THE UNIVERSITY OF SYDNEY SCHOOL OF MATHEMATICS AND STATISTICS Solutions to Problem Sheet for Week MATH9: Differential Calculus (Advanced) Semester, 7 Web Page: sydney.edu.au/science/maths/u/ug/jm/math9/ Lecturer: Daniel Daners Material covered Real valued functions f(, y) of real variables. Natural domain and corresponding range of f(, y). The graph of f(, y). Level curves (more generally, cross-sections). Limits of functions of variables. Outcomes After completing this tutorial you should find the natural domain and corresponding range of a function f(, y); draw level curves, and sketch graphs of functions f(, y) in simple cases; calculate its of functions f(, y), or show that they don t eist. Summary of essential material Functions of two variables. of points Consider a function f D R where the domain D is a subset of R. The set graph(f) = {(, y, f(, y) ) (, y) D } R is called the graph of f and is a surface in R. Given c R the set { (, y) D f(, y) = c } is called the level set or level curve of f at height c. It is typically a proper curve, but can be empty (if c is not in the image of f) or degenerate such as a point or crossing curves. Limits of functions of two variables. We say that f(, y) = l if for every ε > there eists δ > (,y) (a,b) such that < ( a) + (y b) < δ f(, y) l < ε. Analogues of the one-variable it laws, including the Squeeze Law, hold for functions of two-variables. Useful techniques to determine its Use simple inequalities: y + y, sin a a, log ( > ) and others. Convert to polar coordinates: = r cos θ, y = r sin θ, it to (, ) is the same as r independently of θ. Useful ways to show that a it does not eist: Find two paths of approach giving two different its. Often along the aes, the diagonals or some parabola, depending of the structure of the epression of interest. The it along one particular path does not eist. Continuity of functions of two variables. at a point (a, b) in the domain D of f if A real valued function f(, y) of two real variables is continuous f(, y) = f(a, b). (,y) (a,b) Copyright 7 The University of Sydney
Questions to complete during the tutorial. The Taylor polynomial of order n for the eponential function e is T n () = + +! +! + + n n!. Substitute = iθ into this polynomial and show that T n (iθ) = C n (θ) + is n (θ), where C n (θ) and S n (θ) are the Taylor polynomials of order n of the cosine and sine functions respectively. Solution: We know that i =, i = i, i 4 =. If n is even let m = n and if n is odd let m = (n+) this means that T n (iθ) = + iθ θ! iθ! + + θ n in n! = ( θ! + θ4 4! + θ m ) ( )m + i (θ θ (m)!! + θ5 5! + θ m ) ( )m (m )! = C n (θ) + is n (θ) as claimed. This provides another connection between the real and the comple eponential function and motivation for the eponential notation e iθ.. Determine the natural domain and corresponding range of the following functions. (a) (b) f(, y) = y Solution: The domain is D = {(, y) y }. We claim that the range is [, ). For if (, y) D then f(, y) = y, and so range(f) [, ). On the other hand, if a then (a, ) D and f(a, ) = a, and so [, ) range(f). Hence equality holds. f(, y) = tan (y ) Solution: Since tan R ( π, π ), the domain of f(, y) is D = {(, y) } (this is all of R ecept for the y ais). We claim that the image (range) of f is im(f) = ( π, π ). To see this, note that im(f) ( π, π ), because tan R ( π, π ). Now note that f(, y) = tan y, and since tan R ( π, π ) is surjective we see that ( π, π ) range(f). (c) (d) f(, y) = + y y Solution: The domain is D = {(, y) + y and y } = {(, y) y }. These are the points in the first and fourth quadrants on and between the lines y = and y =. Observe that on the line y = in the domain, f(, y) =, which can take all values in [, ). On the line y = in the domain, f(, y) =, which can take all values in (, ]. Thus the range is R. f(, y) = sin ( + y) Solution: The domain is D = {(, y) + y }. These are the points lying in an infinite strip on and between the lines y = and y =. When + y =, f(, y) = sin ( ) = π. When + y =, f(, y) = sin () = π. The sin function is an increasing function and takes all values between π and π as the value of + y moves from to. Therefore the range is [ π, π ].. Sketch the level curves at heights c =, ±, ± for the functions f(, y): (a) y Solution: If f(, y) = c then + y = c. If c = then the level curve is empty (this is because is not in the range of f(, y)). If c = then the level curve consists of the single point (, y) = (, ). For c =,, the level curves are circles with radius,, (respectively).
y c = c = c = c = (b) 4 + y Solution: If c = then the level curve is a single point (, y) = (, ), and if c < then the level curve is empty. So suppose that c >. If f(, y) = c then ( c ) + y ( c) =, and this is the equation of an ellipse. The level curves at heights c =,, are shown in the figure. y c = c = c = (c) y Solution: The level curve f(, y) = c is y = + c. Thus the level curves are parabolas:
y c = c = c = c = c = (d) + y Solution: The level curve f(, y) = c is the straight line + y = c: y c = c = c = c = c = 4. Sketch the level curve of height z = for the function f(, y) = y + y 4. Solution: The domain is D = {(, y) R (, y) (, )}. The level curve is the set of (, y) D such that f(, y) =. Thus y = + y 4, and so (y ) =. Thus = y. We need to remember that (, ) is not in the domain of the function, therefore the level curve is a parabola with the origin removed: 4
y z = 4 5. Consider the function f(, y) = sin( + y ) + y, defined for (, y) (, ). Is it possible to define f(, ) so that f is continuous at (, )? Continuous, as in case of a function of one variable, means that the it eists and is equal to the function value. Solution: Using polar coordinates for and y, (that is, = r cos θ, y = r sin θ), we have sin( + y ) + y Since (, y) (, ) if and only if r, we see that = sin r r. sin( + y ) sin r = =. (,y) (,) + y r r Thus we can define f(, ) = to make f continuous at (, ). 6. Find the it, if it eists, or show that the it does not eist. y + y 5 (a) (,y) (,) y (b) Solution: By it laws, the it is ( + 5) ( ) = 5. (,y) (,) y + y Solution: Let f(, y) = ( y) ( + y ). Approaching (, ) along y = we have f(, ) = =, (c) and the it of this epression as does not eist. Thus f(, y) doesn t eist. (,y) (,) + y (,y) (,) + y Solution: + y (,y) (,) + y = ( + y ) = =. (,y) (,) + y (,y) (,) Etra questions for further practice 7. Find the domains and ranges, and describe the level curves, of the functions defined by: 5
(a) 4 y Solution: The domain is {(, y) + y 4} (a circle and its interior); range = [, ]. Level curves are either empty (at height k < or k > ), a single point (at height k = ) or circles (at heights k where k < ). (b) ( )(y + ) Solution: Domain = R ; range = R; level curves are hyperbolas in shifted first and third quadrants at positive heights and hyperbolas in shifted second and fourth quadrants at negative heights. At height, the level curve is the union of the lines = and y =. 8. Decide whether or not the following its eist. y (a) (,y) (,) + y 4 Solution: Approaching along y = we have Approaching along = y we have f(, ) = + 4 = +. f(y, y) = y4 y 4 =. Therefore the it does not eist. Here is an image of the surface, in which one particular path = y to the origin has been highlighted..5.5.5.5.5.5 (b) y + y sin + y 4 Solution: We use the Squeeze Law: (,y) (,) Since = we have (,y) (,) y + y sin + y 4 (,y) (,) y + y ( + y ) =. + y y + y sin + y 4 =. (c) y (,y) (,) + y 6
Solution: let f(, y) = y. Approaching (, ) along the -ais (that is, y = ), we have + y f(, ) = =, and approaching along the y-ais (that is, = ), we have (d) Therefore the it does not eist. y (,y) (,) + y f(, y) = y y =. Solution: We use the Squeeze Law. Using y + y (triangle inequality) gives: y + y + y + y = + y. Since + y we have (,y) (,) y +y =. Challenge questions (optional) 9. Use the ε, δ definition of the it of a function of two variables to show that (,y) (,) ( + y) =. Solution: We want to show that given any ε >, there eists a δ > such that < ( ) + (y ) < δ + y < ε. Note that the set of points (, y) satisfying the inequality < ( ) + (y ) < δ can be interpreted geometrically as the set of points in the interior of a circle with centre (, ) and radius δ, without the centre itself. We eamine the difference between + y and and try to write this in such a way as to incorporate terms in and y. + y = ( ) + + (y ) + = ( ) + ( ) + (y ) ( ) + + y To guarantee that +y < ε, we need only be sure that each of the three epressions ( ),, y is less than ε. Now as ( ) =, = and y y =, there eists δ > such that < < δ ( ) < ε (for eample, δ = ε ), there eists δ > such that (δ = ε 6), and there eists δ > such that < < δ < ε < y < δ y < ε (δ = ε ). Now choose δ to be the minimum of δ, δ,, δ. Then, whenever (, y) is a point inside a circle with centre at (, ) and radius δ (but not the centre itself), we can be sure that + y < ε. That is, given any ε >, there eists a δ > such that This proves the result. < ( ) + (y ) < δ + y < ε. 7