QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS

Transcription

1 arxiv: v1 [math.gm] QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS Aleander Panfilov stable spiral det A non stable spiral D=0 stable node center non stable node saddle 1 tr A

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3 QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS Aleander Panfilov Theoretical Biolog, Utrecht Universit, Utrecht c

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5 Contents 1 Preliminaries Basic algebra Algebraic epressions Limits Equations Sstems of equations Functions of one variable Graphs of functions of one variable Implicit function graphs Eercises Selected topics of calculus 3.1 Comple numbers Matrices Eigenvalues and eigenvectors Functions of two variables Eercises Differential equations of one variable Differential equations of one variable and their solutions Definitions Solution of a differential equation Qualitative methods of analsis of differential equations of one variable Phase portrait Equilibria, stabilit, global plan Sstems with parameters. Bifurcations Eercises Sstem of two linear differential equations Phase portraits and equilibria General solution of linear sstem Real eigen values. Saddle, node Saddle; λ 1 < 0;λ > 0, or λ 1 > 0;λ < Non-stable node; λ 1 > 0;λ > Stable node; λ 1 < 0;λ < Phase portraits for comple eigen values: spiral, center General ideas on equilibria with comple eigenvalues Center, spiral

6 4 CONTENTS 4.5 Stabilit of equilibrium Eercises Additional concepts (appendi General solution for comple eigen values Sstem of two non-linear differential equations Introduction and first definitions Phase portrait Equilibria Vector field Linearization of a sstem: Jacobian Determinant-trace method for finding the tpe of equilibrium Eercises Graphical methods to stud sstems of differential equations Null-clines Graphical Jacobian Eercises Plan of qualitative analsis and eamples Plan Eamples Eercises Limit ccle Stable and non-stable limit ccles Dnamics of a sstem with a limit ccle How do limit ccles occur? Eample of a sstem with a limit ccle Eercises Historical notes Dictionar Hints: Solution of the initial value problem for a linear sstem Equilibria/derivatives Answers for selected eercises and Formulas lists 109

7 Chapter 1 Preliminaries 1.1 Basic algebra Algebraic epressions Algebraic epressions are formed from numbers, letters and arithmetic operations. The letters ma represent unknown variables, which should be found from solutions of equations, or parameters (unknown numbers on which the solutions depend. Below, we review eamples several basic operations which help us to work with algebraic epress of ions. One of the most basic algebraic operations is opening of parentheses, or simplification of epressions. For that we use the following rule: (a + b (c + d = ac + ad + bc + bd note, that here ac means a c, etc., as in algebra the multiplication is often omitted. Eample (open parenthesis: (4 + a( 3 = a 6a Sometimes we use parentheses to factor epressions: Eample (factor epression: a 4 = ( a 4 = 3 (3 + 1 a 4 In man cases we also need to work with fractions: Eample (the same denominator: a ca 3 = a 3 ca 3 Eample (different denominators: a b + b a = a b a a + b a b b = a a ab + b b ab = a3 +b ab Eample (fractions simplifications: a ca 3a = a c 3 To divide a fraction a b b another fraction c d we just need to multipl it b its inverse d c : a b : c d = a b d c = ad bc, Eample (division: 4 7 : a 5 = 4 7 a 5 = 14a 0 = 10 7a Also note that: + z+d z + d 5

8 6 CHAPTER 1. PRELIMINARIES 1.1. Limits We call A a limit of the function f ( when approaches a, if the value of f ( get closer and closer to A when takes values closer and closer to a. We write it formall as: lim f ( = A (1.1 a In man cases, finding the limit is trivial: we just need to substitute the value of = a into our function: Eample: lim 3 = 3 = 8 lim f ( = f (a (1. a Functions which have such propert are called continuous and most of the functions used in biolog are continuous. However, there are several important eception. The first case, which will be the most important for us, is finding of limit of the function when. Finding such limits is important as it gives an asmptotic behaviour of our sstem when the size of a population becomes ver large. Unfortunatel, there is no such number which we can substitute into our function to find a limit using formula (1.. For functions without parameters, we can guess the limit b substituting large values to (1., e.g. = 10000, 0000, etc, but what to do for functions with parameters? Let us discuss this problem for a special class of functions, which are the most relevant to our course, the so called rational functions f ( = p( g(, where p( and g( are polnomials. In that case we can alwas find the limit using the following propert of the power function: where C is an arbitrar constant and α > 0. C lim α = 0 (1.3 To prove it note that if approaches (becomes larger and larger, the power function α with α > 0 also becomes larger and larger and therefore C α will be closer and closer to zero, thus in accordance with the definition lim C α = 0 To find the limit using this rule we need to do the following: (1 find the highest power of in our epression p( g(, ( divide each term in our function b in that power, and (3 find the limit of each term using propert (1.3. Let us consider three tpical eamples: an Eample (find the limit: lim 3N. N 3 N The highest power is N, division gives: a 0 0 = a Eample (find the limit: lim P ap 3bP 3 cp dp, a,b,c,d 0. an N 3 N N = a 3 N 3 N N 3. N N The limits of the individual terms are:

9 1.1. BASIC ALGEBRA 7 Similar steps give us: ap 3bP3 cp dp = ap P 3 3bP3 P 3 = cp P 3 dp P 3 a P 3b c = 0 3b P P d 0 0 = 3b 0 as we cannot divide b zero and we do not have a finite limit for this function. Eample (find the limit: lim a 3 b +c a 4 b = a 3 4 b 4 + c 4 = a 4 4 b 4 a 3 b +c a 4 b, a,b,c 0. a b + c 4 a b 4 = a 0 = 0 a = 0.. This epression does not have sense Another non-trivial situation occurs when the denominator of our function f ( = p( g( becomes zero for some value of, for eample f ( = 3 for = 3. In this case the formula (1. for limit cannot be used and other more careful analsis is necessar. If using of calculator we substitute some numbers into our function around point 3 we will find the following: if becomes closer and closer to 3 from the left, e.g. = 3.1;3.05;3.01,3.005;etc the function value becomes larger and larger, while if becomes closer and closer to 3 from the right, e.g. =.9;.95;.99,.995;etc the function value is negative and its absolute value also becomes larger and larger. We can formall write it as lim = +, while lim =. However, in a strict sense, as there is no real number for which f ( approaches for 3 3 close to 3 (from either side thus the limit here does not eist. We will use limits for drawing our functions and will see that limits at infinit give us horizontal asmptotes of our graphs, while blow up of functions for some (as in the last eample give us vertical asmptotes Equations An equation is a mathematical relationship involving unknown variables. These unknowns are usuall epressed b letters,, however in biolog we use man other letters (e.g. N, P. T, V, etc., which mabe somewhat related to the name of the species the describe. Solving equations means finding unknown(s such that after substitution in the equation the left and right hand sides will be equal to each other. For eample: equation 16 = 10 has a solution = 3, as 3 16 = 6 16 = 10. The usual wa to solve equations which have unknown variables in the first power onl (linear equations, is to isolate the unknowns: = [known numbers] We can achieve that b using the following rules of equation algebra: (1 we can multipl, or divide both sides of the equation b the same number, and ( we can move numbers/epressions from one to the other side of the equation, b changing their sign. The proof of these rules is trivial. Indeed if two epressions are the same X = Y, then if we multipl (or divide both of them b the same number a, the still will be the same ax = ay. Similarl if X = Y + a, we can add a to the both sides of the equation, which will not change the equalit, but we get: X a = Y + a a, or X a = Y. Thus we see, that we were able to move a from the right hand side to the left hand side of our equation, but it changed its sign as a result. Eample(solve equation: 4 = 4. Solution: + 4 = 4 = = 1

10 8 CHAPTER 1. PRELIMINARIES For a quadratic equation a + b + c = 0 we use the abc formula, which gives us the solutions as 1, = b± b 4ac a. Eample (solve equation: + 1 = Solution: (1+( +1 = = 4; = 0; + 3 = 0 from the abc formula 1, = ± ( 3 = ± 16 = ±4 1 = 1; = 3. Equation ma also contain parameters. A parameter is an unknown number (constant that ma have an value. It is different from the unknown variable, as the parameter is just a constant on which our solution depends. Eample (solve equation: a k P dp = 0, where P in unknown variable and a,k,d > 0 are parameters. Solution: B multipling both sides b P we get ak dp P = 0, or ak = dp P 3, or P 3 = ak d, thus P = 3 ak d. We see that the solution depends on 3 parameters and if someone provides us with their values we will be able to find the solution b substituting the parameter values into the final formula Sstems of equations To solve a sstem of two linear equations we epress one variable via the other and substitute it into the other equation. { + = 5 Eample (solve the sstem of equations: + = 3 Solution: From the second equation we find = 3, so we substitute into the first equation: (3 + = 5; 6 + = 5; = 1; = 1, now substitute this value to = 3 and find = 3 1, thus the solution is =, = 1. Unfortunatel, there are no general rules to solve a sstem of nonlinear equations. The usual practical wa is to start with a more simple equation, tr to obtain from it as much information as possible and then substitute it to the other equation. It is also ver helpful to factor epressions in order to simplif them. { n n Eample (solve the sstem of equations: np = 0 np p = 0 Solution: From the second equation b factoring we find np p = p(n = 0. The product is zero onl if one of the multipliers is zero, thus we have two possibilities p = 0 or n =. If we substitute p = 0 into the first equation we find n n 0 = 0, or n(1 n = 0, thus for p = 0 we have two solutions n = 0, or n = 1; now substitute n = into the first equation: 4 p = 0, 4 8 4p = 0, 4p = 4, thus for n = we found p = 1. Overall, we found the following three solutions of the given sstem (n = 0, p = 0,(n = 1, p = 0,(n =, p = 1. Sstems ma also contain parameters. Eample (solve the sstem: parameters. { an an bnp = 0 np kp = 0, where n, p are variables and a,b,k > 0, are the

11 1.. FUNCTIONS OF ONE VARIABLE 9 Solution: We proceed similarl as in the previous case. From the second equation: np kp = p(n k = 0, thus we have two cases p = 0 or n = k. After substituting p = 0 into the first equation we get: an an 0 = 0, an(1 n 0, thus n = 0, or n = 1; after substituting n = k into the first equation we get: ak ak bkp = 0, ak(1 k = bkp, a(1 k = bp, thus p = a(1 k b. Therefore, we found three solutions: (n = 0, p = 0,(n = 1, p = 0,(n = k, p = a(1 k b. It is eas to see that if we substitute the parameter values a =, b =, k = to these formulas we obtain the solution of the previous problem. Note also, that for sstems with parameters we need to be careful as not all operations are allowed for arbitrar parameter values. In our eample in order to obtain the solution we had to make several divisions b parameters a,b, and k. However we can alwas do that as the parameters are positive numbers (a,b,k > 0 and thus the cannot be equal to zero. Finall note, that we can solve sstems of three and more equations similarl, b subsequent substitutions from one equation to another, etc.. 1. Functions of one variable In science the relationships between quantities are normall epressed using functions. The simplest tpe of functions are functions of one variable. The function of one variable f is a rule that allows us to find the value of a variable (number f from a single variable (number. We denote it as f (. Below are eamples of the most important functions: power functions a, for eample f ( = 1 = ; f ( = = 1. (1.4 polnomials, a c + d, for eample: f ( = (1.5 rational functions f ( = p( g( : trigonometric functions sin, cos, tan: f ( = + 1 (1.6 f ( = sin(; f ( = cos( 1; f ( = tan( (1.7 eponential a and logarithmic function a log( f ( = e +1 ; f ( = 10 log(; f ( = (1.8 etc.

12 10 CHAPTER 1. PRELIMINARIES f( * Figure 1.1: The derivative of a function f ( at point is given b the following limit: f ( = d f d = lim f ( f ( (1.9 The derivative f ( shows the rate of change of a function f ( at a point and has man important applications: If (t is the distance traveled b a car as a function of time t, then d/ gives the velocit of the car. If n(t is the size of a population as the function of time, then dn/ gives the rate of growth of the population. Geometricall, the derivative f ( gives the slope of the tangent line to the graph of the function at the point (fig.1.1. A graph of a line tangent to the function f ( at point (fig.1.1 is given b the following equation: = f ( + f ( ( (1.10 Equation (1.10 is also known as a linear approimation of function f ( at point : Let us check formula (1.10 b approimating the function = + 1 at = 1. We find: f ( = = 3, f = 4, f (1 = 4, hence f ( ( 1. At = 1.1 this approimate formula gives f ( (1.1 1 = 3.4. The eact value is f (1.1 = = 3.4. So the error is just 0.6%. However, if = 0, f ( (0 1 = 1 while the eact value is f (0 = 1. So we see, that the approimate formula works good if is close to onl. Functions with parameters. Functions ma depend not onl on variable(s but also on parameters. We have alread seen the following eample of the function f ( that depends on three parameters a,b,c: f ( = a + b + c (1.11 Equation (1.11 describes a general quadratic polnomial. If we choose, for eample a = 3,b =,c = 1 we will get the function given b equation (1.5. Studing functions with parameters allows us to obtain results for whole classes of functions. We will frequentl use functions with parameters in our course.

13 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 11 This is because biological models usuall depend on man (up to hundreds parameters and in man situations the eact values of these parameters are unknown. One of practical difficulties in working with parameters is that use of calculators is ver limited, because calculators cannot do calculations with unknown quantities. The most valuable methods to stud functions with parameters are direct algebraic computations and analsis of the obtained formulas. In this course we will widel use the graphical methods of representation of function. Let us start with review of the basic function graphs. 1.3 Graphs of functions of one variable Eample of graphs. We usuall represent functions using graphs. To do that we plot the value of the variable along the -ais and the value of the function f ( along the -ais. Let us start first b listing tpical graph shapes that are important in this course. = =p p = 1 a b c Figure 1.: Equation = p produces a horizontal line at the level p (fig.1.. = = 1 =p p a b c Figure 1.3: Equation = p produces a vertical line shifted b p from the -ais (fig.1.3

14 1 CHAPTER 1. PRELIMINARIES = =a (a>0 =3 =0.5 a a =a (a<0 = 1.1 = 0. a b c Figure 1.4: Equation = a + p (linear function produces a straight line with the slope defined b the parameter a: the larger the absolute value of a, the steeper is the slope (fig.1.4. =+ = 1 p =a+p 1 a b c Figure 1.5: The parameter p in = a + p accounts for the vertical shift of the graph fig.1.4. a =a (a>0 = =1. =0.7 = 0.5 = 1.5 a b a c =a (a<0 Figure 1.6: Equation = a produces a parabola, if a > 0 the parabola is opened upward (fig.a,b, and if a < 0 the parabola is opened downward (fig.c. The larger the absolute value of a is, the steeper is the parabola.

15 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 13 =a +c =a +b+c c = b a a b c p =a( p( q q = v b a Figure 1.7: Equation = a + b + c also produces a parabola. Parameter c (fig.1.7a accounts for the vertical shift of the graph. Parameter b accounts for a horizontal shift of the parabola. It is possible to show that the horizontal shift of the parabola is given b a b (fig.1.7b. We can calculate this shift b determining the location of the verte of the parabola which is a point of etremum (maimum or minimum of the function. At this point the derivative of the function to zero (a + b + c = a + b = 0, Thus the coordinate of the verte is given b v = a b, or in other words the (verte of parabola is shifted b v = a b from its central location in (fig.1.7a. Note also, that a parabola ma have up to two points of intersection of the graph with the -ais (zeros of the function. The can be found from the abc formula for roots of the equation a + b + c = 0, and if these roots (p,q are known, the graph can easil be depicted using them (fig.1.7c. Note, that in this case the verte of the parabola is alwas located at the middle between these two roots. = 3 = 3 =a 3 +b +c +d =a( p( q( r p q r a b c Figure 1.8: For a general cubic function = a 3 + b + c + d we have much more possibilities and we will not discuss all of them here. The two basic forms are given b the functions = 3 and = 3 depicted in Fig.1.8a. Important here is the asmptotic behavior of the function at ±. For = 3 we see that goes to + when increases and to when decreases; for = 3 we have the opposite situation. A general graph of = a 3 + b + c + d ma have up to three zeros that can be found from the solution of the equation a 3 + b + c + d = 0, and up to two etrema (fig.1.8b. The etrema are points where the derivative of the function is zero, which in this case results in the following quadratic equation: (a 3 +b +c +d = 3a +b +c = 0. If the zeros of the function (p,q,r are known, the graph can easil be drawn as shown in Fig.1.8c.

16 14 CHAPTER 1. PRELIMINARIES = = 3 = a b c Figure 1.9: Three eamples of graphs of the power function a involving fractional powers are shown in Fig.1.9. If 0 < a < 1 than the graph growth is slower than the function = and is concave downward (in the first quadrant. To draw graph = let us use the graph of parabola = discussed in Fig.g1d5a. If in function = we switch the and we will get =, which is equivalent to = ±. The graph = can be found b switching the and the -ais for the graph of the parabola = in Fig.1.6a and we get a curve depicted in fig.1.9a in which the upper branch corresponds to = (fig.1.9b and the lower branch corresponds to =. Similarl, the graph of the function = 3 (Fig.1.9c can be found b a 90 o rotation of the graph of the function = 3 from Fig.1.8a. = 1 = 1 + b +a a 1 slope a 1 = +a b a b c Figure 1.10: Rational functions p( q( are ver important in theoretical biolog. The graph of the function = 1 (Fig.1.10a has the vertical asmptote ( = 0 and the horizontal asmptote ( = 0. The graph of function = +a 1 + b can be obtained b a shift of the graph = 1 b b units in the (vertical direction and b a units in the (horizontal direction. In this case the vertical asmptote ( at which function goes to infinit is = a, as at this point the denominator in +a 1 equals zero. The horizontal asmptote of this graph is = b, given b lim + b = b. Another rational function = occurs in the classical 1 +a Michaelis-Menten kinetics. Fig.1.10c shows the graph of this function. Because for biological applications and a are alwas considered non-negative ( 0,a > 0, we show the graph in the first quadrant onl. We see that independent of the value of the parameter a the horizontal asmptote is alwas located at = 1, as lim +a +a = 1. The slope of this function at = 0 is given b the function derivative f ( = ( +a = a at = 0, which gives a slope of f (0 = 1 (+a a.

17 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 15 1 λ e t a λ<0 t 1 e λt λ>0 b t π 1 1 c sin( Figure 1.1: Finall in fig.1.1 we show graphs of two other functions that are important in this course e λt and sin(. Note that if t grows the function e λt approaches zero if λ < 0 and diverges to infinit if λ > 0. The function sin( oscillates with a period of π between 1 and +1. π 1 = = 1 + b +a 1 = + a b slope=0 a b c Figure 1.11: Graphs of similar functions involving a second power: = 1 and = 1, are shown in Fig.1.11a,b. +a We see that function = 1 has a graph similar to that of = 1 but located in the first and second quadrants, rather than first and third. One more difference is that function = 1 does not have +a a vertical asmptote, and alwas reaches a maimum at = 0. Function = famous for its ecological applications Hill function = n n +a n +a is an eample of with n =. Its graph (Fig.1.11c has a horizontal asmptote at = 1 (similar to = +a, however, the rate of growth of = for small +a is slower than for = +a : the slope of the tangent line at = 0 here is 0, which can be found from the derivative of this function. Tips on graphs Let us list important rules that ma help to plot graphs of function = f ( with parameters. The graph of the function = f ( + p can be obtained b a vertical shift b p units of the graph of = f (. Eample: In function N+b N N + c, parameter c just shifts the graph of N+b b c units above. Important points of the graph are points at which the graph crosses the -ais (-intercept, given b = f (0, and points where the graph crosses the -ais (zeros of the function, given b f ( = 0. Note, that some graphs do not cross the or the ais and thus do not have -intercepts or zeros. For eample graph of function f ( = 1 (Fig.1.10a does not have finite zeros or -intercepts.

18 16 CHAPTER 1. PRELIMINARIES Eample: For function f (N = N+b N +c, b,c > 0, the -intercept is 0 0+b +c = c. Zeros can be found from N N+b + c = 0, which gives N + c(n + b = 0, or N + cn + cb = 0, or N(1 + c = cb, thus zero is given b the formula N = 1+c cb, which is alwas valid as c > 0. Another important graph feature are asmptotes. To find a horizontal asmptote we need to compute the lim f (. For functions without parameters, ou can tr to compute this limit using calculator b filling in a large numbers 10000, 0000, etc and looking if the function approaches some constant value. For functions with parameter, ou can tr to fill in some reasonable parameter value and tr to find similarl if the asmptote eists, however the best wa here is to find the limit using our plan from section A vertical asmptote is usuall a point where a denominator of a fraction is zero. Not all graphs have asmptotes, for eample graph of function f ( = does not have an vertical or horizontal asmptotes. However, even if the asmptotes are absent it is still useful to understand behavior of the functions at large and show it in the graph. N N N N N + N b Eample: For function N+b N + c we can find lim as N N+b + c = thus this graph has a horizontal asmptote = 1 + c. The vertical asmptote here is at point where N + b = 0, or line N = b. + c = 1 1+ b N + c = c = 1 + c, Several features of the graph can be found from the derivative of the function: a function grows if its derivative f ( > 0, decreases if f ( < 0 and has a local etremum (maimum or minimum if f ( = 0. We do not necessaril need to compute these feature for each graph, but it ma be helpful for some functions. In man applications we will be interested in points of intersection of graphs of two functions f ( and g(. Because at the intersection point functions are equal to each other, such points can be found from the equation f ( = g(. The above mentioned tips are represented graphicall in fig vertical asmptote q(*=0 intercept =f(0 zeros f(*=0 =f(= p( q( horizontal asmptote =lim f( > 8 Figure 1.13: relative etrema df/d=0 Finall let us formulate the main rules for graphing functions with parameters.

19 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 17 Plan for graphing functions with parameters 1 Tr to simplif the function and determine if it belongs to a known class of functions with graphs from (Fig Computer trail: (a Put parameters to reasonable values and plot the graph using a calculator. (b Collect qualitative information such as : number of zeros, eistence of vertical and horizontal asmptotes. (c Var parameter values to see how this changes the shape of the graph. 3 Algebraic approach (note, not all steps ma be possible: (a Find -intercept ( f (0, and zeros of the function ( f ( = 0. (b Find horizontal asmptote from the limit = lim f ( and vertical asmptote(s (for rational function p( q( the are at the points where the denominator becomes zero (q( = 0 ( fig (c Find other special points (e.g. maimum, minimum, etc, if the are important determinants of the graph shape. (d Draw the graph and indicate how the graph shape changes for different parameter values. Eample Plot the graph of the function f ( = a 0 a > 0 c > 0. Find how the graph depends +c on the parameters a and c Solution. 1 We do not need to simplif the function. The function equation has some similarities with graph classes listed above, but does not coincide eactl with an of them. a Let us put a = c = 1 and plot the graph using calculator (fig.1.14a. a The graph (fig.1.14a has the following characteristic features: the -intercept here is = 0, we see one zero of the function = 0. If we fill in large values of, we find that f (1000 = , f (5000 = and f (10000 = , thus we epect to have a horizontal asmptote = 0, we do not see an vertical asmptotes and function has an etremum point (maimum. However, will these features persist for other parameter values? In order to answer that let us perform an algebraic stud. 3a -intercept is f (0 = a 0 0 +c = 0. Zeros of the function are given b a +c = 0, which has onl one solution = 0, as + c 0 for all. 3b The function does not have vertical asmptotes as the denominator cannot be zero ( + c > 0 for all and c > 0. To find the horizontal asmptote let us compute = lim = 0, thus the horizontal asmptote is the -ais. a +c = lim a + c =

20 18 CHAPTER 1. PRELIMINARIES 3c Important point here is the location of the maimum of the function. Let us find it. For that let us find the points where the derivative of the function is zero. The derivative of the function is f ( = a ( +c a = ac a = 0, thus the epression is zero if ac a = 0, or = ±c. For ( +c ( +c 0,c > 0 we have just one solution = c. The value of the function at this etreme point is f (c = ac = a c +c c, thus the maimum is at (c, c a. 3d Let us draw the graph now. Because f (0 = 0 the graph alwas goes through the origin. Then the graph will reach the maimum at (c, c a (Fig.1.14a, smbol 1 and then approaches the ais. If we put all this information together we obtain a qualitative graph shown in fig.1.14b. We see that it qualitativel coincides with the calculator sketch. Now let us find how the graph shape depends on the parameter values. We see that the graph has a bell-shape, with a single maimum at (c, c a. The location of this maimum depends on the parameter c onl, but the maimal value of the function increases if a increases. The solid and the dashed line in Fig.1.14c illustrate how the graph shape changes if a increases while we keep c constant. Alternativel, if we keep the a value constant but increase the value of the parameter c (the solid and the dot-dashed line in fig.1.14c, the location of the maimum shifts to the right, and the maimal value of the function a c decreases. Figure 1.14: 1.4 Implicit function graphs As we know the relation between two variables and can be epressed eplicitl in terms of a function = f ( that gives us the value of if we know the value of. It is also possible that the relation between and is given implicitl as an equation. Such relations are called implicit functions, and their graphs are implicit function graphs. One of the most effective methods to plot such graphs is to tr to solve that implicit equation and rewrite it as one or several eplicit functions. In some cases the relations between and can be plotted directl. Let us consider two eamples: Eample: Draw a graph of the function(s given b equation: + = C Solution: We can either rewrite it as two eplicit functions = ± C and draw the two graphs given b this equation. Alternativel, we can note that + gives a square of the distance from the point (0,0 to the point (,, thus equation + = C gives the points located at a distance C from the origin. That is a circle with a radius C with the center at (0,0 (Fig.1.15a. We will use this graph later in our course in chapter 4 to plot fig.4.8a. Eample: Draw a graph of the function(s given b equation: ar + b R are the variables and a,b,c,d 0 are the parameters. R+c drn = 0, where R,N 0

21 1.5. EXERCISES 19 (, N a d a+b d R Figure 1.15: Solution: Let us factor the equation: ar + b R R + c drn = R(a + b R R + c dn = 0 The product of two numbers is zero if one of these numbers is zero, therefore this equation is equivalent to: R = 0 or, a + b R R + c dn = 0 Graphing of R = 0 is trivial. In order to graph a + b R R+c dn = 0 let us rewrite it as dn = a + b R R+c, or N = d a + d b R a R+c. The horizontal asmptote of this graph is: N = lim R d + d b R R+c = d a vertical asmptote occurs if the denominator of the fraction is zero, i.e. at R = c. However, because a+b + b/d = d. The R 0 and c > 0 this asmptote will be outside the range of our function. Additionall note that this function is similar to the graph of Fig.1.10c, but it is shifted upward b d a. Thus the function graph here contains two branches R = 0 and N = d a + b/dr R+c that are plotted in Fig.1.15b 1.5 Eercises Eercises for section Perform the indicated operations: (a (a b (3 4b + b (a + 3 8b (b 6 r. Find limits: 5r 30r+5 (a lim a+q c + (b lim N an +q b N +c +dn, d 0 3. Solve the equation for the specified variable:

22 0 CHAPTER 1. PRELIMINARIES (a find r in: 3r + 5(r + 1 = 6r + 4 (b find in: + 4 = 4 (c find N in: (b N k N = 0 (d find N in: (b d(1 + N k N = 0, d 0; k 0 (e find N in: ( b 1+N/h dn = 0, b 0; 4. Solve the sstem of equations for the specified variables: { = 5 (a find, in: + = 10 { a + b = 0 (b find, in: c + d = b { (1 + = 0 (c find, in: 4 = 0 { 4 (d find, in: = = 0 { b(1 R (e find R,N in: k d anr = 0 (R δn = 0, a,b,d,k,δ 0; Eercises for section Find the derivative of f (: (a f ( = 1 3 (b f ( = e 5 ; (c = (4 ( + 3 (d = a Eercises for section Without plotting the function find the following information about their graph: find the -intercept and zeros; find horizontal or/and vertical asmptotes (if the eist. (Proof of non-eistence of asmptotes is not required. (a function (n given b = an b n +c, a,b,c > 0 (b function N(R given b N = a r (h + R(1 K R, r,a,h,k 0 7. Sketch graphs of the following functions: (a = 3 6 (b = 3 (c = +a Find how the shape of the graph for, 0 depends on the value of the parameter a > 0.

23 1.5. EXERCISES 1 Eercises for section (a Sketch qualitative graphs of the following implicit functions. (b Find how special points of these graphs (intercepts, zeros, asmptotes depend on parameters. (c If graph contains several lines find their intersection points. Note, all parameters represent positive numbers. (a + 3 = 0 on the plane (b + = 9 on the plane (c = 0 on the plane (d dn(a P = 0 on the NP plane (e dn + N+a NP = 0 on the NP plane (f dr(b R = R+a crn on the RN plane (g br(1 R k dr anr = 0 on the RN plane (h an + P(1 ep + bp = 0 on the NP plane Additional eercises 9. Perform the indicated operations: (a (( ( + 1 (b a b p : 4b a p 10. Solve the sstem of equations for the specified variables: ra(1 K A AB = 0 (a find A,B,C in: AB db BC = 0 r,k,d, f > 0 BC fc = Find the derivative of f (: (a f ( = (b f ( = 1 3 (c f (g = cos( ; (d f = (cos( ; (e = a e b a,b > 0 (f = 5 3 (g = b c a, a,b,c > 0 (h =, d > 0 (i = 1+ d n n +a n, a > 0 1. Solutions of differential equations

24 CHAPTER 1. PRELIMINARIES (a Show that function N(t = A e 3t, where A is an arbitrar constant, is a solution of the differential equation: dn(t = 3N. For that, compute derivative of this function and substitute this derivative and the function itself to the equation and show that the left hand side of the equation equals to the right hand side. (b Show, using the same steps that the function N(t = s(1 e t + Ae t, where a is an arbitrar constant and s is a parameters, is a solution of the differential equation dn(t = s N 13. Assume that (t is an unknown function of t. For f ( listed below find the following derivatives: d f d and d f. (a f ( = 3 (b f ( = e a (c Find the epression for d f for an arbitrar f ((t. 14. Without plotting the function find the following information about their graph: find the -intercept and zeros; find horizontal or/and vertical asmptotes (if the eist. (a function ( given b = (b function ( given b = a : Sketch graphs of the following functions: (a = 4e 3t (b = + 3 (c = +3 b 3 c a,b,c > 0 (d = b + 4. Find how the shape of the graph depends on the value of the parameters a > 0 +a and b > 0. (e = b, Find how the shape of the graph depends on the value of the parameter c > 0 and 3 +c 3 b > 0. (f f (n = rn (1 n k h, k 0 find for which values of the parameter h > 0 the graph touches the n-ais. Tip: draw graph for h = 0 and think about how h affects this graph. 16. Sketch qualitative graphs of the following pairs of implicit functions on the same graph. Find all intersection points. (a an P(1 + ep bp = 0 and bp cn = 0 parameters a,b,c,e > 0 (b rr(1 R/K h+r NR = 0 and NR h+r dn = 0 parameters r,k,h,d > 0

25 Chapter Selected topics of calculus In this chapter we introduce several new notions on calculus and algebra which are important for our course..1 Comple numbers Comple numbers were introduced for the solution of algebraic equations. It turns out that in man cases we can not find the solution of even ver simple quadratic equations. Consider the general quadratic equation: λ + Bλ +C = 0 (.1 The roots of (.1 are given b the well known abc formula: where λ 1 = B ± B 4C = B ± D (. D = B 4C (.3 What happens with this equation if D < 0? Does the equation have roots in this case? Comple numbers help to solve such kind of problems. The first step is to consider the equation Let us claim that (.4 has a solution and denote it in the following wa: where λ = 1 (.4 λ 1 = ±i (.5 i = 1 (.6 Here i is the basic comple number which is similar to 1 for real numbers. Using it we can denote solutions of other similar equations. For eample if λ = 4, λ = 1 4 = 1 4 = i (± = ±i. 3

26 4 CHAPTER. SELECTED TOPICS OF CALCULUS Similarl the equation λ = a, has solutions λ = ±ai. Although we call ai a comple number, it is quite different from usual real numbers. Using comple numbers ai we cannot count how man books we have in the librar, for eample. The onl meaning of i is that i = 1, and ai is just a designation of a root of the equation λ = a. Now we can solve equation (.1 for the case D < 0. If D < 0, then D = i D and λ 1 = B ± i D (.7 Eample. Solve the equation λ + λ + 10 = 0 Solution. λ 1 = ± or λ 1 = 1 + 3i, λ = 1 3i. = ± 36 = ± 6i, (.8 We see, that solution of this equation λ 1, has two parts, one part is just a real number -1, which is the same for λ 1 and λ and the other part, is i times another real number 3 which has opposite signs for λ 1 and λ. This is a general form of representation of comple number. An comple number can be represented in the form: z = α + iβ (.9 where α is called the real part of the comple number z, and β is called the imaginar part of z. The notation for the real part is Rez and for the imaginar part is Imz. In our eample Reλ 1 = 1;Imλ 1 = 3. and Reλ = 1;Imλ = 3. We can work with comple numbers in the same wa as with usual real numbers and epressions. The onl thing which we need to remember, is that i = 1. To add two comple numbers we need to add their real and imaginar parts. For eample z 1 = i,z = 5 + 4i, z 1 + z = (3 + 10i + ( 5 + 4i = i i = + 14i. Similarl, multiplication b a real number results in multiplication of the real and imaginar part b this number z 1 = i; 10z 1 = 10 (3 + 10i = i. Multiplication of two comple numbers is just an eercise in multiplication of two epressions z 1 = i,z = 5 + 4i; z 1 z = (3 + 10i ( 5 + 4i = 3 ( i + 10i ( i 4i = i 50i + 40i = 15 38i 40 (as i = 1 = 55 38i. Similarl (z 1 = (3 + 10i = i + (10i = i + 100i = i 100 = i. Now we can check that λ 1 = 1+3i is a solution of the equation in eample (.8. In fact: λ +λ+10 = ( 1+3i + ( 1+3i+10 = ( 1 + ( 1 3i+(3i +6i+10 = 1 6i 9 +6i+10 = ( i+6i = 0 0i = 0, i.e. left hand side of this equation after substitution of λ 1 = 1+3i equals zero and thus λ 1 = 1 + 3i is the root of this equation.

27 .. MATRICES 5 One more definition. The number z = a ib is called the comple conjugate to the number z 1 = a+ib and is denoted as z 1 = z = a ib. Comple conjugate numbers have the same real parts, but their imaginar parts have opposite signs. Roots of a quadratic equation with negative discriminant D < 0 are comple conjugate to each other. It follows from the formula (.7 λ 1 = B + i D λ = B i D (.10 hence: Reλ 1 = Reλ = B D D ; Imλ 1 = ; Imλ =. (.11 Finall consider two more basic operations. If z = a + ib, then, z = a + b is called the absolute value, or modulus of z. Note, that z = z z, as (a + ib (a ib = a (ib = a + b. We use this trick to introduce division of two comple numbers z z z 1 = z 1 z z So, to divide two comple numbers we multipl the numerator and the denominator b a number which is the comple conjugate to the denominator, and we get the answer in the usual form. Eample 1 + 3i 1 4i = 1 + 3i 1 4i 1 + 4i (1 + 3i(1 + 4i 1 + 3i + 4i + 1i i = 1 + 4i = = = i. Matrices From a ver general point of view a matri is a representation of data in the form of a rectangular table. An eample of a matri composed of numbers is given below: ( A = ( This matri A has two rows and three columns. We will call this a matri of the size 3. In general matri size is defined as number o f rows number o f columns. Even if ou did not have matri algebra in school, ou probabl know at least one matri object, that is a vector. Indeed, a vector is an object which is characterized b its components: two numbers in two dimensions or three numbers in three dimensions. In matri algebra vectors can be represented in two forms: as a column vector, i.e. as n 1 matri (preferred representation, or as a row vector, i.e. as 1 n matri. For eample a vector V with the -component V = and the -component V = 1 can be represented a column or as a row vector as: ( V = or V = ( 1 1 Using matrices we can perform the same operations on large blocks of data simultaneousl. For eample, if we need to multipl all 6 numbers of matri A in (.1 b 4, we can write it as 4A which will mean: ( ( ( A = 4 = = (

28 6 CHAPTER. SELECTED TOPICS OF CALCULUS This operation is called multiplication of a matri b a number. For a general matri this can be written as: ( ( a b λa λb λ = c d λc λd Similarl, addition of matrices is adding the numbers that have the same location. This operation is defined onl for two matrices of the same size: ( ( ( ( A + B = + = = ( For general matrices it can be written as: ( ( a b + c d z w ( a + b + = c + z d + w (.15 Multiplication of matrices is not so trivial. In general matri multiplication is defined as the products of the rows of the first matri with the columns of the second matri. Thus, to fund the element in row i and column j of the resulting matri we need to multipl the ith row of the first matri b the jth column of the second matri. Thus we can multipl two matrices A B onl if the number of columns in matri A equals the number of rows in matri B. For a product of two matrices this gives: ( ( a b c d z w ( a + bz a + bw = c + dz c + dw (.16 From this it follows that multiplication of a matri b a column vector is given b: ( ( ( a b v av + bv = c d v cv + dv (.17 The last equation is useful for representation of linear sstems as can be seen from the following eample. Assume we have a sstem of linear equations: { = 5 (.18 + = 10 we can write the coefficients at and in the left hand side as a square matri: ( 1 A =. 1 We also have two numbers in the right hand side which we can write as a column vector: ( 5 V =. 10 Now if we write and as a column vector: X = (.

29 .3. EIGENVALUES AND EIGENVECTORS 7 we can represent sstem (.18 using matri multiplication (.17 as: ( ( ( 1 5 A X = V ; or = 1 10 ( ( ( 1 1 Indeed, from (.17 we get =, that proves this result (.19 Another ( important matri operation is the determinant of a square matri, which for the matri a b A = is defined as: c d det a b c d = ad cb (.0 The determinant of a matri has man important applications in algebra. For eample using determinants it is possible to find solution of sstem of linear equations (e.g. sstem (.18 in the form of so-called Cramer s rule, which was published b Gabriel Cramer as earl as in mid-18th centur. Cramer s rule is briefl formulated in eercise 7 at the end of this chapter. Now let us consider one of the most important problems in matri algebra: the eigen value problem..3 Eigenvalues and eigenvectors Let us start with a definition: Definition 1 A nonzero vector v and number λ are called an eigen vector and an eigen value of a square matri A if the satisf equation: Av = λv (.1 Eigen vectors are not unique, and it is eas to see that if we multipl it b an arbitrar constant k we get another eigen vector corresponding to the same eigen value. Indeed b multipling (.1 b k we get: kav = kλv or A(kv = λ(kv (. therefore, we can sa that kv is also an eigen vector of (.1 corresponding to eigen value λ. ( ( 1 1 For eample, for matri A =, number λ = 3 and vector v = are an eigen value and 1 1 eigen vector as: ( ( ( ( ( Av = = = = 3 = 3v ( ( ( 1 If we multipl v = b an number, e.g., 5, or etc., we will get new eigen vectors v =, ( 1 5 v = of this matri for λ = 3. You can check it in the same wa as we did in (.3 for a vector ( 5 1 v =. 1

30 8 CHAPTER. SELECTED TOPICS OF CALCULUS Finding eigen values and eigen vectors is one of the most important problems in applied mathematics. It arises in man biological applications, such as population dnamics, biostatistics, bioinformatics, image processing and man others. In our course we will appl it for the solution of sstems of differential equations, which we will consider in chapter 4. ( a b Let us consider how to solve the eigen value problem for a matri A =. For that we need ( c d v to find λ and satisfing: v ( a b Av = c d ( v v ( v = λ v. (.4 We can rewrite it as a sstem of two equations with three unknowns λ,v,v : { a v + b v = λv c v + d v = λv (.5 If we collect all unknowns at the left hand side we will get the following sstem: { ( ( (a λ v + b v = 0 a λ b v or in matri f orm = c v + (d λ v = 0 c d λ v ( 0 0. (.6 This sstem alwas has a solution v = v = 0, however it is not an eigen vector, as in accordance with the definition the eigen vector should be nonzero. In order to find non-zero solutions let us multipl the first equation b d λ, the second equation b b and subtract them. Multiplication gives: Subtraction of the equations results in: { (d λ [(a λ v + b v ] = 0 b [c v + (d λ v ] = 0 (.7 (d λ (a λ v + (d λ b v = 0 b c v + b (d λ v = 0 gives (d λ (a λ v b c v + (d λ b v b (d λ v = 0 or as v 0 we get: [(d λ (a λ b c] v = 0 (.8 (d λ (a λ b c = λ (a + dλ + (ad cb = 0 (.9 This is a quadratic equation with unknown λ and for each particular coefficients a,b,c,d we can find two solutions: λ 1 and λ using the abc formula. Thus we found that the eigenvalue problem for a matri (.4 has solutions for the eigen values λ. In general, for a nn matri that the eigen value problem has n solutions for λ. Equation (.9 is ver important in our course and it has a special name: characteristic equation. In most of the courses on mathematics this equation, however, is written in a slightl different matri form.

31 .3. EIGENVALUES AND EIGENVECTORS 9 To derive it let us recall the definition of the determinant of a matri given in section.: det a b ( a λ b c d = ad cb. Similarl the determinant of matri is: c d λ det a λ c b d λ = (a λ(d λ bc (.30 which coincides with the left hand side of characteristic equation (.9 and thus the characteristic equation can be rewritten as: det a λ b c d λ = 0 (.31 Let us use this approach to find the eigen values of matri A from eample (.3. We get the following characteristic equation: Det 1 λ 1 λ = (1 λ(1 λ = 1 λ λ + λ 4 = λ λ 3 = 0 From the abc formula: λ 1, = ± therefore we found two eigen values λ 1 = 3 and λ = 1. = ± 16 ; λ 1 = 3 λ = 1 Now, let us find eigen vectors. For that let us substitute the found eigen values to the original equation (.6 and solve it for v and v. Let us do it first for a particular eample (.3 for which we have found eigen values λ 1 = 3 and λ = 1. For eigen vector corresponding to eigen value λ 1 = 3 we obtain: { (1 3v + v = 0 v + (1 3v = 0 or { v + v = 0 v v = 0 or { v = v v = v (.3 ( 1 Both equations give the same solution v = v. This means that if v = 1, then v = 1 and a pair 1 satisfies the sstem and thus gives an eigen vector of problem (.3. We can also use an( other value for v. For eample, if we use v = then v will be v = and we get another eigen vector, etc. In general an v = k, and v = k give an eigen vector. We can epress it b the following formula: ( v v ( 1 = k 1 (.33 where k is an arbitrar number. Formula (.33 gives all possible solutions of eq.(.3. It also illustrates a general propert of eigen vectors which we have proven in (., that if we multipl an eigen vector b an arbitrar number k will get also an eigen vector of our matri. Using this propert we can formulate an eas wa to write a formula for all eigen vectors. For that we take an found eigen vector and ( multipl it b an arbitrar number k. ( Note, that( if for problem (.3 we use another found eigen vector v, we can write an answer as = k. At the first glance this formula is different from v (.33. However, it is eas to see that both formulas give the same result: this is because k in (.33

32 30 CHAPTER. SELECTED TOPICS OF CALCULUS ( 1 is an arbitrar constant and an vector given b the formula (.33 with can be obtained using ( 1 the formula k for another value of k. Thus the answer to our problem: to find eigen vectors of ( ( ( ( v 1 v matri (.3 for eigen value λ 1 = 3, can be written as =, or =, or etc. v 1 v These ( vectors give ( particular ( solutions ( of this problem. We can also write a formula for all solutions v 1 v as = k, or = k, or etc. As we discussed above all these answers will be v 1 v correct and equivalent. Similarl we find the eigen vector corresponding to the other eigen value λ = 1: 1. Substitution: { (1 ( 1v + v = 0 v + (1 ( 1v = 0 or { v + v = 0 v + v = 0 or { v = v v = v (.34. Relation between v and v : v = v 3. Eigen vector: use e.g. v = 1, thus v = 1 ( 1 The general form is v = k 1 ( 1 v = 1, where k is an arbitrar number. Note, that in both cases in order to find eigen vectors we could use the first equation onl (see equations (.3 and (.34, and the second equation in both cases did not provide us an new information. It is not a coincidence, and this propert is the basis for the following epress method for finding eigen vectors: Epress method for finding eigen vectors Let us derive a formula for finding the eigen vectors of a general sstem (.5. We assume that we have found eigen values λ 1 and λ from the characteristic equation (.31. To find the corresponding eigen vectors we need to substitute the found eigen values into the matri and solve the following sstem of linear equations (.6: { (a λ1 v + bv = 0 (.35 cv + (d λ 1 v = 0 It is eas to check that if we use for v and v the values v = b and v = a λ 1 it gives the solution of the first equation: (a λ 1 v + bv = (a λ 1 ( b + b(a λ 1 = b(a λ 1 + b(a λ 1 = 0

33 .3. EIGENVALUES AND EIGENVECTORS 31 If we substitute these epressions into the second equation we get: cv + (d λ 1 v = cb + (d λ 1 (a λ 1 = 0 To prove that this epression is also zero, note that (d λ 1 (a λ 1 cb is zero in accordance with the characteristic equation (.9. Therefore v = b and v = a λ 1 give a solution of (.35 which is an eigen vector corresponding to the eigen value λ 1. Similarl we find the eigen vector corresponding to the the eigen value λ. However, this approach does not work if in (.6 both b = 0 and a λ = 0. In this case we can use the second equation cv + (d λ 1 v = 0 and find an eigen vector as v = d λ 1 and v = c. Indeed: cv + (d λ 1 v = c(d λ 1 + (d λ 1 ( c = 0 As in the previous case it is eas to show that this vector satisfies the other (first equation as: (a λ 1 v + bv = (a λ 1 (d λ 1 + b( c = 0 due to (.35. The final formulas are: v 1 = ( v1 v 1 ( b = a λ 1 v = ( v v ( b = a λ (.36 or ( ( v1 d λ1 v 1 = = v v 1 c = ( a b where a,b are the elements of the matri A =. c d ( v v = ( d λ c (.37 Either (.36 or (.37 can be used to find eigen vectors. (Both answers will be valid. If, however, one of the formulas gives a zero eigen vector, we should use the other one to obtain a non-zero vector. ( 1 Let us appl these formulas for the sstem (.3 with matri A = and eigen values λ 1 1 = 3;λ = 1. The eigen vectors can be found from (.36 as: ( v1 λ 1 = 3; v 1 ( = 1 (3 ( = ( v λ = 1; v ( = 1 ( 1 ( = (.38 and from (.37 as: ( v1 λ 1 = 3; = v 1 ( 1 3 = ( ( v λ = 1; v = ( 1 ( 1 ( = (.39 We see that the vectors differ from the vectors found earlier, but it is( eas to find( that the are equivalent. For eample, if we multipl the first vector b 1 we find 1 1 =, thus the same 1 vector which we found earlier in (.33. We also see that formulas (.38 and (.39 give equivalent result. ( Indeed, first ( vectors obtained form (.38 and (.39 are the same. For second vectors note that: 1 =.

34 3 CHAPTER. SELECTED TOPICS OF CALCULUS.4 Functions of two variables A function of two variables f (, describes the rule of finding the value of function f, if we know the values of the variables and. For eample, the area of a right-angled triangle with the sides, and is given b the following function of two variables: f (, = /. Another eample is the rate of growth of a pre population in a tpical ecological predator-pre model: f (, = , where is the pre population and is the predator population. The graph of the function of one variable = f ( is a line on the O-plane. To sketch the graph of the function of two variables f (,, we must use a three dimensional space (,,z: the O-plane for the values of the independent input variables,, and the third ais z for the function output value z = f (,. In such a representation the graph will be a surface in a three dimensional space. Fig..1 shows a graph of the function f (, = plotted b a computer. Figure.1: Derivatives. The net step is the definition of the derivative of f (,. The main idea of finding the derivative of f (, is to fi one variable at a constant value, sa =. After that we will get a function of one variable onl ( f (,. Now, we can find the derivative of f (,, as the usual derivative of a function of one variable. For eample, f (, = Let us fi = =. We get the following function of one variable: f (, = = 6 3. We can easil find the derivative now: d f (,/d = d( 6 3/d = 3. This tpe of derivative is called the partial derivative of f (, with respect to at =. We denote it as f / = = 3 We can find such a derivative at = 3, or at an other value of. In fact for an arbitrar =, f (, = , and f / = = ( / = Here (3 / = 0 as we replaced b a constant and the derivative of a constant is zero. Similarl, ( 3 / = 0, and ( 1.5 / = 1.5, as 1.5 is a constant and the derivative of (k = k. It is generall accepted to make all these differentiations without eplicitl replacing b. We just should keep in mind, that for such a differentiation we treat as a constant. Thus, to find the derivative of f with respect to we just write: f / = ( / = 1.5.

35 .4. FUNCTIONS OF TWO VARIABLES 33 keeping in mind that is considered as a constant and not a variable during this differentiation. This epression is called the partial derivative of f (, with respect to and is denoted as f /. Similarl, we can introduce a partial derivative of f with respect to : f /. To compute it, we fi (treat as a constant and make the usual differentiations with respect to. In our eample it gives: f / = ( / = Here (3/ = 3, ( 3 / = 3, and ( 1.5/ = 1.5 as is fied. Eample. Find z/ and z/ for z = 3 sin Solution z/ = 3 cos, as for / we fi, and (sin/ = cos. Similarl, z/ = ( 3 sin/ = 3 sin, as and hence sin is treated as a constant. * =* =* * Figure.: The geometrical representation of a partial derivative is clear from fig... To compute f / we fi, i.e. assume that has some value =. The condition = geometricall gives a horizontal line on the O plane fig..a, or a line parallel to the -ais. In 3D this line gives a curve on the 3D surface in graph fig..b, which is a 1D function. The partial derivative with respect to for this particular will give us the slope of the tangent line to this 1D function. Thus (see fig.. f / gives the slope of the tangent line in the direction of the -ais or the rate of change of f (, in the direction. Similarl, computing f / we fi, i.e. assume that has some value =. It gives us a vertical line on the O plane fig..a, or a line parallel to the -ais. Thus f / gives the slope of the tangent line in the direction of the -ais, or the rate of change of f (, in the direction. If we consider f (, as a mountain f / gives the slope of the mountain if we climb in the -direction and f / gives the slope of the mountain if we climb in the -direction. Note, that in general at each point on a surface we can draw a tangent line in an direction, and partial derivatives f / and f / give the slopes of two of these possible tangent lines. Note, that the slope of a tangent line an direction can be obtained as a combination of these two slopes. Linear approimation Let us derive a formula for approimating a functions of two variables f (,. Let us assume that we know f (, and its partial derivatives at some point, and that we want to find the value of a function at the close point, (fig..3. Let us move to the point, in two steps. Let us first move from the point, to the point,, i.e. in the -direction, and then from, to