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Lecture 13, 14, and 15

by: AnnMarie

Lecture 13, 14, and 15 Math 240


GPA 3.028
Jonathan B Walters

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This weeks of notes include the information discussed in class. Chapter 4.6 Chapter 4.7 Chapter 10.1 Chapter 10.2
Jonathan B Walters
Class Notes
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This 15 page Class Notes was uploaded by AnnMarie on Wednesday October 14, 2015. The Class Notes belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 38 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.

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Date Created: 10/14/15
46 Modeling with Exponential Functions Exponential Growth and Decav Populations radioactive materials ect Eduation that models Exponential Growth and Decay At A06 where At amount after time A0 initial amount r rate of growthdecay per year time t me Example An initial bacteria count is 500 Rate of growth is 40 per hour A Find the modeling equation B Find the amount of bacteria after 10 hours C How many hours will it take to reach 80000 bacterium Solution a At SOOeO4t b Alo 50060410 AIO 50064 This will be the acceptable answer on the exam since calculators are not allowed C 80000 50060 80000 04t 500 T e Zn160 04t ln160 t T 04 Definition The halflife of radioactive material is the amount of time it takes any given amount of the substance to decay to half of that given amount Example The halflife of Strontium90 is 28 years How long will it take a 60mg sample to decay to 20mg Solution At 606 We know that after 28 years we are left with 30mg 30 606287 E 28739 60 e mg 28 1n5 7 T 28 Using r we can find the time it takes to get 20mg from a 60mg sample At oe g t 20 60 8521 E 2 MI 60 e 28 1 zn5 1713 Ti t 281n3l 1716 Newton s Law of Cooling Tt TS T0 foe kt where Tt Temperature of object after t time TS Temperature of Surrounding area To lnitia Temperature k constant depending on object t me Example A dead body was found at midnight and was measured to be at 92 F One after the body was found the temperature was 90 F Assuming that the person was 986 F when they died If the air temperature was 60 F how long has the person been dead When did the body reach 94 F Solution We know that TS 60 F and T0 986 F To find k 90 60 386e1k 9O 60 3866 i 1k 386 e 14 k k 14 a 92 60 386e1quott 92 60 386e1quot 32 386e1quott i nit 6 386 14 lnt t 14 14 t 074 hrs meaning the body was found about 1115pm b 94 60 386e1nt it 94 60 38661 7386 34 386e1nt i nit 386 e 38 6 14 lnt t 14 14 t 054 hrs meaning it takes about 32 minutes for the body to reach 94F 47 Logarithmic Scales pH Scale The pH value of a solution is a negative common logarithm of the hydrogen ion H1 concentration C measured in moles per liter pH logC or c 10 Example Find the concentration when pH 34 Solution 010 3394 Richter Scale The magnitude M of an earthquake is defined to be M log where l is the intensity of the earthquake S is intensity of a common earthquake 100km away Ratio is what matters Example One earthquake is 33 times more intense than another Find how much larger the magnitude is on the richter scale Solution 12 3311 1 1 M2 M1 log3 log3 331 M2M1 21057T1 X 1E1 M2 M1 log33 Before I begin I would like to say that these notes are not the same as the lecture material that was provided in class on October 16 for l was not in class that day However through communication with Mr Walters who advised that the material that was covered was Chapter 103 and Chapter 104 Chapter 103 Matrixes and Systems of Linear Equations Matrix An m X nmatrix is a rectangular array of numbers with m rows and n columns 6 11 6 12 6 13 am m 6 21 6 22 6 23 a2n r O 6 31 6 32 6 33 a3n W S aml am2 am3 amn n columns used a table to show the relationship of rows and columns in a matrix As you can see a matrix will have m rows depending on how many equations or types you have You will have n rows depending on how many variables you have You can also write a system of linear equations by putting them into matrix form this is known as augmented matrix Example Write the following linear system of equations into augmented matrix form 6x 2y z4 x3y1 7yz5 Solution I first lined the equations up as shown below Then insert coefficients and constants into a matrix 6x 2y z4 6 2 1 4 x3yOz1 1 3 0 1 Ox7yz5 lo 7 1 5 Now that we know how to create augmented matrices We use the augmented matrix to solve Hnearequa ons Elementary Row Operation 1 Add a multiple of one row to another 2 Multiply a row by a nonzero constant 3 Interchange two rows l have included some notation that you will want know to show the symbols of row operations Symbols Description Ri kRj gt R Change the ith row by adding k times rowj to it and put it back into row i le Multiply the ith row by k R lt gt RI Interchange the ith and the jth rows Example Solve the following system of linear equations x y3z4 x2y 2210 3x y5214 Solution 1 1 3 4 1 2 2 10 Firstwrite the system oflinearequations into augmented matrixform l3 1 5 14 RZ Rl JB2 1 1 3 4 lo 2 4 2 42123 1 1 3 4 lo 3 5 6 lo 1 2 1 lo 0 1 3 lo 1 2 1 RZlt gtR3 1 1 3 4 lo 1 2 1 0013 x y3z4 y 22 1 Change the augmented matrix back into linear equation form 23 Using backsubstitution we can find the solution of the system of linear equations y 261 x 73G4 y 61 x 794 y16 x24 y7 x4 2 x2 Thus the solution is 273 for the system of linear equations RowEchelon Form A matrix is in rowechelon form if it meets the following conditions 1 The first nonzero number in each row reading left to right is 1 a This is called the leading entry 2 The leading entry in each row is to the right if the leading entry in the row immediately aboveit All rows consisting entirely of zeros are at the bottom of the matrix 4 Every number above and below each leading entry is zero a This is also known as reduced rowechelon form 9 Solvinq Svstems of Linear Equations with Gaussian Elimination 1 Augmented Matrix Form a Write the system of linear equations into the Augmented Matrix Form 2 RowEchelon Form a Use the elementary row operations to get the augmented matrix into RowEchelon Form 3 BackSubstitution a Use backsubstitution to solve the system of linear equations Example Solve the following system of linear equations using Gaussian Elimination 4x 8y 42 4 3r8y52 11 2xgr122 17 Solution 4 8 4 4 3 8 5 11 2 112 17 12 1 1 zzle 3 8 5 11 I 2 112 17 0510 15 12 1 1 lsz 01 4 7 0 510 15 H 2 1 1 o 010 20 H 2 1 1 1 10R3 m 1 4 7 IO 0 1 2 Write the system of linear equations from the rowechelon form x 2y Z 1 y 42 7 z 2 Using backsubstitution the solution is 3 1 2 Solutions ofA Linear Svstem in RowEchelon Form 1 No Solution gtO C where C is a nonzero constant 2 One Solution gt Each variable is a leading variable 3 Infinitely Many Solutions gtThe last row is all zeros Chapter 104 The Aloebra of Matrices Equality of Matrices The matrices A aij and B bij are equal if and only if they have the same dimensions m gtlt n and corresponding entries are equal that is fori 12 mandj 12 Sum Difference and Scalar Pr0perties of Matrices Let A an and B bij be matrices of the same dimension m gtlt n and let 0 be any real number 1 The sum ofA B is the m gtlt n matrix obtained by adding corresponding entries ofA and B ABh w 2 The difference ofA B is the m gtlt n matrix obtained by subtracting corresponding entries ofA and B ABM 3 The scalar product cA is the m gtlt n matrix obtained by multiplying each entry ofA by c cAcaU ExampleLet A23 B10 C730 D606 O5 31 O15 819 7 I22 carry out each indicated operation or explain why it cannot be performed a AB bCD cCA d5A Solution 23I 10 33I a ABO 5 31 3 6 I7 2 I22 I9 b CD730 606 136 O15 819 804 c CA is undefined because the matrices are not the same 2 3 1045 d 5A5O5 o 25 7 I35 Prooerties of Addition and Scalar Multiplication of Matrices Let A B and C be m x n matrices and let c and d be scalars 1 Commutative Property of Matrix Addition A B B A 2 Associative Property of Scalar Addition ABCABC 3 Associative Property of Scalar Multiplication cdA ch 4 Distributive Property of Scalar Multiplication cd A cA dA cAB cA cB Example Solve the matrix equation 2x A B for the unknown matrix x where A 23 B 4 H 51 Solution 2x AB gt 2xBA gt BA 2x4141 51 13 2x62 44 x 62 x31 44 22 Matrix Multiplication If A an is an m x n and B bu is an n x k matrix then their product is m x k matrix C cu where cij is the inner product of the ith row ofA and the jth column of B We write the product as CAB 1 3 1 5 2 Example Let A 1 O and B O 4 7 calculate if possible the products AB and BA Solution AB13152 10O4 7 AB 10 512 221 10 5020 AB11723 1 5 2 BA is undefined because the dimensions of B and A are 2 X 3 and 2 X 2 105 Inverses No matrix divison but I100HX X 010HYIYI 001HZ Z Identitv Matrices 2 X 2 m X n with 1 on diagnal and O on everything else 3X3 Definition Let A be an m X n matrix If there is an matrix A391 with A39lA ln X m 2 AA391 that we say A391 is the inverse of A Finding Inverses 1 Augment matrix A with the identity of same size 2 Use row operations to reduce LHS left hand side to identity 3 RHS right hand side will be A39l Example Find the Inverse of the following matrix 21 Solution 1st 4510 2nol iR11O 3thR2 2R11 j 4i 0 2301 2 301 0 1 4th2R2 5th 1 3R210 A1gt 5 4 1 1 01 2 2 Example Find the inverse of the following matrix 124 236 3615 Solution 124100 R22Rlll241OO R312410 236010 R33R1012210 01221 3615001 003301 00110 R12R2100320 R22R3100320 320 012210 01041A 141 00110 00110 10 113IIXII4I Now if 1 2 2 y10 wecan solvewith 315z 14 4 1 2 x 4 1 24 161028 2 A39lzll j 2 gt 7 y1 j 2 1022 20 357 3 3 1 z 1 14 141021 3 nd 0 O 103 Determinates and Cramer s Rule Determinates of a 2 X 2 matrix Aab c d detAAa b adbc ICdI Example Find the det of 2 3 5 4 Solution 24 53 8 15 Determinates of a 3 X 3 matrix I 01 a2 03 I A 04 05 06 detA A 0105 19 0608 020409 0507 030408 0507 390708 9 Exam ple Calculate the following matrix 2 1 3 A 7 1 2 detAA 2 4 4 1 28 2 314 1 1 2 4 2 8 1 30 313 1630 39 15 It turns out that A matrix is invertable iff its determinant is not 0 This means A b has a unique solution iff detA 75 0 Cram er s Rule x1 dedAg The solutions to A b is given by Xi detA where 2 x2 and A is the matrix owahose x3 ith folumn has been replace by E 1 IX n Example Use Cramer s Rule to solve the following system of equations xZy10z 6 2xy4z 3 3X 4z7 Solution 1210 6 A214 E3 l3 0 4 7 o 140 28 121003 4 830 2 42 6 2 10 ox 3 1 4 64O212281007 7 0 4 24327o 14 Q 14 l D 42 3 1 6 10 Dy 2 3 4 3 7 4 11228681210149 4o24230 294 By 294 D 42 7 1 2 6 DZ 2 1 3 3 0 7 17O21496O3 21 Q 21 1 D 42 2 The solution is 13 7 12


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