MATH2019 ENGINEERING MATHEMATICS 2CE SESSION II 2004 ...

1 MATH2019 ENGINEERING MATHEMATICS 2 CESESSION II 2004 OUTLINELECTURE NOTEST hese notes are intended to give a brief outline of the course to be used as anaid in learning. They are not intended to be a replacement for attendance atlectures, problem classes or tutorials. In particular, they contain few exam-ples. Since examination questions in this course consist mainly of examples,you will seriously compromise your chances of passing by not attending lec-tures, problem classes and tutorials where many examples will be worked outin MMII School of MATHEMATICS , UNSW1 TOPIC 1 PARTIAL DIFFERENTIATIONP artial derivatives are the derivatives we obtain when we hold constant all butone of the independent variables in a function and differentiate with respectto that of Two VariablesSupposez=f(x, y).Define f x= lim x 0f(x+ x,y) f(x,y) x f y= lim y 0f(x,y+ y) f(x,y) yThese are both functions ofxandyand the usual differentiation rules (prod-uct, quotient etc) f x=fx=zx, f y=fy= are used to denote differentiation with respect to the indicatedvariable.

2 Further f x(x0, y0) =fx(x0, y0) means f xevaluated at the point (x0, y0).Higher-Order Derivatives 2f x2= x( f x)= (fx)x=fxx 2f x y= x( f y)= (fy)x=fyx 2f y x= y( f x)= (fx)y=fxy 2f y2= y( f y)= (fy)y=fyyMixed Derivatives Theorem( )The functionsfxy(theyderivative iffx) andfyx(thexderivative offy)are obtained by different procedures and so would appear to be differentfunctions. In fact, for almost all functions we meet in practical applications,they are identical because of the which saysIff(x, y)and its partial derivativesfx, fy, fxyandfyxare all defined andcontinuous at all points in a region surrounding the point(a, b)thenfxy(a, b) =fyx(a, b).2 This readily extends to higher order derivatives. In particular, if all deriva-tives are continuous then n+mf xn ymcan be used to denote the partial derivativeoff ntimes with respect toxandmtimes with respect toyin any RuleRecall that ifu=f(x) andx=g(t) thendudt=dudxdxdt=dfdxdgdt=f (x)g (t).

3 This readily generalises. Ifw=f(x, y) andxandyare themselves differen-tiable functions oft( the coordinates of a moving point andtis time), thendwdt= f xdxdt+ f of Three or more VariablesIfz=f(x1, x2, x3, ..) then the partial derivative ofzwith respect to anyone variable (call itxi) is obtained by holding all the other variables constantand then differentiating with respect mixed derivatives theoremextends to these RuleThis readily extends to functions of three or more variables. For example, ifw=f(x, y, z) andx, y, zare themselves functions oft,thendwdt= f xdxdt+ f ydydt+ f Rules for Functions defined on SurfacesSupposew=f(x, y, z) andx=x(r, s), y=y(r, s), z=z(r, s) (the lastthree define a surface in 3D space) then f r= f x x r+ f y y r+ f z z rand f s= f x x s+ f y y s+ f z z swhere x retc are taken holdingsconstant and x setc are taken Taylor Series From first year, we know that iffis a function of a single variablex,thenf(x) =f(a) + (x a)f (a) +12!

4 (x a)2f (a) + = n=01n!f(n)(a)(x a)n3 This extends to functions of 2 or more variables. We consider onlyf(x, y).The Taylor Series off(x, y) about the point (a, b) isf(x, y) =f(a, b) + (x a) f x(a, b) + (y b) f y(a, b)+12!{(x a)2 2f x2(a, b) + 2(x a)(y b) 2f x y(a, b)+(y b)2 2f y2(a, b)}+ higher-order Linear ApproximationIfy=f(x) then a reasonable approximation whenxis close tox0isf(x)'f(x0) + (x x0)f (x0)obtained by truncating the Taylor series after the linear term. Geometrically,we are approximating the curvey=f(x) forxnearx0by the tangent to thecurve at (x0, f(x0)).This idea readily extends to functions of two or more variables. All wedo is truncate the Taylor Series after the linear terms. The standard linearapproximation off(x, y) near (x0, y0) is thereforef(x, y)'L(x, y) whereL(x, y) =f(x0, y0) + (x x0)fx(x0, y0) + (y y0)fy(x0, y0).Geometrically, we are approximating the curved surfacez=f(x, y) near(x0, y0) by the tangent plane at (x0, y0, f(x0, y0)).

5 DifferentialsThe expressiondf= f x(x0, y0)dx+ f y(x0, y0)dyis called can think of it as the infinitesimal changedfproduced infby infinitesimal is obtainedfromL(x, y) by replacing x= (x x0) bydxand y= (y y0) EstimationThe differential can be used toestimatechanges infdue to small changesin its arguments. If f=f(x0+ x, y0+ y) f(x0, y0) then f' f x x+ f y f x, f yare evaluated at (x0, y0).4If xand yare known, we just substitute them in. Usually, however, allwe know areboundson xand example, we may only be able tomeasure temperature to that case we have, approximately,| f| f x | x|+ f y | y|.For functions of several variablesf(x1, x2, , xn) f'n k=1 f xk xkand| f| n k=1 f xk | xk|.Leibniz Ruleddx v(x)u(x)f(x, t)dt= v(x)u(x) f xdt+f(x, v(x))dvdx f(x, u(x)) 2 EXTREME VALUESE xtrema for functions of two variablesSuppose we havef(x, y) continuous on some regionR. What are theextreme values off(x, y) ( maxima and minima) and how do we findthem?

6 Definition:The functionf(x, y) has aglobal maximumorabsolute max-imumat (x0, y0) iff(x, y) f(x0, y0) for all (x, y) :The functionf(x, y) has aglobal minimumorabsolute min-imumat (x0, y0) iff(x, y) f(x0, y0) for all (x, y) :The functionf(x, y) has alocal maximumorrelative maxi-mumat (x0, y0) iff(x, y) f(x0, y0) for all (x, y) in some neighbourhood of(x0, y0).Definition:The functionf(x, y) has alocal minimumorrelative mini-mumat (x0, y0) iff(x, y) f(x0, y0) for all (x, y) in some neighbourhood of(x0, y0).Definition:A point (x0, y0) Ris called acritical pointoffiffx(x0, y0) =fy(x0, y0) = 0,or iffis not differentiable at (x0, y0).Definition:A local maximum or minimum is called anextreme can only occur at(i)boundary points ofR5(ii)critical points offSecond Derivative TestIffand all its first and second partial derivatives are continuous in theneighbourhood of (a, b) andfx(a, b) =fy(a, b) = 0 then(i)fhas alocal maximumat (a, b) iffxx<0 andD=fxxfyy f2xy>0 at(a, b).

7 (ii)fhas alocal minimumat (a, b) iffxx>0 andD=fxxfyy f2xy>0at (a, b).(iii)fhas asaddle pointat (a, b) ifD=fxxfyy f2xy<0 at (a, b).(iv)IfD=fxxfyy f2xy= 0 at (a, b) the second derivative test application of these ideas to practical problems will be illustrated inthe lectures. The candidates for maxima and minima are found by looking ati) boundary points, ii) points where one or more of the first partial derivativesfail to exist and iii) points where all the first partial derivatives ideas readily generalise to functions of 3 or more variables althoughthe second derivatives test becomes quite values for parameterised curvesTo find the extreme values of a functionf(x, y) on a curvex=x(t), y=y(t) we find wheredfdt= f xdxdt+ f ydydtiszero. The extreme values are found at(i) Critical points (wheref = 0 orf does not exist).(ii) The endpoints of the parameter extrema and Lagrange multipliersMotivation:Suppose we are asked to find the minimum (or maximum) of afunction subject to a :Find the pointP(x, y, z) on the plane 2x+y z 5 = 0 that liesclosest to the involves finding the minimum of the functionf(x, y, z) = x2+y2+z2subject to the constraint thatx, yandzsatisfyg(x, y, z) = 2x+y z 5 = 06In this simple case, it is easy to use the constraint equation to find an explicitexpression for one of the variables (sayz) in terms of the other two and tothen substitute this intofwhich thus becomes a function of two variablesonly and then to find the extrema offas a function ofxandy.

8 For a morecomplicated constraint, it may not be possible to use the constraint equationto obtain an explicit expression for one of the variables in terms of the othersso a more general procedure is method of Lagrange multipliersTo start off, suppose thatf(x, y) andg(x, y) and their first partial deriva-tives are continuous. To find the local minima and maxima offsubject tothe constraintg(x, y) = 0 we find the values ofx, yand that simultaneouslysatisfy the equations f x g x= 0, f y g y= 0,together withg(x, y) = 0 (1)Justification:We can, in principle, use the equationg(x, y) = 0 to writeyas a function ofxalthough, as indicated above, this may not be possible inpractice. Hence, we may considerfto be a function of a single variablexandlook for points wheredf/dx= 0. Let (x, y) = (a, b) be such a point. But, bythe chain ruled fd x= f xd xd x+ f yd yd x= f x+ f yd yd xThus f x+ f yd yd x= 0 at (x, y) = (a, b)(2)However, sinceg(x, y) = 0,dg/dx= 0 everywhere (including (a, b)).

9 Thus g x+ g yd yd x= 0at (x, y) = (a, b)(3)Thus, eliminatingdy/dxfrom (2) and (3) we obtain f x g y f y g x= 0 at (x, y) = (a, b)which can also be written f x f y g x g y = 0 at (x, y) = (a, b)Hence, the rows of this determinant must be linearly dependent, Thus thereexists a real number such that( f x, f y)= ( g x, g y)These equations, together withg(x, y) = 0, are just (1).7N. quantity is called aLagrange multiplierand the method alsoworks iffandgare also functions ofz. In that case we have the additionalequation f/ z= g/ zto solve. It is also possible to introduce the so-calledLagrangian functionL(x, y, ) =f(x, y) g(x, y)The equations (1) and the constraintg(x, y) = 0 are obtained by setting tozero the first partial derivatives ofL(x, y, ) with respect tox, yand .Lagrange multipliers with two constraintsSuppose we now want to find the maxima and minima off(x, y, z) subjecttog1(x, y, z) = 0andg2(x, y, z) = do this, we introduce two Lagrange multipliers (one for each constraint)and theLagrangian functionfor this situationL(x, y, z, , ) =f(x, y, z) g1(x, y, z) g2(x, y, z)We now need to find the values ofx, y, z, and which simultaneously satisfythe five equations obtained by setting to zero the partial derivatives ofLwithrespect tox, y, z, and.

10 TOPIC 3 - VECTOR FIELD THEORYQ uick Revision of Vector AlgebraScalars are quantities which have only a magnitude (and sign in some cases)such as temperature, mass, time and speed. Vectors have a magnitude anda direction. We will work only in 3D physical space and use the usual right-handedxyzcoordinate system. We denote vector quantities by usingboldsymbols, We leti,j,kbe the three unit vectors parallel to thex, yandzaxes respectively. If a pointPhas coordinates (p1, p2, p3) andQhascoordinates (q1, q2, q3) then the vector PQfromPtoQhas componentsa1=q1 p1, a2=q2 p2, a3=q3 p3anda= PQ=a1i+a2j+a3k= OQ OP. The length ofais|a|= a21+a22+ position vector of a typical point with coordinates (x, y, z)is usually writtenr=xi+yj+ 0i+ 0j+ is the vector all of whose components are zero,and is not to be confused with the scalar 0. All the usual rules apply, forexamplea+b= (a1+b1)i+ (a2+b2)j+ (a3+b3)k8a+0=aca=ca1i+ca2j+ca3k a= ( 1)a= a1i a2j a3ka+b=b+a(a+b) +c=a+ (b+c) =a+b+ca+ ( a) = (a+b) =ca+cbInner or Dot or Scalar Product of Vectorsa b=a1b1+a2b2+a3b3=|a||b|cos where (0 ) is the angle a=|a|2and the dot product of two (non-zero) vectors is 0 if andonly if they are orthogonal ( = 2).