P1 Calculus II Partial Differentiation & Multiple Integration

Transcription

1 /1 P1 Calculus II Partial Differentiation & Multiple Integration 4 lectures, MT 217 David Murra Rev November 2, 217 david.murra@eng.o.ac.uk Course Page dwm/courses/1pd Introduction So far in our eplorations of the differential and integral calculus, ou have considered onl functions of one variable: = f (), and all that. But ou do not have to look too hard to find quantities which depend on more than one independent variable. The pressure in an oil field will depend on latitude, longitude and depth, and no doubt will change over time p = p(θ, φ, d, t). What is the change of pressure when these variables change? How might one integrate quanities over the complete oil field to find, sa, the ield? This set of lectures answers the questions of how our notions of differentiation and integration can be etended to cover multiple variables. Sllabus (from Weblearn) Partial differentiation: the chain rule and simple transformations of first-order (not second-order) partial differential coefficients. Multiple integrals and their evaluation, with applications to finding areas, volumes, masses, centroids, inertias etc. (ecluding line and surface integrals and using spherical and clindrical coordinate sstems onl). At the end of this course students will: Be familiar with the basic operations of calculus of functions of several variables. Understand and perform partial differentiation on a function of multiple variables. Be able to use the chain rule on partial differentials and simple transformations of 1st order partial differential coefficients. Understand the concept of a Jacobian and its use Be able to evaluate multiple integrals and use this to find areas, volumes, masses, centroids, moments of inertia, etc.

2 /2 Lecture Content 1. The partial derivative. First and higher partial derivatives. Total and partial differentials, and their use in estimating errors. Testing for total (or perfect) differentials. Integrating total differentials to recover original function. 2. Relationships involving first order partial derivatives. Function of a function. Composite functions, the Chain Rule and the Chain Rule for Partials. Implicit Functions. 3. Transformations from one set of variables to another. Transformations as old in terms of new and new in terms of old. Jacobians. Transformations to Plane, spherical and polar coordinates. What makes a good transformation? Functional dependence. Shape. 4. Double, triple (and higher) integrals using repeated integration. Transformations, the use of the Jacobian. Plane, spherical and polar coordinates. Tutorial Sheets The tutorial sheet associated with this course is 1P1H Calculus 2: Partial Differentiation and Multiple Integrals Reading James, G. (21) Modern Engineering Mathematics, Prentice-Hall, 3rd Ed., ISBN: (paperback). Stephenson, G. (1973) Mathematical Methods for Science Students, Longman Scientific & Technical, 2nd Ed., ISBN: (paperback). Stephenson s book is a suitable introductor tet for our first ear. It is divorced from applications, and is now looking a bit dull. Good for basic problem bashing though. Kreszig, E. (1999) Advanced Engineering Mathematics, John Wile & Sons, 8th Ed., ISBN: (paperback). Course WWW Pages Pdf copies of these notes (in colour), copies of the lecture slides, the tutorial sheets, corrections, answers to FAQs etc, will be accessible from dwm/courses/1pd Onl the notes and the tute sheets get put on weblearn.

3 Lecture 1 Total and Partial Derivatives and Differentials 1.1 Revision of continuit and the derivative for one variable Functions and ranges of validit Suppose we want to make a real function f of some real variable. We certainl need the function recipe, but we should also specif a range R of admissible values for for which is mapped onto b = f (). We might have = f () = 2, with R being the interval 1 1. Note that with this definition it is meaningless to ask what is f ( 2)?. Often, however, the range is not specified: the assumption then is that the range is such as to make real. For eample, if = (1 + ) 1/2 we would assume that 1 <. (Those of ou eperienced in writing functions in a computer language will know the importance of setting proper ranges for input variables. Without them, a program ma crash.) Continuit for a function of one variable Suppose we have a value = a. We can define a neighbourhood of -values around = a b requiring a < η or, equivalentl, ( a) 2 < η 2. A function = f () is said to be continuous at = a if, for ever positive number ɛ (however small), one can find a neighbourhood a < η in which f () f (a) < ɛ. (1.1) Another wa of epressing this is: lim f () = f (a). a It should not matter whether tends to a from below or from above. (1.2) You will recall from previous lectures that: The sum, difference and product of two continuous functions is continuous. The quotient of two continuous functions is continous at ever point where the denominator is not zero. 1

4 1/2 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS f() f(a) <ε <ε η a Figure 1.1: Neighbourhood for continuit for a function of one variable The derivative of a function of one variable You will also recall that the derivative is defined as [ ] d f ( + δ) f () f () = lim. (1.3) d δ δ f () is differentiable if this limit eists, and eists independent of how δ. η f(+ δ) f() δ Tends to tangent as δ tends to zero δf= f(+ δ) f() + δ Figure 1.2: Geometrical interpretation of the derivative.

5 1.2. MOVING TO MORE THAN ONE VARIABLE 1/3 1.2 Moving to more than one variable The big changes we will encounter take place when going from n = 1 variables to n > 1 variables. So for much of the time we can keep the page uncluttered b dealing with functions of onl n = 2 variables, as in f = f (, ). As illustrated in Fig. 1.3, functions of two variables are convenientl represented graphicall using the Cartesian aes O z. The function representation is a surface, as opposed to a plane curve for a one variable function. It is of course progressivel harder to represent functions of more than two variables. 3 2 f(,) f(,) (a) (b) f(,).5 f(,) (c) (d) Figure 1.3: Surface plots of (a) , (b) ( + )( ), (c) ep[ ( )/1] sin(2) cos(4), and (d) ( ) 1/2. (The Matlab code for such plots is given as an Appendi.)

6 1/4 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS Continuit for functions of several variables A function f = f (, ), defined in some region R, is continuous at a point (, ) = (a, b) in R if, for ever positive number ɛ (however small), it is possible to find a positive δ such that for all points in the neighbourhood defined b ( a) 2 + ( b) 2 < η 2 (1.4) we have f (, ) f (a, b) < ɛ. (1.5) Or equivalentl, as before, lim f (, ) = f (a, b). (1.6) (,) (a,b) Note that for functions of more variables f ( 1, 2, 3,...) the neighbourhood would be defined b ( 1 a) 2 + ( 2 b) 2 + ( 3 c) < η 2. The 2-variable case is illustrated in Figure 1.4. f(,) f(a,b) ε ε project onto surface circle, radius at (,)=(a,b) η Figure 1.4: Neighbourhood for continuit for a function of 2 variables 1.3 The partial derivative The etension of the idea of continuit to functions of several variables was direct. Etending the notion of the derivative is not quite as simple the slope or gradient of the f (, ) surface at (, ) depends on which direction ou move off in. The ke idea is to consider the gradient in a particular direction and the obvious directions are those along the ais and the ais. Now, if one wants to move off from (, ) in the direction, one has to keep fied, and vice versa.

7 1.3. THE PARTIAL DERIVATIVE 1/5 The partial derivative of f (, ) wrt, then wrt ( ) [ ] ( ) f f ( + δ, ) f (, ) f = lim ; δ δ = lim δ [ ] f (, + δ) f (, ) The subscript indicates that is being kept constant, and then similarl for Geometrical interpretation of the partial derivative δ (1.7) Figure 1.5 shows the geometrical interpretation of the partial derivatives of a function of two variables. f(,) = constant plane f Slope is f f(,) = constant plane f Slope is f Figure 1.5: Interpreting partial derivatives as the slopes of slices through the function A possibl useful shorthand Given that the list of variables is known, and the one being varied is eplicit, writing the the held constant subscripts after the derivative is unnecessar. The subscripts are often omitted, but ou ma wish to leave them in until ou are more familiar with the subject. Another commonl used shorthand (but one which ou, like me, might find unhelpful when writing b hand) is f = ( f ) and f = ( f ). (1.8)

8 1/6 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS More than two variables If we are dealing with a function of more variables, we keep all but the one variable constant. Eg for f ( 1, 2, 3,...) we have The partial derivative again... ( ) [ ] f f (1, 2, [ 3 + δ 3 ], 4,...) f ( 1, 2, 3, 4,...) f 3 = = lim 3 δ 3 δ Finding partial derivatives is eas The definition indicates that actuall doing partial differentiation is eactl the same as normal differentiation with respect to one variable, while all the others are treated as constants. Eamples. Q: Find the first partial derivatives of f (, ) = A:First assume is a constant, then : f = 2 3 and f = (1.9) Q: Find the 1st partial derivatives of f (, ) = e ( 2 + 2) sin( 2 ) A: f = e ( 2 + 2) [ 2 sin( 2 ) + 2 cos( 2 )] (1.1) f = e ( 2 + 2) [ 2 sin( 2 ) + 2 cos( 2 )] (1.11) Q: If f (, ) = ln(), derive an epression for f f in terms of f. A: f (, ) = ln() + ln() (1.12) f = 1/ and f = 1/ (1.13) f f = 1/ = e f (,). (1.14) 1.5 Higher partial derivatives? Although ou do not need to consider complicated relationships involving 2nd order derivatives, ou do need to know how to find them. It is ver obvious! f and f are (well, probabl are) perfectl good functions of (, ). An eample of the notation used is: 2 f 2 = f = f (1.15)

9 1.5. HIGHER PARTIAL DERIVATIVES? 1/7 Eample. f (, ) = f = 2 3 f = f = 2 3 f = (1.16) (1.17) (1.18) (1.19) (1.2) But we should also consider and f = f = 2 f : in this case, 6 2 (1.21) f = f = 2 f : in this case, 6 2. (1.22) Spook! In this case f = f is that alwas true? A meatier random eample: f (, ) = e ( 2 + 2) sin( 2 ) (1.23) f = e ( 2 + 2) [ 2 sin( 2 ) + 2 cos( 2 )] (1.24) f = e ( 2 + 2) [ 2 sin( 2 ) + 2 cos( 2 )] (1.25) f = e ( 2 + 2) [ 4 2 cos() + 2 cos() 2 3 sin() (1.26) 2[ 2 sin() + 2 cos()]] = e ( 2 + 2) [sin()[ ] + cos()[ ]] and f = e ( 2 + 2) [ 2 3 cos() + 2 cos() 2 3 sin() (1.27) 2[ 2 sin() + 2 cos()]] = e ( 2 + 2) [sin()[ ] + cos()[ ]] Hmm. So the are equal in this case too. When both sides eist, and are continuous at the point of interest... The differential operators are equivalent: 2 = 2 This result has an interesting consequence for higher partial derivatives. (1.28)

10 1/8 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS Eample. Q: For a general function f show that 3 f 2 [ 3 ] 4 f 2 [ 3 f ] 5 = (1.29) A: Using the ordering result 3 f 2 = 2 f = 3 f = and thus the equation is p.p 4 p 5 which is indeed zero. 3 f, (1.3) So the order of higher partials is unimportant but a sensible ordering can save time! Eample. Q: Find 3 t [ ( 5 + ) cosh(cosh( 2 + 1/)) + 2 t ]. (1.31) A: Thoughtless Method. Grind awa blindl differentiating with respect to then then t. This ma take a fortnight because the functions of and are moderatel unpleasant. You will be cross when ou remember that... A: Thoughtful Method. The result of... can be obtained much more quickl b differentiating with respect to t first. 1.6 A stern Warning. Partial derivatives are not fraction-like. Although total derivatives have fraction-like qualities, this is NOT the case with partial derivatives. Eample. Q: Find d/d given = u 1/3, u = v 3 and v = 2. A: d d = d du du dv dv d = 1 3 u 2/3.3v 2.2 = 2 (1.32) which ou can check b finding = 2 eplicitl. Eample. But but but! Suppose we were asked: Q: Given the perfect gas law pv = RT, determine ( ) ( ) ( ) p V T. V T p (1.33)

11 1.7. TOTAL AND PARTIAL DIFFERENTIALS 1/9 A: You would guess +1 of course. But WRONG! If pv = RT then p = RT/V, V = RT/p and T = V p/r. Thus ( ) ( ) ( ) ( ) ( ) ( ) p V T RT R V = V T p V 2 = 1. (1.34) p R In fact, we will be able to show after studing implicit functions that if we have an function f (,, z) =, then z z = 1. Partial derivatives, unlike total derivates, do not behave as fractions. Although we CAN write down epressions like df =... We CANNOT write down f =.... If ou ever think that ou are dividing partials, THINK AGAIN! 1.7 Total and partial differentials First, note that a differential is different from a derivative. Differentials are about the following. Suppose that we have a continuous function f (, ) in some region, and both f and f are continuous in that region. How much does the value of the function change as one moves infinitesimal amounts d and d in the and directions? The amount, df, is the total differential how do we epress it? Given small changes in and it is eas enough to write the change in f as: δf = f ( + δ, + δ) f (, ). (1.35) B adding in two cancelling terms this can be rewritten as df = [f ( + d, + d) f (, + d)] + [f (, + d) f (, )]. (1.36) But recall that [ ] f f ( + δ, ) f (, ) = lim δ δ so that df = & f = lim δ [ ] f (, + δ) f (, ) δ (1.37) ( ) +δ ( ) f f δ + δ. (1.38)

12 1/1 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS where ( ) f +δ means ( ) f evaluated at + δ. But for an function g(, ) ( ) g g(, + δ) g(, ) + δ (1.39) so that ( ) +δ f ( ) f + ( 2 ) f δ (1.4) and δf f ( 2 ) δ + f δδ + f δ. (1.41) In the limit as δ and δ tend to zero, we reach The total or perfect differential of f (, ) df = f f d + d (1.42) The total differential df is the sum of the partial differentials f f d and d. f Function surface f+df f d +d f f d +d Figure 1.6: Total differential as the sum of partial differentials. Remember that d and d are infinitesimals.

13 1.7. TOTAL AND PARTIAL DIFFERENTIALS 1/ Using the total differential Eample. Q: A material with a linear temperature coefficient α is made into a block of sides,, z measured at some temperature T. The temperature is raised b a finite δt. (a) Derive the new volume of the block. (b) Then let δt dt and find the change using the total differential. A: (a) Eactl: V + δv = (1 + αδt )(1 + αδt )z(1 + αδt ) = V (1 + αδt ) 3. (1.43) A: (b) The volume of the block is V = z. Using the epression for the total differential dv = zd + zd + dz (1.44) = z(αdt ) + z(αdt ) + z(αdt ) (1.45) = 3V αdt (1.46) Thus V + dv = V (1 + 3αdT ). This answer using the total differential is eact in the limit as the small δt in part (a) tends to zero and becomes dt in part (b) [**] A little aside... not on the sllabus! Note that our epression 1.42 is an eact one for df in the limit as d and d tend to zero. If δ and δ are just small rather than infinitesimall small then δf f f δ + δ. (1.47) Recalling the epression for Talor s epansion in one variable, ou ma guess that this is a Talor s epansion to 1st order in two variables. In other words f ( + δ, + δ) f (, ) + f f δ + δ. (1.48) As we are alread off piste, we might as well see how it continues! To 2nd order f ( + δ, + δ) f (, ) 1st : + f f δ + δ 2nd : + 1 [( ) ( ) ( ) ] 2 f (δ) 2 2 f 2 f + 2 (δ)(δ) + (δ) 2 2! 2 2 3rd : +... think binomial!... (1.49)

14 1/12 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS When is an epression a total differential? Suppose we are given some epression p(, )d+q(, )d. Can we determine whether it is the total differential of some function f (, )? Now, if it is, df = p(, )d + q(, )d. (1.5) But then we must have that p(, ) = f and q(, ) = f (1.51) and using f = f p = 2 f = q = 2 f (1.52) That is the test is The t.d. test: p(, )d + q(, )d is a total differential iff p = q. (1.53) Eample. Show that there is no function having continous second partial derivatives whose total differential is (d d). Set p = and q = 2 2. Then p = q = 4. (1.54) Recovering the function from its total differential Suppose we found p(, )d + q(, )d to be total differential using the above test. Could we recover the function f? To recover f we must perform the reverse of partial differentiation. As f / = p(, ): f = p(, )d + g() + K 1 (1.55) where g is a function of alone and K 1 is a constant. You can see that we need the g() because when we differentiate with respect to it vanishes. As far as is concerned, g() is a constant of integration.

15 1.7. TOTAL AND PARTIAL DIFFERENTIALS 1/13 Similarl, f = q(, )d + h() + K 2 (1.56) We now need to resolve the two epressions for f, and this is possible, up to a constant K, as the following eample shows. Eample. Let f = 3 + sin sin but pretend we do not know it. Instead we are asked Q: Determine whether ( 3 + cos sin )d + (3 2 + sin cos + 6)d (1.57) is a perfect differential and, if so, of what function f? A: Testing whether p/ = q/, we find that ( 3 + cos sin ) = cos cos (1.58) (3 2 + sin cos + 6) = cos cos The are the same, so it is a perfect differential. Integrating ( 3 + cos sin ) over and (3 2 + sin cos + 6) over we find: and f = 3 + sin sin + g() + K 1 (1.59) f = 3 + sin sin h() + K 2. (1.6) Comparing and resolving these epressions we have So f = 3 + sin sin + + g() + K 1 (1.61) f = 3 + sin sin + h() K 2 f = 3 + sin sin K 1. (1.62) Thus we have indeed recovered the original function, up to a constant of integration. We would need some etra piece of information to recover this sa the value of the function at a particular point.

16 1/14 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS Given p(, )d + q(, )d Test whether: p/ = q/. If good: Integrate p(, ) wrt, remembering g() is a const of integration Integrate q(, ) wrt, remembering h() is a const of integration Resolve the two epressions. A.1 [**] For interest: Plotting Functions of several variables In lecture 1 we noted that functions of 2 variables were relativel eas to visualize as surfaces, whereas those in more variables were considerabl more difficult, simpl because we live in a 3D world. Time is an eception: a method of plotting n variables can alwas be converted to a video where time is the n + 1th variable. The Figure shows the surface plot and the level contour plots of the function f (, ) = f(,) The following is some simple Matlab code to give a surface plot of a function.

17 1.7. TOTAL AND PARTIAL DIFFERENTIALS 1/15 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % surface plot script % % vectors of grid and positions = [ -2:.1:2]; = [ -1:.1:1]; % turn these in arra suitable for surface plotting [X,Y] = meshgrid (,); % find the function values at each point in mesh Xsq = X.*X; Ysq = Y.*Y; Z =.6* Xsq.* Xsq +.1* Xsq.* Ysq - 2.*( Xsq - Ysq ) +5; % initial graphics set (gca, FontSize, 18); colormap ( default ); brighten (); % now plot surface surfc (X,Y,Z); % label aes label ( ); label ( ); zlabel ( f(,) ); % save as postscript ( vector graphics ) print ( - depsc, surfplot.eps ); % save as png image print ( -dpng, surfplot.png ); A.2 [**] For interest: Slice plots Another sort of contour plot is a slice plot. Here one traces out the function f (, o ) for several fied values of (Eg, o = 1, 2,...), and/or the function f ( o, ) for several fied values of o. This sort of plot is commonl used to displa transistor characteristics. Figure 1.7: A slice plot through transistor characteristics

18 1/16 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS A.3 [**] For interest: Finding maima, minima and saddle points Stationar points in a function of two variables occur when f = f =, but further tests are required to determine whether these points are minima, maima, or saddle points. The required conditions are summarized in the following table. Tests for stationar points f = f Q = [f 2 f f ] f (or f ) Tpe = < > Minimum = < < Maimum = > Irrelevant Saddle Eample. From the surface plot of the earlier function f (, ) = it appears that there is a saddle at (, ) and two minima at (±1.something, ). Let s derive from the table. f = = ( ) (1.63) f = (1.64) f = = ( ) (1.65) f = (1.66) =.4 (1.67) f Requiring f = gives = alwas, as ( ) >. As =, requiring f = gives = or = ± 4/2.4 = ± 5/ So we have the three stationar points. What tpe are the? (, ) Q = [f 2 f f ] f (or f ) Tpe ( 5/3, ) [7.2(5/3) 4][.2(5/3) + 4] < [7.2(5/3) 4] > Minimum ( 5/3, ) [7.2(5/3) 4][.2(5/3) + 4] < [7.2(5/3) 4] > Minimum (, ) ( 4)(4) > Irrelevant Saddle So all is as epected.

19 1.7. TOTAL AND PARTIAL DIFFERENTIALS 1/ Where does the table of conditions come from? A function has a maimum at (a, b) if f (a + δ, b + δ) f (a, b) < (1.68) for arbitraril small δ and δ. Geometricall we see that slices through the function in both and directions ehibit maima. Figure 1.8: Similarl a function has a minimum at (a, b) if f (a + δ, b + δ) f (a, b) > (1.69) for arbitraril small δ and δ. Geometricall we see that slices through the function in both and directions ehibit minima. How do we compute the stationar values. One might guess b looking for points where f / and f / are both zero. Indeed this is the case. For f (, ) to be stationar we require the total differential to be zero, ie: df = f f d + d =. (1.7) Because d and d are independent, this gives rise to the condition that f / = and f / =. But how to decide whether the point is a maimum or minimum? Again one might guess that this will have something to do with second derivatives. Using Talor s Theorem we have f (a + δ, b + δ) = f (a, b) + [(δ)f (a, b) + (δ)f (a, b)] (1.71) + 1 2! [(δ)2 f + 2(δ)(δ)f + (δ) 2 f ] +...

20 1/18 LECTURE 1. TOTAL AND PARTIAL DERIVATIVES AND DIFFERENTIALS But the first order derivativens f, f are zero, so that, to the second order in small quantities: f (a + δ, b + δ) f (a, b) = 1 2 [(δ)2 f + 2(δ)(δ)f + (δ) 2 f ]. (1.72) Now the conditions for maima and minima above (eqs 1.68 and 1.69) tell us that we should be interested in the sign of the rhs of equation 1.72 or, equivalentl, the sign of: [(δ) 2 f + 2(δ)(δ)f + (δ) 2 f ]. (1.73) This epression can be rewritten in two was: Either Or 1 { [(δ)f + (δ)f ] 2 (δ) 2 [f 2 f f ] } f (1.74) 1 { [(δ)f + (δ)f ] 2 (δ) 2 [f 2 f f ] } f (1.75) It is clear that in general the sign depends on the actual δ and δ under consideration, that is, on how ou move off from the point where f = f =. But what we can sa unequivocall is that if Q = [f 2 f f ] < then the term in {...} is positive. So then the sign depends on f, or equivalentl f. (In fact, we have just shown that, when Q <, SIGN(f ) = SIGN(f ).) So, when Q < and f < (or equivalentl f < ) the point is a maimum when Q < and f > (or equivalentl f > ) the point is a minimum. What about when Q >? Then the sign reall does depend on how ou move off from the point where f = f =. This is a saddle point. The function appears to have a maimum if ou move in one direction and a minimum if ou move in another.