.81E-06 1.24E-06 1.91E-06 x2 3 2 2 2 2 1.9375 1.90625 1.90625 1.898438 1.898438 1.896484 1.895508 1.895508 1.895508 1.895508 1.895508 1.895508 1.895508 1.8955 1.895496 1.895494
14
Newton-Raphson Method
Gradient f '(x) f(x) f(x1)
x2
x1 f ( x) x1 x 2 f (x1 ) = f (x1 ) f (x1 ) f (x1 )
Gradient = f (x) =
( x1 x 2 )
x 2 = x1
The iterative procedure for the Newton/Raphson method can be presented as 1. Take initial estimate x1 2. Compute f(x1) and f '(x1) 3. Determine x 2 = x 1 f (x 1 ) f ' (x 1 )
4. Set x1 = x2 and repeat. Example f(x) = 2 sin x x = 0 f '(x) = 2 cos x 1 Newton Raphson Method I x f(x) 0 1.5 0.49499 1 2.076558 -0.32695 2 1.910507 -0.0248 3 1.895622 -0.00021 4 1.895494 -1.5E-08 5 1.895494 0 6 1.895494 0 f'(x) -0.85853 -1.96895 -1.66643 -1.63829 -1.63805 -1.63805 -1.63805 x-f(x)/f'(x) 2.076558 1.910507 1.895622 1.895494 1.895494 1.895494 1.895494
15
The Derivative of a Function and Differentiation The derivative of a function measures the rate of change of the function with respect to the independent variable. dy The derivative of a function y = f(x) is expressed as (dy by dx) where y is the dependent dx variable and x is the independent variable. The word dependent is used as we are usually trying to find the response of y as x varies. Thus the value of y depends on the value of x. Some common derivatives that are of interest to engineers are: Velocity Cooling which is the rate of change of distance with respect to (w.r.t.) time v = The rate at which something cools dx dt
dT dt Strain The strain in a rod is given by the rate of change of the displacement w.r.t. distance du = dx A feature of derivative is that it measures continuous change. So the velocity is not expected to stay constant with time or the strain to remain uniform through the rod. We will start our study of derivatives and the process of finding the derivatives (differentiation) by looking at the gradient of a function. A gradient of a function is a measure of how much the curve is changing and is therefore a derivative.
f(x0+h) f(x0 x0 x0+h
An approximation to the gradient can be obtained by considering the value of a function a small distance h away from x0 f ( x0 + h ) f ( x0 ) The gradient at x0 h This provides us with what is called a forward difference approximation for the derivative of the function at x0. It can be seen that as h is made smaller and smaller the approximation will improve. definition of a derivative can be obtained from this process. df f (x + h) f (x) = f ( x ) = lim h 0 dx h The
16
Table of derivatives of common functions Function constant xn e ax ln x sin ( a x b ) cos ( a x b ) tan ( a x b ) sec ( a x b ) cosec ( a x b ) cot ( a x b ) sinh ( a x b ) cosh ( a x b ) tanh ( a x b ) Derivative 0 n x n 1 a e ax 1x a cos ( a x b ) a sin ( a x b ) a sec 2 ( a x b ) a sec ( a x b ) tan ( a x b ) a cosec ( a x b ) cot ( a x b ) a cosec 2 ( a x b ) a cosh ( a x b ) a sinh ( a x b ) a sech 2 ( a x b )
Implicit Differentiation An equation in the form y = 2x2 +5 is said to be an explicit equation in that y is directly stated as a function of x. However it is not always possible or desirable to express equations in this form and they may be presented in an implicit form. e.g. x2 + y2 = 9 dy by finding the derivatives of both sides w.r.t. x dx
For implicit relationships we determine Thus for x2 + y2 = 9
d ( x2 ) dx
+
d ( y2 ) dx
=
d9 dx
To evaluate the derivative of y2 we use the chain rule letting u = y2. du dy dy = 2y dx dy dx dx dy Thus 2x + 2y =0 dx = d ( y2 )
17
Higher-order Derivatives dy of the function y = f(x) is itself a function of x and may also be differentiated. dx The derivative of a derivative is called the second derivative and there is no need to stop there. The derivative Diff y f(x) dy dx f ( x) first derivative d y by d x Diff dy dx 2 f ( x ) second derivative d squared y by d x squared
2
Diff d3 y dx 3 f ( x ) third derivative etc
Modelling Often when we observe a physical phenomenon and attempt to describe the behaviour in mathematical terms we end up with equations that contain derivatives. These equations are referred to as differential equations. The solution of these differential equations is a important goal for engineers. Consider the case of a heated liquid, Newton observed that the rate at which it cooled was proportional to its excess temperature over that of its surroundings. Thus if is the temperature of the liquid, s that of its surroundings then d = k ( s ) dt where k is the constant of proportionality. This is a differential equation. Solving this will allow the relation of temperature with time to be explicitly defined. But that exercise will have to wait to a future date.
18
Taylors Series Approximations of Functions x 2 x3 x 4 x5 + + + + .... 2! 3! 4! 5!
ex = 1 + x +
Taylor's Approximation for exp(x)
25
Exp(x) 4 terms 1+x+x^2/2
20 15
f(x) 10 5 0 -4 -2 0 x 2 4
sin x = x
x3 x5 x7 x9 + + + .... 3! 5! 7! 9!
Taylor's Approximation for sin x
1.5 1 0.5 f(x) -3 0 -2 -1 -0.5 -1 -1.5 x
sin x 7 terms x-x^3/3!
0
1
2
3
19
Integration The indefinite integral or the anti-derivative We have looked at calculating the derivative f ( x ) of a function f (x). However it may be required The process of determining f (x) from f ( x ) is called anti-differentiation or integration. Example:- Find the anti-derivative of f (x)=x2 One such function is 1 F ( x ) = x3 3 dF ( x ) 1 since = ( 3x 2 ) = x 2 dx 3 Another such function is 1 F ( x ) = x3 + 2 3 dF ( x ) 1 since = ( 3x 2 ) + 0 = x 2 dx 3 1 In fact, if C is any constant the function F ( x ) = x 3 + C is also an anti-derivative of x2. 3 Suppose that f (x) is a function whose anti-derivatives are F (x) + C. The standard way to express this fact is to write
to reverse the process. We are given the derivative f ( x ) and must determine the function f (x).
f ( x ) dx = F ( x ) + C
The symbol is called the integral sign. The entire notation integral.
f ( x ) dx
is called the indefinite
20
In the same way as a table can be drawn up for derivatives the reverse table can be constructed for integrals. Function f(x) F(x) = f (x)dx a (constant) ax + C n 1 n +1 x , except n = x +C 1 n +1 1 ln x + C if x > 0, ln ( x ) + C if x < 0 x 1 ax eax e +C a ln x sin ax 1 cos ax + C a cos ax 1 sin ax + C a tan ax 1 ln sec ax + C a cosec ax 1 ln cosec ax cot ax + C a sec ax 1 ln sec ax + tan ax + C a cot ax 1 ln sin ax + C a sinh ax 1 cosh ax + C a cosh ax 1 sinh ax + C a tanh ax 1 ln cosh ax + C a coth ax 1 ln sinh ax + C a 1 1 x tan 1 + C 2 2 x +a a a 1 1 1 x sin +C a a a2 x2 1 1 xa ln +C 2 2 x a 2a x + a
More complicated integrals can, like differentiation, be accomplished by applying a set of rules of integration. However applying these rules is more problematic than for differentiation and indeed 2 many functions cannot be integrated analytically at all. For example the integral e x dx cannot be
found. Before we explore further the techniques of integration, lets view integration from a different perspective. 21
Numerical Integration Numerical Integration usually involves splitting the interval [a, b] into equal sub-intervals and approximating the area of each sub-interval Trapezoidal rule
x0 a
x1
x2
x3
x4
x5 b
Subdivide [a, b] at x1, x2, x3, xn-1 with x0 = a and xn = b with h = xi -xi-1 for i = 1, 2 n. Area of one segment is approximated by f ( x i ) + f ( x i +1 ) A = h 2
f(xi) xi h
f(xi+1) xi+1
Thus for the complete interval b h f ( x ) dx 2 f ( x 0 ) + 2f ( x1 ) + 2f ( x 2 ) ... + 2f ( x n 1 ) + f ( x n ) a h { f ( x 0 ) + f ( x n ) } + 2 { f ( x1 ) + f ( x 2 ) ... + f ( x n 1 ) } 2
Or writing f(xi) = yi b h f ( x ) dx 2 { y0 + yn } + 2 { y1 + y2 ... + yn 1} a Simpson's rule Simpson's rule allows for curvature.
yo
y1
y2
-h 0 h 2 Approximate the curve by polynomial y = Ax + Bx + C passing through the points (-h, y0), (0, y1) and (h, y2).
22
y 0 = Ah 2 Bh + C y1 = C y 2 = Ah 2 + Bh + C y 0 y1 = Ah 2 Bh y 2 y1 = Ah 2 + Bh Adding y 0 2y1 + y 2 = 2Ah 2 y0 2y1 + y 2 2h 2 Subtracting y0 y 2 = 2Bh A= B= y 2 y0 2h
h
y dx = ( Ax
h
h
h
2
+ Bx + C ) dx
h
Ax 3 Bx 2 = + + Cx 2 3 h Ah 3 Bh 2 Ah 3 Bh 2 + + Ch + + Ch 3 2 3 2 Ah 3 = 2 + Ch 3 =
Substitute in for A and C gives
h
y dx = ( y
=
h
0
2y1 + y 2 )
h + 2y1h 3
h ( y0 2y1 + y 2 + 6y1 ) 3 h = ( y 0 + 4y1 + y 2 ) 3 This provides a formula to approximate the area of a segment. Divide [a, b] into an even number ba 2n intervals of width h = . 2n
y2n a x0 x1 x2 Then using the above formula for successive pairs. x3 b x2n
23
f ( x ) dx 3 ( y
a
b
h
0
+ + 4y1 y 2 ) +
h ( y2 + 4y3 + y 4 ) + ... 3
h ( y 2n 2 + 4y 2n 1 + y 2n ) 3 h { ( y 0 + y 2n ) + 4 ( y1 + y3 + ... + y 2n 1 ) 3 + 2 ( y 2 + y 4 + ... + y 2n 2 ) } + Example Use the Trapezium rule and Simpson's rule to evaluate i 0 1 2 3 4 5 6 7 8 9 10 x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 First + last 1
1
2
dx for 10 strips. x Odd Even 0.953463 0.912871 0.877058 0.845154 0.816497 0.790569 0.766965 0.745356 0.725476
0.707107 4.139458 3.29395
Summation 1.707107 By the Trapezium rule 2 dx 0.1 x = 2 { ( y0 + y10 ) + 2 ( y1 + y2 + ... + y9 ) } 1
= 0.05 { 1.707107 + 2 ( 4.139458 + 3.29395 ) }
= 0.8286962 By Simpson's rule 2 dx 0.1 x = 3 { ( y0 + y10 ) + 4 ( y1 + y3 + y5 + y7 + y9 ) + 2 ( y2 + y4 + y6 + y8 ) } 1 0.1 { 1.707107 + 4 ( 4.139458) + 2 ( 3.29395) } 3 = 0.828428 Checking against analytical solution. 2 2 1 dx x = x 2 dx 1 1 = 1 = 2x 2 = 2 1
2
(
2 1 = 0.828427
)
24
Moments of Inertia/Second Moment of Area Definition Axis mi or Ai ri
The moment of inertia of a body about a given axis is defined as I = mi ri2
i
Where mi is the mass of the ith particle and
i
means sum over all particles or small masses
which make up the body. Similarly the second moment of area is defined as: I = Ai ri2
i
In this case the body is planar and density has been omitted. Whether moment of inertia or second moment of area is required depends on which application is being considered. The moment of inertia/second moment of area is often expressed as mk2 or Ak2 where k is called the radius of gyration. Example Find the moment of inertia/second moment of area of the uniform circular loop about an axis through the centre and perpendicular to the plane of the hoop.
M or A r
All elements are a distance r from the axis. I = Mr 2 or I = Ar 2 and the radius of gyration is k = r. Example Find the moment of inertia/second moment of area of the uniform circular lamina of density kg / m 2 about an axis through the centre and perpendicular to the plane of the lamina.
25
a
r
r
For the circular element I = 2rr r 2 or I = 2 r 3 dr
0 a
2rr r 2 2 r 3 dr
0 a
or or
r = 2 4 0
4
a
r4 2 4 0 a 4 2 A = a 2 Ak 2 where k = a 2
a
a 4 or 2 Now M = a 2 or = I = mk 2 or
Perpendicular axis theorem y xi ri O I x = mi yi2
i
mi yi x
and
I y = mi x i2
i
Now consider Ir i.e. the moment of inertia about the axis through 0 and perpendicular to the plane of the lamina. I r = mi ri2 = mi ( x i2 + yi2 ) = mi x + m i yi2
2 i i i i i
= Iy + Ix Example 26
Find the second moment of area of a circular lamina about the axis through the centre and lying in the plane of the lamina. y
a
x
From the previous example a 4 Ir = 2 And due to symmetry Ix = Iy The perpendicular axis theorem states that Ir = Ix + I y 2I x = a 4 2 a 4 a Ix = and k = 4 2
Example Find the second moment of area of a rectangle about its major axis. I = b x x 2 I = b x 2 dx
d 2
d b = x2 2 d 2 3 2b d 3 = 3 8 d 2
b
x
x d
=
bd 3 12
27
Parallel Axis Theorem y' a y x y C.A b x' It is sometimes necessary to determine the second moment of area about an axis parallel to the centroidal axis. 2 I x = ( y + b ) dA = ( y 2 + 2by + b 2 ) dA
A A
A
x
= y 2 dA + 2b y dA + b 2 dA
A A A
But y dA = 0
A
since it is the first moment about a centroidal axis I x = I x + b 2 A and similarly I y = I y + a 2 A Example: Find the second moment of area of the I-Section about its major axis. B
T Part2 D Part1 t
D T 2
I xx = I part1 + 2I part 2
3 bT 3 D T 2 t ( D 2T ) = +2 + BT 12 12 2 I for web I for flange found using parallel axis theorem
28
EXERCISES
Polynomials
1. Use a spreadsheet to sketch the graphs of the functions 2 a) y = 2x 3 + 3x 2 12x + 32 ; b) y = ( x 1) 3 1 Find the inverse function (if it is defined) of the following functions a) f (x) = 2x 3 2x 3 b) f ( x ) = x+4 f ( x ) = x2 +1 c) If f(x) does not have an inverse function, suggest a suitable restriction that will allow the definition of an inverse function. Obtain the formulae for the linear function such that a) f(0) = 3 and f(2) = -1; b) f(-1) = 2 and f(3) = 4 Find the formulae for the quadratic functions f(x) such that a) f(1) = 2, f(2) = 3 and f(4) = 5 b) f(-1) = 1, f(1) = -1 and f(4) = 2 Complete the squares for the following functions and determine if they are irreducible quadratic functions. a) x 2 + 2x + 3 b) 4x 2 12x + 9 c) 6 4x 3x 2 d) 3x 1 5x 2 Factorise the following polynomial functions a) x 3 2x 2 11x + 12 b) x 4 + x 2 2 c) 2x 4 + 5x 3 x 2 6x Express the following rational functions as the sum of a polynomial function and a proper rational function x2 + x +1 x5 x 4 x + 1 a) f ( x ) = , b) f ( x ) = ( x + 1) ( x 1) x2 + x +1 Express as a single fraction 1 2 x 1 +2 a) ; x x 2 x +1 Express as partial fractions 2x 1 a) ( x + 1) ( x 2 ) c) 1
3 2
2.
3. 4.
5.
6.
7.
8.
b)
x 3x + 3x 1 x x x + 1
3 2
1
9.
b) d)
x 1
( x + 1) ( x 2 )
x2 + x 1
2
( x + 1) ( x 2 + 2x + 2 )
1
(x
2
+ 1)
2
29
EXERCISES
Indices, Logarithms, Circular and Hyperbolic Functions
1. Simplify the following a) 4 x 42x 2. b) e 2 x e 4x c) ( 1 + e 2x )
2
d)
e5 x 8e 4x
3.
Express the following in terms of log a, log b and log c ab a 2 a) log ( a b c ) b) log c) log 3 c bc Simplify 1 1 x a) exp ln b) e 2 ln x 2 1 + x Use your calculator to find a) log 2 3.66 Solve the equation b) log 3.4 0.293
4.
5. 6.
2 log a x log a ( x 1) = log a ( x 2)
Convert the following to degrees a) 3.721 rad b) 7 6 rad c) 11 12 rad Find the value of each of the following 6 5 a) cos18 b) sin c) tan 5 2 x prove that 2 Hence solve the equation 2sin x cos x = 1 Given that t = tan sin x = 10 3
7.
d) cosec
e) cot 500
8.
2t 1 t2 and cos x = 1+ t2 1+ t2
9.
Express as a product of sines and cosines sin 3 + sin , cos cos 2 Express in the form r sin ( ) 3 sin cos ; sin cos
1 b) cos ( 0.5 )
10.
11.
Evaluate 1 a) sin ( 0.5 )
1 c) tan 3
(
)
12.
Solve the equation 5 cosh x + 3sinh x = 4
1 Show that tanh x =
13.
1 1+ x ln 2 1 x
( 1 < x < 1)
30
EXERCISES
Non-linear Equations
1. Use the bisection method to determine the specified roots of the following functions. f (x) = 2x 3 + 3x 3 given the root lies between x = 0 and x = 1. f (x) = cos x + x given the root lies between x = -1 and x = 0. f (x) = e x + x 2 given the root lies between x = 1 and x = 2. 2. Use the Newton/Raphson method to determine the root or roots of the following functions f (x) = sin x + 3x e x , root near 0.0 f (x) = 3sin x x f (x) = tan x e x , the least positive and the least negative root . 3. A semicircle is bounded by its diameter with a line drawn from one end of this diameter across the semicircle to bisect the area. It can be shown that if the line makes an angle with the diameter that 2 + sin 2 = 4. . Solve this equation to obtain . 2
43 A sphere of density and radius r has a weight of r g . When floating in water the 3 depth to which it sinks is h, the volume of the spherical segment below the surface then being 1 ( 3rh 2 h 3 ) . Archimedes principle states that when an object floats the weight of 3
fluid displaced must equal the weight of the object. Hence determine h if the density of the sphere of radius 1m is 400kg/m3 and that of the water 1000kg/m3.
31
EXERCISES
Complex Numbers
1. Express in the form x + jy a) ( 4 j7 ) ( 2 + j3) 2. Find z such that zz + 3 ( z z ) = 13 + j12 3. 4. 5. Write the complex number 10 60 in cartesian form. Given z = 2 - j2 is a root of 2z 3 9z 2 + 20z 8 = 0 find the remaining roots of the equation. Express the following complex numbers in polar form. a) 3 - j4 6.
b)
4 + j3 2 j
b) -3 - j4
c) -j4
Given z = e j 4 and z = e j 3 find 1 2 a) The modulus and arguments of z1z 2 and 2
2 b) The real and imaginary parts of z1 + jz 2 . 2 z1 . z2
7. 8.
Express z =
( 2 j) ( 3 j2 )
3 j4
in the form x + jy and also in polar form.
Show that by using Euler's formula that e j + e j e j e j cos = and sin = 2 2j and hence show that cos jx = cosh x and sin jx = jsinh x
32
EXERCISES
Differentiation I
1. Find from first principles the derivatives with respect to x of 1 x+3 x (1 - x)
a)
b)
2.
Differentiate the following functions with respect to x, giving the answers in their simplest forms: a) (x + 1) (2x - 5) (2x + 3) (x 2 + 2) 2 ( x 2 + x 1 ) 3 b) x +1 (x + 2x + 2)
2
c)
(3x + 5)10
d)
e)
(1 - 2x) 3
3
x2 + 1
3.
Find derivatives with respect to x of the functions 1+ x , (iii) x n log e x , (iv) x n e ax , ( v) log e (1 + x 2 ), ( vi) x 1/x , 1- x
(i) e
x
, (ii) loge
(vii) log e {loge(x2 + 1)}, (viii) (sin x)tan x.
4.
The technique of logarithmic differentiation is useful when we need to differentiate a cumbersome product. The method involves taking the natural logarithm of the function to be differentiated. Use logarithmic differentiation to find dy as a function of x and y when dx
a)
y = xy
b)
(cos x)y = (sin y)x.
33
5.
Find derivatives with respect to x of (i) cosech x, (ii) sech x, (iii) cosh (v) x sinh 2x, (vi) x tanh 1 x , (vii) logesinh x.
1
x , (iv) tanh 1 x ,
6.
Assuming y is a differentiable function of x find y when a) b) x3 + 2x2y xy2 + 2y3 = 4, x6 + 2x3y xy7 = 10.
7.
Find y by implicit differentiation when x2 + y2 = 10. Solve for y in terms of x such that the particular function is satisfied by (3, -1). Show that y evaluated for the particular function satisfies the general form for y.
8.
If y = e 2 x cos 4x , prove that
d2 y dy + 4 + 20y = 0 . 2 dx dx
9.
If y = e x sin x , show that dy d2 y x = 2e sin(x + ) , = 2e x cos x 2 dx 4 dx dy when dx x = 3cos , y = 4sin x = a ( sin ) , y = a ( 1 cos )
10.
Use parametric differentiation to obtain a) c) x = 4t, y = t 2 x = 2t, y = t 2 + 2t + 1 b) d)
34
EXERCISES
Differentiation II
1. 2. 3. Determine the fourth degree Taylor polynomial approximation to the function f(x) = cos 2x at x = 0. Find the Maclaurin series for (1+x)n. Note the series obtained is known as the binomial series. Maclaurin series is Taylor Series when x0 = 0. Obtain the following Maclaurin expansions: a) b) 4. 1 2 cos 2 x = 1 x 2 + x 4 x 6 + .... 3 45 1 1 2 3 2 x = 1 + x ln 2 + ( x ln 2 ) + ( x ln 2 ) + .... 2 6
Derive the Taylor series expansions for: a) b) ex about x = 1 2 about x = 2 . (1 + x)
5.
= 0.03491 and tan 45 = 1, use four terms of the Taylors series to estimate 90 tan 43 to 4d.p. Given that The radius of a sphere is measured to be 50 cm with a measured error of 0.02cm. Estimate 4 the error in the computed volume of the sphere. The volume V of a sphere is r 3 where r 3 is the radius of the sphere. Show that there are three points of inflexion on the graph of f(x) = 3x5 + 10x4 10x3 60x2 + 5 and find the ranges of x over which the graph is concave downwards.
6.
7.
8.
Sketch the following function roughly for all values of x 4x3 + 3x2 18x + 9
9.
Find the relative maxima and minima of the following function: (x 2)2 (x 7)
10.
The rate r at which a chemical reaction proceeds depends on the quantity x of a chemical and is given by r = k(a x)(b + x). Determine the maximum rate. 35
11.
A cylinder has a radius and height h with the sum of the radius and height being 2m. Determine the radius giving the maximum volume.
12.
The shape of a hole bored by a drill is a right circular cone surmounting a cylinder: the h height of the cylinder is h, its radius r and the semi-vertical angle of the cone is tan -1 ( ) . r Show that for fixed depth H of the hole the volume removed is a maximum if h = H (1 + 7 )/6.
13.
A vessel in the form of a right circular cone with its axis vertical and vertex upward is being filled at a uniform rate with liquid through a hole at the vertex. Show that when the vessel is 8 1 full the surface of the liquid is rising 4 times as fast as when it is full. 9 9 hr 1 HINT: Set V, the volume of water, equal to V0 r 2 h . 3
36
EXERCISES
Integration I
1. Integrate the following functions with respect to x: 2 3 1 a) x 4 3 b) 2 c) x + x x Evaluate the following indefinite integrals: a) 3. 1 1 3 + 2 2 d b) ax 2 + bx 1 + c dx x 3
2.
Evaluate the following definite integrals: a) 2 2 x 1 x 4 dx 1
2
b)
x ( x 1) ( x 2 ) dx
0
4
4.
If ( 1 + x 2 )
dy = 1 , find the general value of y. dx dy = 2x 1 . If the curve passes through the dx
5.
The gradient of a curve at any point is given by point x =1, y = 1, find the equation of the curve.
6.
A particle is moving along a straight line with acceleration (2 + 3t) m/s2 at time t seconds. At zero time its distance from the origin is 5m; at time t =1 its velocity is 10m/s. Where is it at time t = 1? Integrate the following expressions with respect to x: a) (2 x ) 3 b)
7.
( 2x 1)
3
c)
1 5x 7
2 d) sec ( 2x 1)
8.
Evaluate the following definite integrals:
6
a)
0
sin 2 d 3
23
b)
0
du 4 + 9u 2
c)
0
2
dx 4 x2
9.
Integrate the following expressions with respect to x: x x3 a) b) 1+ x2 4 2x 7x + 5 5 c) 2 d) ( x 3) ( x 2 + 4 ) x + x 6 Evaluate the following integrals: 37
10.
2
a) d) 11
0
2 sin 2 x dx
b) e)
2 cos ( x ) dx 0 2
3
c)
2sin 3x cos x dx
0
4
sin x cos x dx
0
sin m sin n d,
0
where m and n are integers and mn.
Integrate the following functions with respect to x: x2 2 4 a) x cos ( x 1) b) c) sin x cos 4 x ( x3 + 8) d) sin 3 2x e) x3 1 + x8 f) 4 x2
12
Integrate a) x sin x Evaluate
2 a) x cos x dx 0
b) log e x
x c) e cos x dx
13
2x b) x e dx 0
1
1 c) cos x dx 0
1
10
14
Use the trapezoidal and Simpsons rules to evaluate the following table:x 9.00 y 0.1111 9.25 0.1081 9.50 0.1053
y dx
9
when y is given in terms of x by 10.00 0.1000
9.75 0.1026
15 16
Use eleven ordinates and Simpsons rule to find an approximate value of
dx . x 1
2
Use Simpsons rule and eleven ordinates to obtain an approximate value of the definite
1
integral 17
0
2
dx (1 x 2 )
. Compare your result with the exact value of the integral.
The ordinates of a curve y = f(x) are given by the table x0 1 2 3 4 5 6 y 0.0 2.0 2.5 2.3 2.0 1.7 1.5 Use Simpsons rule to estimate the volume generated when the area bounded by this curve and the ordinates at x = 0 and x = 6 revolves about the x-axis.
38
EXERCISES
Integration II
1. Find the area enclosed by the curve y = 4x x 2 , the x-axis and ordinates at x = 0 and x = 6. 2. Find the area bounded by y 2 = 16x and y = 3x . 3 Find the area between the curves a 7 y = x 8 and a 7 x = y8
4. From the point P, with coordinates (18,12) on the curve y2 = 8x, a line PN is drawn perpendicular to the x-axis until the point N lies on the x-axis. Find the distances from the axes to the centre of mass of a lamina of uniform density bounded by PN, the x-axis and the curve. 5. Find the position of the center of mass of a lamina of uniform density bounded by the curve y2 = x3, the x-axis and ordinates at x = 1 and x = 4. 6. The density of a rod varies as the square of the distance from one end. If the length of the rod is L, find the distance of the center of mass from this end. 7. Find the second moment of area about the y-axis of the lamina bounded by the curve ay = x(a x) and the x-axis. 8. Find the second moment of area of a uniform lamina in the form of an isosceles triangle of height h about a line through its vertex parallel to its base b. 9. Find the moment of inertia of a rod in which the line-density varies as the square of the distance from one end about an axis through that end perpendicular to the rod. 1 10. Find the volume of the solid formed by the rotation of the curve y = cos x from x = 0 to x 2 = about the x axis. 11. Water is poured into a hemispherical bowl whose axis is vertical and radius is a. At the instant when measured from the waters surface to the lowest point of the bowl is x, the area of the waters surface is given by S ( x ) = x ( 2a x ) Find the volume of the water at this instant.
39
Answers to some questions Polynomials 1. a)
y = 2x^3+3x^2-12x+32
150 100 y -6 -4 -2 50 0 -50 -100 x 0 2 4
b)
y = (x-1)^(2/3)-1
2 1.5 1 0.5 y 0 -0.5 0 -1 -1.5
1
2
3
4
5
6
x
4y + 3 3+ y ; b) x = ; c) x = y 1 for y 1 y2 2 1 5 2 2 3. a) y = 3 - 2x; b) y = x + . 4. a) y = x + 1; b) y = x 2 x 2 2 5 5 2 2 2 22 3 2 5. a) (x + 1) + 2, irreducible; b) 4 x , reducible; c) 3 x + , reducible; 3 9 2 2. a) x =
2 3 11 x + d) 5 , irreducible 10 100 6. a) (x 1)(x 4)(x + 3); b) (x2 + 2)(x + 1)(x 1); c) x(x 1)(x + 2)(2x + 3) x2 + x +1 x+2 x5 x4 x +1 3x = 1+ 7. a) ; b) = x 3 2x 2 + x + 1 2 2 ( x + 1) ( x 1) ( x + 1) ( x 1) x + x +1 x + x +1
5x 2 + x 2 2 x 2 4x + 2 8. a) ; b) 3 x ( x 2)( x 2 + 1) ( x 3x 2 + 3x 1)( x 3 x 2 x + 1) x 1 2 2 1 1 1 x+ 1 2x 1 11 = + + = 2 =+ 9. a) ; b) 2 2 ; c) 2 ( x + 1) ( x + 2x + 2) ( x + 1) ( x + 2x + 2) ; ( x + 1) ( x 2) ( x + 1) ( x 2) ( x + 1) ( x 2) 9 ( x + 1) 9 ( x 2) 3 ( x 2) d) x2 + x 1
( x + 1)
2
2
=
1 x 2 + ( x + 1) ( x2 + 1) 2
2
Indices, Logarithms, Circular and Hyperbolic Functions 2 ex 1. a) 43x; b) e6x; c) ( 1 + e2x ) = 1 + 2e2x + e4x ; d) 8 1 ab a 1 11 2. a) log a + log b + 2 log c; b) log = loga log b logc ; c) log 3 = log ab log c3 = log(ab) 3log c = log a + log b 3log c bc 2c 2 22 2 2 3. a) 1 x ; b) x2. 4. a) 1.87184; b) 1.00311. 5. x = . 6. a) 213.2; b) 210; c) 165 3 1+ x 7. a) 0.951056; b) 0.587785; c) ; d) 1.1547005; e) 1.191753 3 9. sin 3 + sin = 2 sin 2 cos ; cos cos 2 = 2 sin sin 2 2 2 1 10. Therefore 3 sin cos = 2sin ; Therefore sin cos = 2 sin . 11.a) ; b) ; c) . 12. x=ln 6 3 3 2 6 4 40
1
Non-linear Equations 1. f (x) = 2x 3 + 3x 3 : 0.735139; f (x) = cos x + x : 0.73908; f (x) = e x + x 2 : 1.841405 2. f(x) = sin x + 3x e x , root near 0.0 : 0.360422; f (x) = 3sin x x : 0 and 2.278863; f(x) = tanx ex ,theleast positiveandtheleast negativeroot : 1.306327, 3.09641 3. 0.415856. 4. 0.865862 Complex Numbers 1. a) 29 j2; b) 1 + j2. 2. 3 + j2 or 3 + j2. 3. 5 + j5 3 . 4. 2 + j2, 5. a) 5ej5.35589; b) 5ej4.06889; c) 4 e j 2 6. a) z1z 2 modulus 1, argument 2 7. 8 1 j , 5 5 65 j6.15883 e 5
2 5 z1 5 33 , modulus 1, argument ; b) ,j 12 z 2 6 22
Differentiation I 1 1 1. a) 3 ; b) (1 x ) 2 2 ( x + 3) 4 x 2 + 12 x + 1 x 2 + 2x ( x 2 + 2)(2 x 5 + 12x 3 7 x 2 6) 9 2. a) ; b) 2 ; c) 30(3x + 5) ; d) ; (2 x + 3) 2 ( x + 2 x + 2) 2 x 2 ( x 2 + x 1 ) 4
1 e) (1 2 x ) 2 ( x 2 + 1) 3 (13x 2 2 x + 9) 3 x1 / 2 2x 2x 2x e 3. ; ; xn1 (n ln x + 1); eaxxn-1(n + ax); ; x1/x2 (1 + ln x); 2 ; 2 2 1/ 2 ( x + 1) ln( x 2 + 1) 1 x 1+ x 2x (sin x)tan x (1 + sec2x ln sin x) y tan x + ln sin y y2 4. a) ; b) ln cos x x cot y x (1 y ln x ) 1 1 x 5. coth x cosech x; tanh x sech x; ; ; 2x cosh 2x+sinh 2x; + tanh1x; coth x 2 2 1 x2 x 1 1 x x 3x 2 + 4 xy y 2 6 x 5 + 6x 2 y y 7 6. a) 2 . 7. ; y = (10 x2)1/2 2 ; b) 3 6 y 2 x 2xy + 6 y 2 x 7 xy t 4 sin 10. a) ; b) cot ;c) t + 1; d) 2 3 1 cos 1 2
Differentiation II n ( n 1) x 2 n ( n 1) ( n 2 ) x 3 2 2 x 2 24 x 4 26 x 6 n 1. cos2x = 1 + + ... . 2. ( 1 + x ) = 1 + nx + + + ... 2! 4! 6! 2! 3! ( x 1) 2 ( x 1) 3 2 ( x 2) ( x 2) 2 ( x 2) 3 1 + ( x 1) + ; b) 1 + + ... + + + ... 4. a) e 2 3 2! 3! 3 3 3 3 6. 200. 7. < x < 2 and 1 < x < 1
41
8.
60 50 (-3/2,117/4) x 40 30
20 (-1/4,109/8) x 10 x 0 -3 -2 -1 -10 -20 -30 0
(1,-2) x 1
2
3
9. (2, 0) is a local maximum and (16/3, -500/27) is a local minimum. 10.
ab 4 . 11. m 2 3
Integration I 7 3 1 1 a b c x3 3 1. a) x 3 +C; b) +C; c) x1 + 2x + C. 2. a) 2 2 + C; b) x 2 + x 3 + x4 +C x 2 3 4 2 3 7 4 3. a) ; b) 16. 4. tan1x + C. 5. y = x2 x + 1. 6. 13 m 3 1 1 2 1 5x 7 + C; d) tan(2x1) +C 7. a) (2 x)4 + C; b) 2 + C; c) 4(2 x 1) 4 5 2 1 8. a) ; b) ; c) 4 24 2 x2 1 x3 x2 +C; 9. a) ln(1 + x2) + C; b) 2 x 4 ln | 4 2x | +C ; c) ln x+3 2 6 2 1 x d) 2 ln | x 3 | ln(x 2 + 4) + tan 1 + C 2 2 1 9 1 10. a) ; b) + sin 2; c) ; d) ; e) 0 2 24 8 4 1 1 1 1 cos 5 x 11. a) sin(x2 1) + C; b) + C; c) + C; d) cos 2x + cos3x + C; 3 3 9( x + 8) 2 2 6 5
2 1 x 1 4 1 x e) tan x + C; f) 2 sin + x 1 + C 4 2 2 e2 12. a) x cos x + sin x + C; b) x ln x x + C; c) (sin x + cos x) + C. 13. a) 2; b) e 2; c) 1 2 14. Trapezoidal rule 0.1053875; Simpsons rule 0.105375. 15. 0.693146 16. 0.523599, exact value . 17. 74.8432 6
Integration II 64 128 7 635 1275 3 a4 bh 3 ,y = 1. . 2. . 3. a2. 4. x = 10.8, y = 4.5 . 5. x = . 6. L. 7. . 8. 3 81 9 217 496 4 20 4 2 2 x 3ML 9. . 10. . 11. x2 a 3 5 2 42