DOC PREVIEW
UIUC MATH 241 - Worksheet_021314_sol

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Thursday, February 13 ∗ Solutions ∗ Taylor series, the 2ndderivative test, and changing coordi-nates.1. Consider f (x, y) = 2 cos x − y2+ ex y.(a) Show that (0,0) is a critical point for f .SOLUTION:∂ f∂x|(0,0)= (−2sin x + yex y)|(0,0)= 0 and∂ f∂y|(0,0)= (−2y + xex y)|(0,0)= 0(b) Calculate each of fxx, fx y, fy yat (0,0) and use this to write out the 2nd-order Taylor ap-proximation for f at (0,0).SOLUTION:The second order Taylor approximation of a function f (x, y) at (0, 0) is given by T2(x, y) =f (0,0)+ fx(0,0)x + fy(0,0)y +(fxx(0,0)/2)x2+(fy y(0,0)/2)y2+ fx y(0,0)x y. For this problemwe have fxx= −2cos x + y2ex y, fy y= −2+ x2ex y, and fx y= ex y+ x yex y. So fxx(0,0) = −2 =fy y(0,0) and fx y(0,0) = 1. Also f (0, 0) = 3. So the second order Taylor approximation for fat (0,0) is g (x, y) = 3 − x2− y2+ x y.2. Let g (x, y) be the approximation you obtained for f (x, y) near (0,0) in 1(b). It’s not clear fromthe formula whether g , and hence f , has a min, max, or a saddle at (0,0). Test along severallines until you are convinced you’ve determined which type it is. In the next problem, you’llconfirm your answer in two ways.SOLUTION:Let’s test a general line y = mx which goes through (0,0) as x → 0. Then g (x,mx) = 3 − x2−m2x2+mx2= 3−(1−m+m2)x2. The polynomial 1−m +m2is always positive (it opens upwardand has its global minimum at m = 1/2 where 1−m+m2> 0). So g (x,mx) is always a downwardopening parabola. This suggests that (0,0) is a relative maximum.3. Consider alternate coordinates (u,v) on R2given by (x, y) = (u − v, u + v).(a) Sketch the u- and v-axes relative to the usual x- and y-axes, and draw the points whose(u, v)-coordinates are: (−1,2), (1,1), (1,−1).SOLUTION:If we express u and v in terms of x and y we get u = 1/2(x + y) and v = 1/2(y − x). So theu-axis is given in x and y coordinates by all multiples of the vector (1,1) and the v-axis isgiven by all multiples of the vector (−1,1). The two axes and the points are shown below.H-1,2LH1,1LH1,-1Luv(b) Express g as a function of u and v, and expand and simplify the resulting expression.SOLUTION:3 − x2− y2+ x y = 3 − (u − v)2− (u + v)2+ (u − v)(u + v) = 3 − (u2− 2uv + v2) − (u2+ 2uv +v2) + u2− v2= 3− u2− 3v2.(c) Explain why your answer in 3(b) confirms your answer in 2.SOLUTION:This is an elliptic paraboloid (in uv coordinates) opening downward with maximum at(0,0,3), so it confirms that (0,0) is a local maximum ( (0,0) goes to (0,0) under the trans-formation, so this reasoning makes sense).(d) Sketch a few level sets for g . What do the level sets of f look like near (0,0)?SOLUTION:The level sets are sketched for g = 2.7,2.8, 2.9 on the left and for f = 2.7,2.8,2.9on the right. The level sets for g are ellipses that approximate the level sets of f close to(0,0). The ellipses shrink as they get closer to g(x, y) = 3, which consists of the singlesolution (x, y) = (0,0).(e) It turns out that there is always a similar change of coordinates so that the Taylor series ofa function f which has a critical point at (0,0) looks like f (u, v) ≈ f (0, 0) + au2+ bv2. Infact this is why the 2ndderivative test works.Double check your answer in 2 by applying the 2nd-derivative test directly to f .SOLUTION:The Hessian fxxfy y− (fx y)2is (−2)(−2) − 12= 3 > 0 at (0,0) and fxx(0,0) = −2 < 0. So f hasa relative maximum at (0,0) as suspected.4. Consider the function f (x, y) = 3xey− x3− e3y.(a) Check that f has only one critical point, which is a local maximum.SOLUTION:fx= 3ey− 3x2and fy= 3xey− 3e3y. fy= 0 only if x = e2yand fx= 0 only if ey= x2.Solving these simultaneously we see that x must satisfy (x2)2= (ey)2= x, so x = 0,−1,or 1. But x = e2y> 0 so the only critical point is x = 1, y = 0. Calculating, we see thatfxx(1,0) = fy y(1,0) = −6 and fx y(1,0) = 3. So the Hessian fxxfy y− ( fx y)2= 36 − 9 = 27 > 0at (1,0). Since fxx(1,0) < 0, the second derivative test tells us that f (1,0) = 1 is a localmaximum.(b) Does f have an absolute maxima? Why or why not?SOLUTION:f does not have an absolute maximum. For instance if we take the trace curve y = 0 weget f (x,0) = 3x − x3− 1, which is unbounded as x → ∞. Absolute maxima and minima areonly guaranteed over a closed and bounded set in the domain. The plane R2is closed butnot bounded, so there is no guarantee that a continuous function will achieve an absolutemaximum or minimum over


View Full Document

UIUC MATH 241 - Worksheet_021314_sol

Documents in this Course
Notes

Notes

9 pages

16_05

16_05

29 pages

16.6

16.6

43 pages

16_07

16_07

34 pages

16_08

16_08

12 pages

16_09

16_09

13 pages

exam1

exam1

10 pages

exam2

exam2

7 pages

exam3

exam3

9 pages

15_03

15_03

15 pages

15_04

15_04

13 pages

15_04 (1)

15_04 (1)

13 pages

15_05

15_05

31 pages

15_10

15_10

27 pages

15_07

15_07

25 pages

15_08

15_08

12 pages

15_09

15_09

24 pages

15_10_B

15_10_B

8 pages

16_04

16_04

17 pages

14_01

14_01

28 pages

12_06

12_06

12 pages

12_05

12_05

19 pages

12_04

12_04

26 pages

Lecture1

Lecture1

31 pages

Lecture 9

Lecture 9

41 pages

Lecture 8

Lecture 8

35 pages

Lecture 7

Lecture 7

40 pages

Lecture 6

Lecture 6

49 pages

Lecture 5

Lecture 5

26 pages

Lecture 4

Lecture 4

43 pages

Lecture 3

Lecture 3

29 pages

Lecture 2

Lecture 2

17 pages

m2-1

m2-1

6 pages

-

-

5 pages

Load more
Download Worksheet_021314_sol
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Worksheet_021314_sol and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Worksheet_021314_sol 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?