Q1 /10 Q2 /10 Q3 /10 Q4 /10 Q5 /10 Q6 /10 Q7 /10 Q8 /10 Q9 /10 Q10 /10 Total /100

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1 Midterm Maths Calculus III July 23, 2012 Name: Solutions Instructions You have the entire period (1PM-3:10PM) to complete the test. You can use one half-page for formulas, but no electronic devices may be used. In particular, you may not use calculators or cell phones. Cell phones should be on silent and out of sight. Except for the true/false question, you must show all of your work. Partial credit will be awarded, but correct answers without supporting work will not get credit. Make sure that your final answer is clearly labelled as such. If you need more space, you may continue on the other side of the page, or use the last page of the exam. Be sure to indicate the number of the problem if you do so. For grading purposes only (do not fill in) Q1 /10 Q2 /10 Q3 /10 Q4 /10 Q5 /10 Q6 /10 Q7 /10 Q8 /10 Q9 /10 Q10 /10 Total /100

2 Q1: Find the determinant of the following matrix A = Perform row operations to transform A into an upper triangular matrix: det A = (R 1 R 2 ) = (R 3 3R 1 + R 3, R 4 2R 1 + R 4 ) = (R 3 R 2 + R 3, R 4 3R 2 + 2R 4 ) = (R 3 R 4 ) = 1 (12) = 6. 2

3 Q2: Find the inverse of the following matrix A = (R 1 R 2 ) /2 1 1/2 3/ /2 0 1/2 5/ R 3 4R 4 + R 3 R 2 R 4 + R 2 R 1 R 4 + R 1 (R 2 R 3 + R 2 ) (R 1 R 2 + 2R 1 ) ( R1 1 2 R 1 R R 2 ). So A 1 = 1/2 1 1/2 3/2 1/2 0 1/2 5/

4 Q3: Suppose A is a 2 2 matrix such that A ( ) 4 = 5 ( ) 4 and A 5 ( ) 5 = 2 6 ( ) 5. 6 What is A? ( ) ( ) 4 5 and are eigenvectors of A corresponding to the eigenvalues 1 and respectively. Therefore A = ( ) ( ) ( ) = ( ) ( ) 6 5 = 5 4 ( )

5 Q4: The set of solutions to the following system of equations is a subspace of R 4 (you don t need to prove this): Find a basis for the set of solutions. 2w + x y 2z = 0 w + x + y z = 0 3w 2x + 3z = 0 Row reduce the augmented matrix to reduced row-echelon form: (R 1 R 2 ) ( ) R2 2R 1 + R 2 R 3 3R 1 + R 3 ) R 3 R 2 + R 3 R 1 R 2 + R 1 R 2 R 2 So the free variables are y and z. Let y = s, z = t. Then w 2y z = 0 and x + 3y = 0 give w = 2y + z = 2s + t and x = 3y = 3s, so w 2s + t 2 1 x y = 3s s = s t 0 0. z t 0 1 Therefore a basis is ,

6 Q5: Circle either true or false (you do not need to give reasons) (a) T/F : The set V = {f(x) C(R) : f(1) = 1} is a subspace of C(R), the vector space of continuous functions f : R R. (b) T/F : If 0 is an eigenvalue of an n n matrix A, then rank(a) < n. (c) T/F : If A is a 5 5 matrix and rank(a) = 3, then the system Ax = b always has infinitely many solutions. (d) T/F : If a square matrix has an eigenvalue with multiplicity greater than 1, then it is not diagonalizable. (e) T/F : If A and B are square matrices of the same size with B symmetric, then ABA is symmetric. (f) T/F : If a diagonalizable matrix A satisfies det(a λi) = λ 3 λ 2 + λ + 1, then A 2 = I. ( ) ( ) (g) T/F 1 1 : If and are both eigenvectors of matrices A and B, then 2 1 AB = BA. (h) T/F : The differential equation y sin(x)y = cos(2x) has a unique solution with initial conditions y(0) = 2, y (0) = 1. (i) T/F : If y p is a particular solution to a second order nonhomogenous linear DE and y p (0) = y p(0) = 0, then y p (x) = 0 for all x. (j) T/F : If y 1 and y 2 are solutions of a nonhomogenous linear DE, then 2y 1 y 2 is also a solution.

7 Q6: Find the general solution to y (3) + 3y 4y = 0. Auxilliary equation: 0 = λ 3 + 3λ 2 4 = (λ 1)(λ 2 + 4λ + 4) = (λ 1)(λ + 2) 2, so the roots are λ = 1 (mult 1) and λ = 2 (mult 2). Therefore the general solution is y = µ 1 e x + µ 2 e 2x + µ 3 xe 2x, µ 1, µ 2, µ 3 R.

8 Q7: Find a particular solution to y 3y + 2y = 2xe 2x. Guess: y p = (A + Bx)e 2x. However, 2 is a root (of multiplicity 1) of the auxilliary equation λ 2 3λ + 2 = 0, so replace y p by y p = x(a + Bx)e 2x = (Ax + Bx 2 )e 2x. y p = (A + 2Bx)e 2x + (2Ax + 2Bx 2 )e 2x = (A + (2A + 2B)x + 2Bx 2 )e 2x, y p = (2A+2B+4Bx)e 2x +(2A+(4A+4B)x+4Bx 2 )e 2x = (4A+2B+(4A+8B)x+4Bx 2 )e 2x, so 2xe 2x = y p 3y p + 2y p = (A + 2B + 2Bx)e 2x. Therefore A + 2B = 0, 2B = 2, which implies B = 1, A = 2, and a particular solution is y p = ( 2x + x 2 )e 2x.

9 Q8: Solve the initial value problem on (0, ) x 2 y 3xy + 13y = 0, y(1) = 0, y (1) = 6. This is a Cauchy-Euler problem. Auxilliary equation: 0 = λ(λ 1) 3λ + 13 = λ 2 4λ + 13 = (λ 2) 2 + 9, so λ = 2 ± 3i. Therefore the general solution is Since y = µ 1 x 2 cos(3 ln x) + µ 2 x 2 sin(3 ln x) = x 2 [µ 1 cos(3 ln x) + µ 2 sin(3 ln x)]. y = 2x[µ 1 cos(3 ln x) + µ 2 sin(3 ln x)] + x 2 [ 3µ 1 sin(3 ln x)x 1 + 3µ 2 cos(3 ln x)x 1 ] = x[(2µ 1 + 3µ 2 ) cos(3 ln x) + ( 3µ 1 + 2µ 2 ) sin(3 ln x)], the initial conditions imply µ 1 = 0, 2µ 1 + 3µ 2 = 6. Thus the solution is y = 2x 2 sin(3 ln x).

10 Q9: An experiment was carried out to ascertain the damping constant of high-fructose corn syrup. When a 2kg mass was attached to a spring, and both were submerged in a vat of high-fructose corn syrup, the resulting motion was found to be critically damped. When the mass was released with an upwards velocity of 5m.s 1 from a point 2m below the equilibrium position, it passed through the equilibrium position after 1 second. What is the damping constant of high-fructose corn syrup? [Hint: λ = ρ 2m, where ρ is the damping constant.] Since the motion is criticallly damped, the equation of motion is of the form x(t) = (µ 1 + tµ 2 )e λt. Since x (t) = µ 2 e λt λ(µ 1 + tµ 2 )e λt = (µ 2 λµ 1 tλµ 2 )e λt, the initial conditions imply that µ 1 = 2 and µ 2 λµ 1 = 5. After 1 second the mass was at the equilibrium position, so µ 1 + µ 2 = 0. Therefore µ 2 = µ 1 = 2, so 2 2λ = 5, which implies λ = 3/2. Therefore ρ = 2mλ = 2(2)(3/2) = 6 (in N.s.m 1 ).

11 Q10: Find the general solution to the linear system This is X = AX, where x 1 = 2x 2 x 3, x 2 = x 2, x 3 = 2x 1 + 4x 2 + 3x A = Eigenvalues of A: λ = 0 1 λ λ = (1 λ) λ λ = (1 λ)[λ2 3λ+2] = (λ 1) 2 (λ 2). So eigenvalues are 1 (mult. 2) and 2 (mult. 1). Eigenspace for λ = 1: (A I 0) = so there are two free variables, k 2 = s and k 3 = t, so k 1 = 2k 2 k 3 = 2s t, hence 2s t 2 1 K = s = s 1 + t 0. t Thus 1 and 0 are linearly independent eigenvectors corresponding to 0 1 λ = 1. Eigenspace for λ = 2: (A 2I 0) = / so there is one free variable, k 3 = s, and k 1 = 1/2k 3 = 1/2s, k 2 = 0, hence 1/2s 1/2 K = 0 s = s 0 1.,,

12 Extra page 1 Thus 0 is an eigenvector corresponding to λ = 2. 2 Therefore the general solution is x x 2 = µ 1 1 e t + µ 2 0 e t + µ 3 0 e 2t, µ 1, µ 2, µ 3 R. x