Math 107: Calculus II, Spring 2015: Midterm Exam II Monday, April Give your name, TA and section number:

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1 Math 7: Calculus II, Spring 25: Midterm Exam II Monda, April 3 25 Give our name, TA and section number: Name: TA: Section number:. There are 5 questions for a total of points. The value of each part of each question is stated. 2. Do not open our booklet until told to begin. The exam will be 5 minutes long. 3. You ma not use phones, calculators, books, notes or an other paper. Write all our answers on this booklet. If ou need more space, ou can use the back of the pages. 4. Unless specified otherwise, ou must show ALL our working and explain our answers clearl to obtain full credit! 5. Read the questions carefull! Make sure ou understand what each question asks of ou. Please read the following statement and then sign and date it: I agree to complete this exam without unauthorized assistance from an person, materials, or device. Signature: Date: Question Points Score Total: i

2 . Let X be a discrete random variable with range {,, 2, 3, 4} and probabilit mass function P (X ) 9, P (X ) 2 9, P (X 2) 3 9, P (X 3) 2 9, P (X 4) 9. For the following questions, carr out the computations and leave our answer as an irreducible fraction. (a) (2 points) Find E(X) and var(x). Show all our work! E(X) 4 k P (X k) k var(x) (k 2) 2 P (X k) k ( 2) ( ) (b) (8 points) Find P ( X 3). Show all our work! P ( X 3) P (X ) + P (X 2) + P (X 3) Alternativel, ou could do P ( X 3) P (X ) (X 4) (c) ( points) We measure X four times independentl. What is the probabilit that X 2 for exactl two of the four measurements? Show all our work! We have 4 trials (the 4 measurements), and we can think of the outcome X 2 as success and X 2 as failure. There are ( 4 2) was of getting 2 success in 4 trials, the probabilit of each success is P (X 2) 3 9 3, and the probabilit of failure is P (X 2) Thus the probabilit of getting X 2 exactl 2 out of 4 times is: ( ) 4 ( 2 3 )2 ( ) NOTE: This question was the same as the one in the practice slides, just with different numbers.

3 2. Suppose that a continuous random variable X has distribution function { e x for x > ; f(x) otherwise. (a) (5 points) Find P (X 5). Show all our work! P (X 5) 5 f(x)dx dx + 5 e x dx ( e x ) 5 e 5. (b) (5 points) Find E(X). Show all our work! since f(x) for x. Thus, E(X) E(X) lim b b xe x dx xe x dx xe x dx lim b ( xe x e x ) b B L Hopital, Also, lim b ( be b e b + ) lim b be b lim b e b. lim b be b lim b b e b lim b e b. lim b e b, so E(X). NOTE: This question was also in the practice slides, with e 2x instead of e x, and it was part of a long homework question ou had to solve where ou had e ax for a parameter a. 2

4 3. (2 points) The following pictures, in some order, denote the image of the vector v transformations given b the matrices. ( ) A ; A 2 ( ) ; A 3 ( ) ; A 4 ( ) ( ) 4 under the Match the pictures with the corresponding matrix b writing A i for the correct value of i in the line next to the picture. You do not have to show an work. I am sorr, I don t have the pictures. (a) A 2, because ( ( ) 4 ) ( ) 4 (b) A 4, because ( ( ) ( 4 ) 4) (c) A 3, because ( ( ) ( 4 ) 4) (d) A, because ( ( ) ( 4 4 ) ) 3

5 4. (8 points) Find the general solution to the differential equation d ( ). dx Show all our work! You will be graded on the completeness and qualit of our solution. First note that the constant functions (x) and (x) satisf the differential equation, therefore the are solutions. Now, assume that (x),, then we can divide b ( ) to get (x)( (x)) (x). Integrating we get, Now we can use the substitution of variables (x)( (x)) (x)dx dx. (x), d (x)dx, to get ( ) dx. NOTE: You did not lose points if ou didn t write out the substitution explicitl and separated variables to get ( ) dx. However, if ou wrote ( ) dx. that is a crucial mistake that shows profound misunderstanding. Alex was ver generous in grading that, and didn t subtract nearl as man points as I would have. If ou have written that, please think about it carefull. dx dx means (x), the derivative of, which is a function of x. You cannot split up the dx and multipl b it - that s not a number and that makes no sense. I know some high school books sa that, but I told ou that s dumming down the procedure for obtaining the answer and ou should not bu that. Please think carefull about how we derived the separation of variables we are using. I was sad to see some students still write that. We use partial fraction decomposition to compute d d ( ) + d ln ln + C, for C an arbitrar real constant Thus ln ln x + C for C an arbitrar real constant. So, ln x + C 4

6 for C an arbitrar real constant. So, ec e x for C an arbitrar real constant (so e C can range through all positive real numbers). So, i.e., ±ec e x, Cex, for C an arbitrar non-zero constant (since ±e C ranges through all nonzero real numbers as C ranges through all the real numbers.) Now we have to solve for : ( )Ce x ( + Ce x ) Ce x, so the general solution of our differential equation is (x) Cex, for an C, and the constant functions (x) and (x). + Cex NOTE: This is just a particular case of the logistics equation, with all constants being equal to. NOTE: It was important to not forget the trivial solutions. I emphasized that in class even when we had an initial value condition - we alwas checked the constant solutions first that we disregard when we separate variables (if we had an initial constraint that the didn t satisf, we dismissed them.) Also, that s wh I put the question on the review slides where the solution turned out to be precisel the constant solution. 5

7 5. Let A ( ) 3. 2 (a) (5 points) State the definition of what it means for a scalar λ to be an eigenvalue of A. A scalar λ is an eigenvalue of A if there exists a nonzero vector v such that Av λv. NOTE: That s it. That s all ou had to write. NONZERO is the essential part of this definition. If ou don t sa that, then Av λv is triviall satisfied b the zero vector for an real number λ, so an real number would be an eigenvalue. I stressed this over and over again in class. (b) (5 points) Find the eigenvalues of A (start from the definition, carr out and explain all the steps in our argument.) Show all our work! You will be graded on the completeness and qualit of our solution. ( x A scalar λ is an eigenvalue of A if there exists a nonzero vector such that Av λv. ) Now Av λv Av λv Av λiv (A λi)v [( ) ( )] ( ) 3 x λ 2 ( ( ) ( 3 λ x 2 λ) ) ( ) Thus ( ) λ ( is ) an eigenvalue if and onl if the last equation has a nonzero solution, i.e., if there is x which satisfies it. ( ) 3 λ This equation has a nonzero solution if and onl if the matrix 2 ( ) λ x if the inverse existed, then we could multipl b it on both sides to get be the unique ( solution. ) 3 λ The matrix 2 λ we get so is NOT invertible ( ) and that would ( ) 3 λ is not invertible if and onl if det. Solving this, 2 λ (3 λ)( λ) 2, λ 3 or. NOTE: The essential part was to realize that we need a NONZERO solution to ( ( ) ( 3 λ x, 2 λ) ) 6

8 and to know that that onl happens if the coefficient matrix is not invertible, and that is wh we need its determinant to be. Unless ou explicitl said these things, our solution is not complete. NOTE: When ou distribute and pull out v from Av λiv, ou need to keep it on the right: (A λi)v. Note that ( ) ( ) x 3 2 is not even defined!! And also, it s not true that (A B)C CA CB, even if the multiplications are defined! What is true is that (A B)C AC BC, but we don t have that AC CA or BC CB. 7

Math 106: Calculus I, Spring 2018: Midterm Exam II Monday, April Give your name, TA and section number:

Math 106: Calculus I, Spring 2018: Midterm Exam II Monday, April 6 2018 Give your name, TA and section number: Name: TA: Section number: 1. There are 6 questions for a total of 100 points. The value of