REVIEW PROBLEMS FOR MIDTERM I MATH 2373, SPRING 2015 ANSWER KEY

REVIEW PROBLEMS FOR MIDTERM I MATH 2373, SPRING 2015 ANSWER KEY Problem 1 Standing in line at the supermarket I see Alice, Bob and Carol ahead of me in the express check-out lane. Alice buys 2 bags of chips, 3 diet sodas, 1 bag of circus peanuts and spends $5.54. Bob buys 3 bags of chips, 1 diet soda, 3 bags of circus peanuts and spends $7.13. Carol buys 1 bag of chips, 1 diet soda, 1 bag of circus peanuts and spends $2.97. Find the price for a bag of chips, for a diet soda and for a bag of circus peanuts. You may use your calculator for the numerical work. Use the x A 1 b method of solution. Identify A, x and b clearly. A x x x price of chips y price of diet soda z price of circus peanuts 2 3 1 3 1 3 1 1 1 x y z x y z 2 3 1 3 1 3 1 1 1 5.54 7.13 2.97 1 5.54 7.13 2.97 Alice s bill Bob s bill Carol s bill.79.89 1.29. b Remark. Another way to solve these equations, different from the method we asked for in this problem, is the following: and hence rref 2 3 1 5.54 3 1 3 7.13 1 0 0.79 0 1 0.89, 1 1 1 2.97 0 0 1 1.29 price of chips price of diet soda price of circus peanuts.79.89 1.29. Date: Feb. 18, 2015. 1

2 REVIEW FOR MT1: ANSWER KEY Solve the system of equations Problem 2 2x y +5w 1 2x +z +w 3 using the standard method taught in class, which means: write out the augmented matrix for the system, reduce the augmented matrix to reduced row echelon form, and present the final answer in the format of Example 7 on pp. 139-140 (see the next-to-last displayed line in the example). Do your work entirely by hand. Explain each row operation using notation like R 2 R 2 3R 3, as in the textbook. The augmented matrix of the system is 2 1 0 5 1 2 0 1 1 3. Applying R 1 R 1/2, we get 1 1/2 0 5/2 1/2 2 0 1 1 3 Applying R 2 R 2 2R 1, we get 1 1/2 0 5/2 1/2 0 1 1 6 4 Applying R 2 R 2, we get 1 1/2 0 5/2 1/2 0 1 1 6 4 which is in row echelon form. Finally, applying R1 R 1 R 2 /2, we get 1 0 1/2 1/2 3/2, 0 1 1 6 4 which is in reduced row echelon form. The last matrix above corresponds to the pair of equations.. x +z/2 +w/2 3/2 y z 6w 4 The variables z and w are free since they appear in columns without pivots in the reduced row echelon matrix. Let Then z s and w t., x s/2 t/2 + 3/2 and y s + 6t 4.

REVIEW FOR MT1: ANSWER KEY 3 The final answer written in the officially acceptable manner is x s/2 t/2 + 3/2 1/2 1/2 y z s + 6t 4 s s 1 1 + t 6 0 + w t 0 1 where s and t may take any (real) values. 3/2 4 0 0, Problem 3 The following matrix is not quite in row echelon form. 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 1 0 1 1 0 0 0 1 0 0 0 (i) Reduce the matrix to row echelon form. (At most two more steps are required.) Clearly identify which matrix is supposed to be in row echelon form. (ii) Then reduce the matrix all the way to reduced row echelon form. The last matrix you write down should be exactly in reduced row echelon form. Furthermore, every step of row-reduction should be explained by using notation like R 2 R 2 3R 1, as in the textbook. (We are only asking for you to do row operations. You are not being asked to solve any equations.) 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 1 0 1 1 0 0 0 1 0 0 0 R4 R 4 R 3, R4 R 4 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 1 0 1 1 0 0 0 0 0 1 1 (given matrix) (row echelon form) R3 R 3 R 4, R2 R 2 R 4, R1 R 1 R 4 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 R2 R 2 R 3, R1 R 1 R 3, R1 R 1 R 2 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 (reduced row echelon form)

4 REVIEW FOR MT1: ANSWER KEY Problem 4 Expand the following determinant along the third row. You need not evaluate the resulting 3 by 3 determinants, just get all the numbers in the right places, and get all the signs right. 2 1 6 7 3 4 1 0 12 6 0 8 5 11 4 10 12 1 6 7 4 1 0 11 4 10 checkerboard: ( 6) + + + + + + row 3 + + 2 6 7 3 1 0 5 4 10 + 0 8 2 1 6 3 4 1 5 11 4 Solve the initial value problem Problem 5 t dy dt 4y + t5 sin t and y(1) 2. We start by rewriting the equation in the standard form and then calculate in standard fashion: dy dt 4 t y t4 sin t ( ) exp 4 dt/t exp ( 4 log t) t 4 d dt t 4 y sin t t 4 y cos t + C 2 y(1) cos(1) + C C 2 + cos(1) t 4 y cos t + 2 + cos(1) y t 4 cos t + (2 + cos(1))t 4.

REVIEW FOR MT1: ANSWER KEY 5 Problem 6 Solve the initial value problem dy dt y + 1 and y(0) 3. 2t + 8 dy y + 1 dt 2(t + 4) log y + 1 1 log t + 4 + C 2 y + 1 e C t + 4 1/2 y + 1 C 1 (t + 4) 1/2 (C 1 ±e C, t + 4 > 0) 4 y(0) + 1 4 1/2 C 1 C 1 2 y + 1 2(t + 4) 1/2 y 2(t + 4) 1/2 1. using old math professor s trick We always get rid of the absolute value bars the same way. So instead of repeating those moves every time, we organize the calculation as follows: dy y + 1 dt 2(t + 4) log(y + 1) 1 log(t + 4) + ln C 2 y + 1 C(t + 4) 1/2 4 y(0) + 1 4 1/2 C C 2 y + 1 2(t + 4) 1/2 y 2(t + 4) 1/2 1. It is understood we will not attempt to evaluate the square root by plugging in negative numbers. The old math professor s trick is just a shorthand to skip over some steps that are always the same when you are getting logarithms on one or both sides of the equation after integrating. Problem 7 Find k such that the following system of homogeneous equations has infinitely many solutions: 3 4 7 2 3 5 1 k 2 x y z 0 0 0.

6 REVIEW FOR MT1: ANSWER KEY It is the same thing to solve the equation 2 3 5 3 4 7 1 k 2 0 because in general, for a square matrix A, the system of equations A x 0 has infinitely many solutions if and only if det A 0. Using basketweave to evaluate the left side we get the equation 2 4 ( 2) + 3 7 1 + ( 5) ( 3) k ( 5) 4 1 3 ( 3) ( 2) 2 7 k 0 which simplifies to Thus the solution is k 7. Find a, b, c, d, w, x, y and z. 2 1 0 0 0 0 1 5 2 0 0 0 0 2 10 6 0 0 0 0 6 8 2 0 0 0 0 2 2 1 0 0 0 0 1 1 7 + k 0. Problem 8 1 1 1 2 3 6 d 1 2 4 6 12 12 2 4 9 w x 27 3 6 y z 41 41 a 12 27 41 83 83 b 12 c 41 83 84 You must do all the work by hand, and justify the steps. Note: Okay, I admit that I went nuts making up this puzzle. 0 0 0 6 8 2 0 1 1 2 3 a b 0 0 0 0 0 1 1 12 24 2a 12 2a a 6. 1 1 2 3 a b a + b b 6.

REVIEW FOR MT1: ANSWER KEY 7 0 0 0 0 0 1 1 0 2 1 0 0 0 0 0 1 5 2 0 0 0 0 1 5 2 0 0 0 3 6 w z 41 41 6 12 x 41 41 83 2 4 9 y 27 c d 12 27 41 83 84 (watch out: next pitch is a curveball!) 1 0 2 10 6 0 0 0 0 0 0 2 2 1 27 + c c 27. 2d + 12 d 6. 3 + 30 2w 27 2w w 27/2 6 + 60 2x 54 2x x 27. 2 4 9 y 27 c 0 81 6y y 27 2 3 6 w z 41 41 8 + 90 6y 82 6y Finally, here s the puzzle with the blanks all filled in: 2z+82 41 2z+41 z 41 2.

8 REVIEW FOR MT1: ANSWER KEY 2 1 0 0 0 0 1 5 2 0 0 0 0 2 10 6 0 0 0 0 6 8 2 0 0 0 0 2 2 1 0 0 0 0 1 1 1 Problem 9 1 1 2 3 6 6 1 2 4 6 12 12 2 4 9 27/2 27 27 3 6 27/2 41/2 41 41 6 12 27 41 83 83 6 12 27 41 83 84 (i) Find the general solution of the autonomous differential equation y (1 + y)(5 y). (ii) Also find the solution which satisfies the initial condition y(0) 7. (iii) Also find the equilibrium solutions. (iv) Determine the stability of each equilibrium solution. (v) Sketch the direction field for the differential equation well enough to justify your determination of stability. First find the general solution. Start by separating variables dy (1 + y)(5 y) dt Partial fractions: 1 (1 + y)(5 y) A y + 1 + B y 5 Multiply on both sides by (y + 1)(y 5) to get 1 A(y 5) + B(y + 1) Plug in y 1 to get A 1/6. Plug in y 5 to get B 1/6. We can rewrite the separated differential equation as 1 dy 6 y + 1 1 dy 6 y 5 dt Clear denominators: dy y + 1 dy y 5 6 dt Integrate on both sides and use the old math professor s trick: Solve for y: log(y + 1) log(y 5) 6t + ln C y + 1 ( ) y 5 Ce6t y + 1 yce 6t 5Ce 6t y(1 Ce 6t ) 5Ce 6t 1 y 5Ce6t + 1 Ce 6t 1

REVIEW FOR MT1: ANSWER KEY 9 Next we solve the indicated initial value problem. Plug t 0 in earlier line marked with ( ) to get 7 + 1 7 5 4 C. Thus y 20e6t + 1 4e 6t 1 is the solution satisfying y(0) 7. Equilibrium solutions are y 1 and y 5. (The numbers 1 and 5 are just the roots of (1 + y)(y 5) 0.) Looking at the dfield8 plot, we can see that y 5 is stable and y 1 is unstable. Using dfield8 here was overkill. Any sketch of the direction field which establishes where the slopes are positive and where negative and is halfway legible is good enough. 10 Figure 1. Direction field y (1 + y) (5 y) 8 6 4 y 2 0 2 2 1 0 1 2 3 4 5 6 t Problem 10 Find the least squares line y mx + b for the following data points. (1, 4), (3, 8), (4, 9), (6, 12). Do so by first expressing the two equations for the unknowns m and b in matrix form. You can do the numerical work with your calculator.

10 REVIEW FOR MT1: ANSWER KEY (also see the remark immediately following). The quick (and easy-to-remember) way to write down the equations is like this: The equations that we wish would hold exactly are these: 1 1 3 1 4 1 6 1 m b These wished-for equations would hold were the line y mx + b to pass through all four data points exactly. But of course there is no such line since we live in the real world. To do the best we can (as measured by the sum of the squares of the errors) we multiply by the transpose of the coefficient matrix to get a system of equations we can actually solve: 1 1 4 1 3 4 6 3 1 m 1 3 4 6 1 1 1 1 4 1 8 b 1 1 1 1 9 6 1 12 4 8 9 12 After multiplying out using MATLAB we get 62 14 m 136 14 4 b 33. Finally, 1 m 62 14 136 1.5769, b 14 4 33 2.7308 so the desired line fit to the data by the method of least squares is y 1.5769x + 2.7308..

REVIEW FOR MT1: ANSWER KEY 11 Figure 2. Least squares line and data points 15 y1.5769 x+2.7308 10 y 5 0 0 1 2 3 4 5 6 7 8 9 10 x (We just included this figure as an illustration. It is not an official part of the answer to this line-fitting problem.) Remark. One is not obliged always to use the same letters and arrangments of letters in order solve problems. The same idea can be logically and correctly developed using various choices of notation. But I do understand that it might have caused some students confusion that I did not reproduce the textbook notation exactly. Sorry about that. To follow the book exactly I should look for a least squares line in the form y k + mx. The equations I need to solve then take the form 1 1 1 1 1 3 4 6 1 1 1 3 1 4 1 6 k m After multiplying out using MATLAB we get 4 14 k 33 14 62 m 136 Finally, k 4 14 m 14 62 Thus the least squares line is 1 33 136 1 1 1 1 1 3 4 6 y 2.7308 + 1.5769x,. 2.7308 1.5769 same as before. (Half a dozen of one, six of the other.). 4 8 9 12

12 REVIEW FOR MT1: ANSWER KEY Problem 11 The rate at which the amount of a radioactive substance changes is proportional to the amount present. At a certain time, a sample of a certain radioactive substance contains 40 mg. The same sample only contains 25 mg of the substance 50 years later. (i) How much of the substance will remain after 120 years? (ii) What is the half-life of the substance? Start with the differential equation. Let y denote the amount of the substance in milligrams and let t be time in years measured from the time observations of the substance began. We are given that dy ky, y(0) 40, y(50) 25. dt Here k is a negative constant we need to determine. The general solution of the differential equation is y Ce kt. We have C 40 because y(0) 40. We have 40e 50k 25 because y(50) 25. Thus e 50k 25/40 5/8 and hence y 40(5/8) t/50. Some students may prefer to solve for k, so we present that route as well. From, say, the equation e 50k 5/8 above we get k ln(5/8) 50 0.009400. We then have in various ways the (same!) conclusion y(120) 40(5/8) 120/50 40 exp( (0.009400)(120)) 12.947mg. which answers part (i) of the question. To answer part (ii) of the question we have to calculate the half-life T of the substance. This is determined, say, by the equation y(t ) 40(5/8) T/50 40 2 20 40e 0.009400T or equivalently 1 2 (5/8)T/50 e ln(5/8) 50 T e 0.009400T. Solving the latter by precalculus-level manipulation of logarithms, we have T 50 ln(1/2) ln(5/8) ln(1/2) 73.738 years. 0.0094 I solved this problem two ways but in practice of course you only have to pick one way of solving and stick with it.

REVIEW FOR MT1: ANSWER KEY 13 Problem 12 Initially an 800 gallon tank contains 400 gallons of water in which is dissolved 50 lbs of salt. Brine containing 4 lbs of salt per gallon enters the tank at the rate of 5 gallons per minute. The well mixed brine leaves the tank at the slower rate of 3 gallons per minute. (i) Find an expression for the number of pounds of salt in the tank at time t, until the tank overflows. (ii) How much salt is in the tank at the moment the tank overflows? Let y y(t) be the amount (lbs) of salt in the tank at time t (min). We have to solve the initial value problem y(0) 50 and dy dt 4 lbs gal 5 gal y lbs min 400 + (5 3)t gal 3 gal. min }{{}}{{} rate in rate out After simplifying algebraically we get y(0) 50 and dy dt + 3y 400 + 2t 20. We now use the integrating factor method to finish up. ( ) ( ) 3 dt 3 dt exp exp (t + 200) 3/2 400 + 2t 2 200 + t d dt (t + 200)3/2 y 20(t + 200) 3/2 (t + 200) 3/2 y 8(t + 200) 5/2 + C 200 3/2 50 8 200 5/2 + C C 200 3/2 50 8 200 5/2 4, 384, 062. y 8(t + 200) 4, 384, 062/(t + 200) 3/2. The tank overflows when 400 + 2t 800 and hence t 200 minutes. Plugging in t 200 we get y(200) 8(200 + 200) 4, 384, 062/(200 + 200) 3/2 2652 pounds of salt in the tank at the moment of overflow. Note: On an exam, to answer just part (i) of the question, we would be okay with you writing, say, C 200 3/2 50 8 200 5/2 without simplifying further. Whether or not to simplify C further would depend on what further questions you might be asked.

14 REVIEW FOR MT1: ANSWER KEY Problem 13 A steel ball with an initial temperature of 60 o F is placed in an oven which is maintained at 420 o F. After 4 minutes the temperature of the ball is 180 o F. (i) Find an expression for the temperature of the ball at time t. (ii) Find the surface temperature of the ball after 12 minutes. (iii) Find the time at which the ball will have a temperature of 350 o F. Start with the differential equation. Let y y(t) be the temperature in degrees Fahrenheit of the ball at time t in minutes after being placed in the oven. We know that dy k(420 y) dt by Newton s law of heating, where k > 0 is a constant we have to determine using the information given in the problem. We are also given that y(0) 60 and y(4) 180. We solve using the integrating factor method. We start by rewriting the differential equation as dy dt + ky 420k. The integrating factor is ekt. We then have d dt ekt y 420ke kt e kt y 420e kt + C y 420 + Ce kt. To get the constant C we use the initial value information: Thus we have y(0) 60 420 + C C 360. y 420 360e kt. To answer part (i) of the question, we are going to replace e kt by an expression in which the only variable is t. One way to do that is to solve for k and another way is to use algebra to get such an expression. We will show both ways, and we will show below how to pursue both ways to answer parts (ii) and (iii) of the question as well. We have 180 y(4) 420 360e 4k e 4k 240 360 2 3 e kt ( ) 2 t/4. 3 It follows that ( ) 2 t/4 y 420 360, 3 which answers part (i) of the question. We can also solve the equation e 4k 2 3 for k, obtaining k 1 ( ) 2 4 ln 0.10137. 3

REVIEW FOR MT1: ANSWER KEY 15 Then we get another (equivalent) answer for part (i), namely We then have y(12) 420 360 y 420 360e 0.10137t. ( ) 2 12/4 420 360 exp( 0.10137 12) 313.3. 3 which answers part (ii) of the question both ways. Finally, to find the time T asked for in part (iii) we have to solve the equation y(t ) 420 360 This can be rewritten 7 36 ( ) 2 T/4 420 360 exp( 0.10137T ) 350. 3 ( ) 2 T/4 exp( 0.10137T ). 3 Solving by the usual precalculus methods, we get T 4 ln(7/36) ln(2/3) ln(7/36) 16.155 minutes, 0.10137 which answers part (iii) of the question (in both ways). Remark. Just so there is no misunderstanding: I solved the problem two different ways. When you are solving problems like this you only have to pick one of the ways and then stick with it. Find the determinant Problem 14 1 1 1 1 1 0 0 1 0 3 0 1 2 1 2 1 by hand. Use whatever combination of methods seems the easiest. Explain each step of your calculation by a word or phrase.

16 REVIEW FOR MT1: ANSWER KEY 1 1 1 1 1 0 0 1 0 3 0 1 (use the row operation R4 R 4 + 2R 1 ) 2 1 2 1 1 1 1 2 1 0 0 1 0 3 0 1 (expand on third column) 0 3 0 1 1 0 1 0 3 1 (expand on first column) 0 3 1 3 1 3 1 3 3 6. Problem 15 Find the inverse of the following matrix. 2 1 3 1 1 1 2 2 1 You must show all work. You must use the textbook notation for row operations to explain each step. Check at least three times that you wrote out the correct starting matrix for your calculation. 2 1 3 0 1 0 1 1 1 1 0 0 2 2 1 0 0 1 1 1 1 1 0 0 0 1 1 2 1 0 0 0 1 2 0 1 1 0 0 5 1 2 0 1 0 4 1 1 0 0 1 2 0 1, R 2 R 2 + 2R 1, R 3 R 3 2R 1, R 3 R 3, R 2 R 2 + R 3, R 1 R 1 R 3, R 1 R 1 R 2, hence: 2 1 3 1 1 1 2 2 1 1 5 1 2 4 1 1 2 0 1.

Solve the system of equations REVIEW FOR MT1: ANSWER KEY 17 Problem 16 5w z 7, 4z 2y + w 9, 2y + w z 17, x + y + z + w 14 for z using Cramer s Rule. You are expected first to rewrite the system of equations neatly, and then to express z as a ratio of 4-by-4 determinants. You do not have to evaluate any determinants. The emphasis is on demonstrating that you know the pattern of Cramer s Rule. I will write w, x, y, z in alphabetical order throughout the whole solution. Any order is okay, but once chosen, you have to stay committed to that choice of order to the bitter end. Note that the equations were written sloppily. You have to begin by rewriting the equations neatly, with each variable placed in its own column. 5w z 7 w 2y +4z 9 w +2y z 17 w +x +y +z 14 Having written things out neatly, I can read off the required ratio of determinants easily: z 5 0 0 7 1 0 2 9 1 0 2 17 1 1 1 14 / Problem 17 Find the inverse of the following matrix: 4 1 17 5 5 0 0 1 1 0 2 4 1 0 2 1 1 1 1 1 Express all the entries of the inverse matrix as fractions, not as decimals. Solve this problem entirely by hand. Show all your work. 4 1 17 5 1 5 1 17 4 4 1 17 5 5/3 1/3 17/3 4/3.