3.2 Limits and Continuity of Functions of Two or More Variables.

Transcription

1 0 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES 3. Limits and Continuit of Functions of Two or More Variables Elementar Notions of Limits We wish to etend the notion of its studied in Calculus I. Recall that when we write f () = L, we mean that f can be made as close as we want to a L, b taking close enough to a but not equal to a. In this process, f has to be defined near a, but not necessaril at a. The information we are tring to derive is the behavior of f () as gets closer to a. When we etend this notion to functions of two variables (or more), we will see that there are man similarities. We will discuss these similarities. However, there is also a main difference. The domain of functions of two variables is a subset of R, in other words it is a set of pairs. A point in R is of the form (, ). So, the equivalent of a will be (, ) (a, b). For functions of three variables, the equivalent of a will be (,, z) (a, b, c), and so on. This has a ver important consequence, one which makes computing its for functions of several variables more diffi cult. While could onl approach a from two directions, from the left or from the right, (, ) can approach (a, b) from infinitel man directions. In fact, it does not even have to approach (a, b) along a straight path as shown in figure 3.4. With functions of one variable, one wa to show a it eisted, was to show that the it from both directions eisted and were equal ( f () = f ()). Equivalentl, when the its from a a + the two directions were not equal, we concluded that the it did not eist. For functions of several variables, we would have to show that the it along ever possible path eist and are the same. The problem is that there are infinitel man such paths. To show a it does not eist, it is still enough to find two paths along which the its are not equal. In view of the number of possible paths, it is not alwas eas to know which paths to tr. We give some suggestions here. You can tr the following paths: 1. Horizontal line through (a, b), the equation of such a path is = b.. Vertical line through (a, b), the equation of such a path is = a. 3. An straight line through (a, b),the equation of the line with slope m through (a, b) is = m + b am. 4. Quadratic paths. For eample, a tpical quadratic path through (0, 0) is =. We will show how to compute its along a path in the net sections. While it is important to know how to compute its, it is also important to understand what we are tring to accomplish. Like for functions of one variable, when we compute the it of a function of several variables at a point, we are simpl tring to stud the behavior of that function near that point. The questions we are tring to answer are:

2 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.03 Figure 3.4: Possible paths through (a, b) 1. Does the function behave "nicel" near the point in questions? In other words, does the function seem to be approaching a single value as its input is approaching the point in question?. Is the function getting arbitraril large (going to or )? 3. Does the function behave erraticall, that is it does not seem to be approaching an value? In the first case, we will sa that the it eists and is equal to the value the function seems to be approaching. In the other cases, we will sa that the it does not eist. We have the following definition: Definition 3..1 We write f (, ) = L and we read the it of (,) (a,b) f (, ) as (, ) approaches (a, b) is L, if we can make f (, ) as close as we want to L, simpl b taking (, ) close enough to (a, b) but not equal to it. Remark 3.. It is important to note that when computing f (, ), (,) (a,b) (, ) is never equal to (a, b). In fact, the function ma not even be defined at (a, b), et the it ma still eist. While (a, b) ma not be in the domain of f, the points (, ) we consider as (, ) (a, b) are alwas in the domain of f. Remark 3..3 There are several notation for this it. The all represent the same thing, we list them below.

3 04 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES 1. f (, ) = L (,) (a,b). f (, ) = L a b 3. f (, ) approaches L as (, ) approaches (a, b). We now look at how its can be computed. 3.. Finding Limits Using the Numerical Method We tr to estimate or "guess" if a it eists and what its value is b looking at a table of values. Such a table will be more complicated than in the case of functions of one variable. When (, ) (a, b), we have to consider all possible combinations of a and b. This usuall results in a square table as the ones shown below. Eample 3..4 Consider the function f (, ) = sin( + ) +. Use a table of values to "guess" f (, ). We begin b making a table of values of f (, ) for (, ) close to (0, 0) Looking at the table, we can estimate the it along certain paths. For eample, each column of the table gives the function values for a fied value. In the column corresponding to = 0, we have the values of f (, 0) for values of close to 0, from either direction. So we can estimate the it along the path = 0. In fact, the column corresponding to = b can be used to estimate the it along the path = b. Similarl, the row = a can be used to estimate the it along the path = a. The diagonal of the table from the top left to the bottom right correspond to values =. It can be used to estimate the it along the path =. The other diagonal, from top right to bottom left corresponds to =. So, it can be used to estimate the it along the path =. Looking at the table, it seems that the it along an of the paths discussed appears to be 1. While this does not prove it for sure, as there are man more paths to consider, this gives us an indication that it might be. We can then tr to use other methods we will discuss in the net sections to tr to show the it is indeed 1. It turns out this it is indeed 1.

4 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.05 Eample 3..5 Consider the function g (, ) = +. Use a table of values to "guess" g (, ). We begin b making a table of values of g (, ) for (, ) close to (0, 0) Looking at the table as indicated in the previous eample, we see that the it along the path = 0 appears to be 1 while the it along the path = 0 appears to be 1. This proves g (, ) does not eist. Eample 3.. Consider the function h (, ) = to "guess" h (, ) Use a table of values We begin b making a table of values of h (, ) for (, ) close to (0, 0) Looking at this table as indicated in the previous eamples, it appears that the it along the paths = 0, = 0, = and = is 0. However, as we will see in the net section, this it does not eist. In this case, the table would have given the wrong indication. In conclusion, we see that tables do not provide as good an answer as in the case of functions of one variable. The can be helpful when the it does not eist, if the table shows two paths leading to a different it. However, since the number of paths we can see on the table is ited, the will not, in general tell us for sure if a it eists. The can still be used to get an idea of whether the it might eist and what it might be. Given a function, and a it to compute, if one does not have an idea of what this function does, looking at a table of values might help to point the person in one direction. Usuall, solving a problem is easier if one has an idea of what the answer might be. So, while the

5 0 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES use of such tables is more ited than in the case of functions of one variable, these tables are not useless Finding Limits Using the Analtical Method Computing its using the analtical method is computing its using the it rules and theorems. We will see that these rules and theorems are similar to those used with functions of one variable. We present them without proof, and illustrate them with eamples. Theorem 3..7 (Properties of Limits of Functions of Several Variables) We list these properties for functions of two variables. Similar properties hold for functions of more variables. Let us assume that L, M, and k are real numbers and that hold: (,) (a,b) f (, ) = L and 1. First, we have the obvious its If c is an constant,. Sum and diff erence rules: 3. Constant multiple rule: 4. Product rule: 5. Quotient rule: provided M 0. (,) (a,b) (,) (a,b) = a (,) (a,b) = b c = c (,) (a,b) [f (, ) ± g (, )] = L ± M (,) (a,b) [kf (, )] = kl (,) (a,b) [f (, ) g (, )] = LM (,) (a,b) (,) (a,b) g (, ) = M, then the following [ ] f (, ) = L g (, ) M. Power rule: If r and s are integers with no common factors, and s 0 then r r [f (, )] s = Ls (,) (a,b) r provided Ls is a real number. If s is even, we assume L > 0.

6 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.07 Theorem 3..8 The above theorem applied to polnomials and rational functions implies the following: 1. To find the it of a polnomial, we simpl plug in the point.. To find the it of a rational function, we plug in the point as long as the denominator is not 0. Eample 3..9 Find (,) (1,) + Combining the rules mentioned above allows us to do the following Eample Find (,) (1,) + = 1 + (1) () (,) (1,1) 4 + = + 4 = Combining the rules mentioned above allows us to do the following (,) (1,1) 4 + = = 1 Remark Like for functions of one variable, the rules do not appl when "plugging-in" the point results in an indeterminate form. In that case, we must use techniques similar to the ones used for functions of one variable. Such techniques include factoring, multipling b the conjugate. We illustrate them with eamples. Eample 3..1 Find 3 3 We cannot plug in the point as we get 0 in the denominator. We tr to rewrite the fraction to see if we can simplif it. 3 3 Eample Find ( ) ( + + ) = ( = + + ) = 0 Here, we cannot plug in the point because we get 0 0, an indeterminate form. Since this is a fraction which involves a radical, we multipl b the conjugate.

7 08 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES We get: = ( ) ( + ) ( ) ( + ) ( ) ( + ) = ( + ) = 1 = Limit Along a Path We have mentioned several times above how important taking the it along a specific path might be. In particular, one wa to prove that f (, ) (,) (a,b) does not eist is to prove that this it has different values along two different paths. We now look at several eamples to see how this might be done. In general, ou need to remember that specifing a path amounts to giving some relation between and. When computing the it along this path, use the relation which defines the path. For eample, when computing the it along the path = 0, replace b 0 in the function. If computing the it along the path =, replace b in the function. And so on... Make sure that the path ou select goes through the point at which we are computing the it. Eample Consider the function f (, ) = find (,) (1,0) The goal is to tr to You ma remember from Calculus I that in man cases, to compute a it we simpl plugged-in the point. If ou tr to do this here, ou obtain 0 0 which is an indeterminate form. It does not mean the it does not eist. It means that ou need to stud it further. We will do this b looking at the it along various paths. As mentioned in the introduction, some obvious paths we might tr are the path = 1 and = Limit along the path = 0. First, we find what the function becomes along this path. We will use the notation to mean along the =0 path = 0 and = 0. We have: (,) (1,0) along = to mean + 1 = =0 (,) (1,0) + 1 = along the path

8 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.09 Also, note that along the path = 0, is constant hence (, ) (1, 0) can be replaced b 1. Therefore (,) (1,0) along =0 + 1 = 1 0 = z Limit along the path = 1. We have: + 1 = = = = 1 Hence, (,) (1,0) along =1 + 1 = 0 1 = 1

9 10 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES 4 4 z Conclusion: The its are diff erent, therefore (,) (1,0) eist. + 1 does not Eample Consider the function f (, ) = +. The goal is to tr to find +. As mentioned in the introduction, some obvious paths we might tr are the path = 0 and = 0. Note that we can also combine both computations (finding what the function is along the path and finding the it). 1. Limit along the path = 0. Along this path, we have along =0 + = + = = 1 = 1 =0

10 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES z Limit along the path = 0. Along this path, we have along =0 = 0 + = = = 1 = 1 + =0

11 1 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES z Conclusion: The its are diff erent, therefore + eist. Eample 3..1 Prove that + does not eist. does not First, we tr the it along the paths = 0 and = 0. The user will check that both its are 0. Net, we tr along the path =. We get + = + along = We obtained a diff erent it. So, Eample Prove that = (,) (0,0) + = (,) (0,0) 1 = (,) (0,0) = = does not eist. does not eist. You will recognize this function, it is the function in the third table we did earlier. From the table, it appeared that the it along the paths = 0, = 0,

12 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.13 and = was 0. Yet, ou were told the it did not eist. We prove it here. The reader will check that if we compute the it along the paths = 0, = 0, =, we obtain 0 ever time. In fact, the it along an straight path through (0, 0) is 0. The equation of such a path is = m. Along this path, we get 4 + = 4 + along =m =m m 3 = (,m) (0,0) 4 + m = 0 m 3 = 0 = 0 ( + m ) m + m However, we will get a different answer along the path =. 4 + = 4 + This proves that along = = (, ) (0,0) = = 0 = 1 does not eist. = 3..5 Additional Techniques to Find Limits: Change of Coordinates and Squeeze Theorem or Sandwich Theorem Sometimes, changing coordinates ma be useful. Consider the eample below. Eample Using polar coordinates, find Recall that the relationship between a point in polar coordinates (r, θ) with r 0 and rectangular coordinates (, ) is = r cos θ = r sin θ

13 14 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES From which, we can see that and + = r cos θ + r sin θ = r ( cos θ + sin θ ) = r = r 3 cos 3 θ + r 3 sin 3 θ = r 3 ( cos 3 θ + sin 3 θ ) Also, saing (, ) (0, 0) is equivalent to saing r 0 +. Hence, we have: ( r 3 cos 3 θ + sin 3 θ ) + = r 0 + r = r ( cos 3 θ + sin 3 θ ) r 0 + = 0 There is also an equivalent of the squeeze theorem. Suppose we are tring to find f (, ) given f (, ) and we suspect the it might be L. (,) (a,b) Theorem Suppose that f (, ) L g (, ) for ever (, ) inside a disk centered at (a, b),ecept mabe at (a, b). If g (, ) = 0 then f (, ) = L. (,) (a,b) (,) (a,b) The diffi cult with this theorem is that we must suspect what the it is going to be. This is not too much of a problem. If ou have tried a table of values and found that along all the paths the table allows ou to investigate, the it is the same, or if ou have tried to compute the it along different paths and have found the same value ever time. Then, ou might suspect the it eists and is the common value ou have found. It is this value ou would tr in the squeeze theorem. The second diffi cult is finding the function g. This is done using approimation of the initial function f. How it is done depends on f. We illustrate how to do it with a few eamples. Eample 3..0 Find f (, ) for f (, ) = +. The reader will check that computing this it along various paths such as = 0, = 0, = gives 0. So, ou might start suspecting the it eists and is 0. We now use the squeeze theorem to tr to prove it. In other words, we need to find a function g (, ) such that f (, ) 0 g (, ) and g (, ) = 0.

14 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.15 To find g, we proceed as follows: f (, ) 0 = = = = = since we can make a fraction bigger b making its denominator smaller. Thus, we have If we let g (, ) =, we see that theorem, f (, ) = 0. Eample 3..1 Find f (, ) 0 = (,) (1,0) + = g (, ) = 0. Thus, b the squeeze ( 1) f (, ) for f (, ) = ln. ( 1) + The reader will verif that the it along the paths = 1, = 0, = 1 is alwas 0. So, we suspect the it we want might be 0. We now use the squeeze theorem to tr to prove it. In other words, we need to find a function g (, ) such that f (, ) 0 g (, ) and we proceed as follows: f (, ) 0 = f (, ) ( 1) ln = ( 1) + = ( 1) ln ( 1) + ln g (, ) = 0. To find g, Since ln 0 as (, ) (1, 0), it follows b the squeeze theorem that f (, ) = 0. (,) (1,0)

15 1 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES There is another version of the squeeze theorem. As before, we suppose we are tring to find f (, ) given f (, ). (,) (a,b) Theorem 3.. If g (, ) f (, ) h (, ) for all (, ) ( 0, 0 ) in a disk centered at ( 0, 0 ) and if g (, ) = h (, ) = L then f (, ) = L. (,) ( 0, 0) (,) ( 0, 0) (,) ( 0, 0) Here, the diffi cult is to find the two functions g and h which satisf the inequalit and have a common it. We illustrate this with an eample. Eample 3..3 Does knowing that 4 4 cos 4 4 cos help ou with finding? If we divide the inequalit we have b, then we will have an inequalit involving the function for which we want the it. If the two outer functions in our new inequalit have the same it, then we will be done. Dividing each side of the given inequalit b which is positive (hence preserves the inequalit) gives us 4 4 cos that is 4 4 cos We compute. We cannot just plug in the point because we get 0 0. We will einate the absolute value b considering cases. case 1: > 0. In this case, = hence = ( ) = = = ( )

16 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.17 case : < 0. In this case, = hence = ( + ) = = = ( ) + in conclusion: = and since =, we are e- actl in the situation of the squeeze theorem. We conclude that. 3.. Limits with Maple To compute f (, ), use (,) (a,b) 3..7 Continuit it( f (, ), { = a, = b} ); Like in Calculus I, the definition of continuit is: Definition 3..4 A function f (, ) is said to be continuous at a point (a, b) if the following is true: 1. (a, b) is in the domain of f.. f (, ) eists. (,) (a,b) 3. f (, ) = f (a, b) (,) (a,b) Definition 3..5 If a function f is not continuous at a point (a, b), we sa that it is discontinuous at (a, b). Definition 3.. We sa that a function f is continuous on a set D if it is continuous at ever point in D. Thus, if we know that a function is continuous at a point, to find the it of the function at the point it is enough to plug-in the point. We now review rules and theorem which allow us to determine if a function is continuous at a point, or where a function is continuous. 4 4 cos =

17 18 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES Theorem 3..7 The following results are true for multivariable functions: 1. The sum, diff erence and product of continuous functions is a continuous function.. The quotient of two continuous functions is continuous as long as the denominator is not Polnomial functions are continuous. 4. Rational functions are continuous in their domain. 5. If f (, ) is continuous and g () is defined and continuous on the range of f, then g (f (, )) is also continuous. We now look at several eamples. Eample 3..8 Is f (, ) = continuous at (0, 0)? Where is it continuous? f (, ) is a polnomial function, therefore it is continuous on R. In particular, it is continuous at (0, 0). Eample 3..9 Where is f (, ) = + continuous? f is the quotient of two continuous functions, therefore it is continuous as long as its denominator is not 0 that is on R {(0, 0)}. Eample Where is f (, ) = 1 continuous? As above, f is the quotient of two continuous functions. Therefore, it is continuous as long as its denominator is not 0. The denominator is 0 along the parabola =. Therefore, f is continuous on { (, ) R }. ( ) Eample Find where tan 1 is continuous. + Here, we have the composition of two functions. We know ( that ) tan 1 is continuous on its domain, that is on R. Therefore, tan 1 will be continuous where + is continuous. Since + is the quotient of two polnomial functions, therefore it will be continuous as long as( its denominator ) is not 0, that is as long as. It follows that tan 1 + is continuous on { (, ) R }. + Eample 3..3 Find where ln ( + 1 ) is continuous. Again, we have the composition of two functions. ln is continuous where it is defined, that is on { R > 0}. So, ln ( + 1 ) will be continuous as long as + 1 is continuous and positive. + 1 is continuous on R, but + 1 > 0 if and onl if + > 1, that is outside the circle of radius 1, centered at the origin. It follows that ln ( + 1 ) is continuous of the portion of R outside the circle of radius 1, centered at the origin.

18 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.19 { Eample Where is f (, ) = + if (, ) (0, 0) 0 at (0, 0) continuous? Awa from (0, 0), f is a rational function alwas defined. So, it is continuous. We still need to investigate continuit at (0, 0). In an earlier eample, we found that + at (0, 0). did not eist. Therefore, f is continuous everwhere ecept { Eample Where is f (, ) = + if (, ) (0, 0) 0 at (0, 0) continuous? Awa from (0, 0), f is a rational function alwas defined. So, it is continuous. We still need to investigate continuit at (0, 0). In an earlier eample, we found that + that f is continuous everwhere. = 0. Therefore, f is also continuous at (0, 0). It follows Remark If the function of the eample we just did had been defined such that f (0, 0) = 1, then it would not have been continuous at (0, 0) since the value of the it at (0, 0) would not be the same as the value of the function Problems 3 1. Find Find (,) (3,4) Find sec tan (,) (0, π 4 ) 4. Find (,) (0,ln ) e e 5. Find sin sin. Find (,) (1,0) Find + (,) (1,1) + 8. Find (,) (1,1) Find 10. Find (,) (,0) 4 b first rewriting the fraction. b first rewriting the fraction. b first rewriting the fraction. b first rewriting the fraction.

19 0 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES ( ) Find P (1,3,4) z 1. Find P (3,3,0) ( sin + cos + sec z ) 13. Find P (π,0,3) ze cos 14. At what points in the plane is f (, ) continuous? (a) f (, ) = sin ( + ). (b) f (, ) = ln ( + ). 15. At what points in the plane is g (, ) continuous? (a) g (, ) = sin 1 (b) g (, ) = + +cos. 1. At what points in space if g (,, z) continuous? (a) f (,, z) = + z. (b) f (,, z) = At what points in space if g (,, z) continuous? (a) g (,, z) = sin 1 z. (b) g (,, z) = 1 +z B considering different paths, show that does not eist B considering different paths, show that + 0. B considering different paths, show that + 1. B considering different paths, show that + has it 0 along ever straight line approach-. Show that f (, ) = 4 + ing (0, 0). does not eist. does not eist. does not eist. 3. Does knowing that tell us anthing about 1 3 < tan 1 tan 1 < 1

20 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES.1 4. Does knowing that sin 1 1 tell us anthing about sin 1 5. Consider the function f (, ) = +. Suppose we want to find m (a) Along the path = m, the function becomes 1+m. Substituting m = tan θ, show how the value of f varies with the line s angle of inclination.so, the value does depend solel on θ. (b) Use the formula obtained in part a to show that the it along the line = m varies from 1 to 1 depending on the angle of approach.. Use polar coordinates to find Use polar coordinates to find + 8. Use polar coordinates to find tan 1 + +Y 3..9 Answers 3 1. Find Find + 1 (,) (3,4) 3. Find sec tan (,) (0, π 4 ) = = (,) (3,4) sec tan = 1 (,) (0, π 4 ) f (, ). 4. Find (,) (0,ln ) e (,) (0,ln ) e = 1

21 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES e 5. Find sin e sin = 1 sin. Find (,) (1,0) +1 sin (,) (1,0) + 1 = 0 7. Find + (,) (1,1) + 8. Find (,) (1,1) Find 10. Find (,) (,0) 4 b first rewriting the fraction. + = 0 (,) (1,1) b first rewriting the fraction. + = 1 (,) (1,1) 1 b first rewriting the fraction. + = b first rewriting the fraction. (,) (,0) 4 = 1 4 ( ) Find P (1,3,4) z ( 1 P (1,3,4) ) = 19 z 1 1. Find P (3,3,0) ( sin + cos + sec z ) ( sin + cos + sec z ) = P (3,3,0) 13. Find P (π,0,3) ze cos P (π,0,3) ze cos = 3

22 3.. LIMITS AND CONTINUITY OF FUNCTIONS OF TWO OR MORE VARIABLES At what points in the plane is f (, ) continuous? (a) f (, ) = sin ( + ). f (, ) is continuous on R. (b) f (, ) = ln ( + ). f is continuous for all (, ) ecept (0, 0). 15. At what points in the plane is g (, ) continuous? (a) g (, ) = sin 1. g (, ) is continuous for all (, ) not on the or -aes. + +cos. (b) g (, ) = g is alwas continuous. 1. At what points in space if f (,, z) continuous? (a) f (,, z) = + z. Continuous at all points since it is a polnomial. (b) f (,, z) = + 1 f is continuous outside the clinder of radius 1 along the z-ais. 17. At what points in space if g (,, z) continuous? (a) g (,, z) = sin 1 z. g is continuous as long as z 0. (b) g (,, z) = 1 +z 1. g is continuous as long as + z B considering different paths, show that does not eist. + Just follow the instructions. 19. B considering different paths, show that + Just follow the instructions. 0. B considering different paths, show that + Just follow the instructions. 1. B considering different paths, show that + Just follow the instructions. does not eist. does not eist. does not eist.. Show that f (, ) = 4 + has it 0 along ever straight line approaching (0, 0). Just follow the instructions (recall that a line through (0, 0) is of the form = m).

23 4 CHAPTER 3. FUNCTIONS OF SEVERAL VARIABLES 3. Does knowing that 1 3 < tan 1 tan < 1 tell us anthing about 1 tan Yes, 1 = 1 4. Does knowing that sin 1 1 tell us anthing about Yes, sin 1 = 0 sin 1? 5. Consider the function f (, ) = +. Suppose we want to find m (a) Along the path = m, the function becomes 1+m. Substituting m = tan θ, show how the value of f varies with the line s angle of inclination. We obtain m = sin θ 1 + m So, the value does depend solel on θ. (b) Use the formula obtained in part a to show that the it along the line = m varies from 1 to 1 depending on the angle of approach. This follows from the fact that 1 sin θ 1.. Use polar coordinates to find = 0 7. Use polar coordinates to find The it does not eist Use polar coordinates to find tan 1 + +Y tan Y = π? f (, ).