Comprehensive Exam Solutions: Calculus & Linear Algebra (Jan 27, 2012)

Solutions to the Calculus and Linear Algebra problems on the Comprehensive Examination of January 27, 2012 Solutions to Problems 1-4 are omitted since they involve topics no longer covered on the Comprehensive Examination. 5. [15 points] Evaluate the following integrals: Z 1Z 1 (a) x3 sin(y 3 ) dy dx. 0 x2 Solution: We need to switch the order of integration. The region of integration lies between y = x2 and y = 1 for 0 ≤ x ≤ 1. This is the same as the region √ between x = 0 and x = y for 0 ≤ y ≤ 1. Thus Z 1Z 1 Z 1 Z √y 3 3 x sin(y ) dy dx = x3 sin(y 3 ) dx dy 0 x2 Z0 1 0 Z 1 √ y 3 1 4 1 = x sin(y ) dy = 4 y 2 sin(y 3 ) dy 4 0 0 0 Z 1  1 1 1 = 12 sin(u) du = 12 − cos(u) 0 0

1 1 = 12 − cos(1) − (− cos(0)) = 12 1 − cos(1) , where we used the substitution u = y 3 , du = 3y 2 dy.
R (b) C (y 2 + 6y)dx + cos(y 2 ) + 2x(y + 1) dy, where C is some circle of radius 3 in the xy-plane, oriented counterclockwise. Solution: Just apply Greens’s Theorem, letting D be the domain enclosed by C: ZZ Z
2 2 (2(y + 1) − (2y + 6)) dA (y + 6y)dx + cos(y ) + 2x(y + 1) dy = C ZDZ ZZ −4 dA = −4 = D dA D You can do this integration if you like, but the area of a circle of radius 3 is just 9π so the answer is −36π p 6. [10 points] Find the volume of the solid above the cone z = x2 + y 2 and below the sphere x2 + y 2 + z 2 = z. Note that, while the sphere is not centered at the origin, using spherical coordinates still works nicely for this problem. Solution: Above the cone z 2 = x2 +y 2 just means, for example, z ≥ x on the xz-plane; in spherical coordinates, clearly tan(φ) ≤ 1 so φ ≤ π/4. We can also convert the other constraint to spherical: x2 + y 2 + z 2 ≤ z ⇒ ρ2 ≤ ρ cos(φ) ⇒ r ≤ cos(φ), so: 1 ZZZ V = dV Z V 2π Z π/4 Z cos(φ) ρ2 sin(φ) dρ dφ dθ = 0 Z 0 π/4 = 2π 0 0 1 3 ρ sin(φ) 3 cos(φ) dφ ρ=0 Z 2π π/4 = cos3 (φ) sin(φ) dφ 3 0 √  π/4  π π  2 4 2π  1 4 − cos (φ) =− −1 = = 3 4 6 2 8 0 where we used the substitution u = cos(φ), du = − sin(φ) dφ. 7. [12 points] Consider the function ( f (x, y) = xy x2 +y 2 0 if (x, y) 6= (0, 0), if (x, y) = (0, 0). (a) Is f continuous at (0, 0)? Justify your answer. Solution: f is not continuous. Take lim(x,y)→(0,0) f (x, y) along the line y = x: 1 x2 = lim = 1/2 6= 0 = f (0, 0) 2 2 x→0 2 x→0 x + x lim If lim(x,y)→(0,0) f (x, y) exists then it must be 1/2 because that’s the result of approaching along y = x, but then the limit does not equal the actual value of f at (0, 0) so f is not continuous. (Alternatively, it is sufficient to show that lim(x,y)→(0,0) f (x, y) DNE by approaching along, for example, the y-axis. That would give a limit of 0, which 6= 1/2.) (b) Find fx (0, 0) and fy (0, 0). Solution: Use the definition of partial derivative. Note that since f is symmetric with respect to x and y, fx = fy so only one calculation is needed: fy (0, 0) = fx (0, 0) = lim h→0 h(0) h2 +02 h 0 =0 h→0 h3 = lim (c) Is f differentiable at (0,0)? Justify your answer. Solution: Will not need to prove differentiability in the new comps. At any rate, though, f is obviously not differentiable at (0, 0) because it isn’t even continuous. 2 8. [10 points] Assume that the temperature in degrees Celsius at a point (x, y) on the circle x2 + y 2 = 4 is given by T (x, y) = x2 + 4x − y 2 + 12. Find the points on the circle at which the temperature is highest and lowest, and state the temperature at each of these points. Solution: We use Lagrange multipliers here to find the max and min of T with the constraint g(x, y) = x2 + y 2 = 4. To find the points to test, we solve the system ∇T (x, y) = λ∇g(x, y) (so 2x + 4 = λ(2x) and −2y = λ(2y)) and g(x, y) = 4. We have two cases here. If y = √ 0, then clearly x = ±2. If y 6= 0, then λ = −1 so x = √ −1, which in turn gives us y = ± 3. So we just plug in the points (±2, 0) and (−1, ± √3) to get the max and min. We find that (2, 0) gives us the max T of 24, and (−1, ± 3) gives us the min of 6. 9. [10 points] Let C be a 3 × 5 real-valued matrix. Answer the following questions about C and briefly justify your answers: (a) Can the columns of C be linearly independent? Solution: No. The columns are ∈ R3 so no more than 3 columns can be linearly independent. (If 3 columns of C are linearly independent, then they form a basis for R3 so they span, among other things, the fourth and fifth columns of C.) (b) Does the equation Cx = 0 have a unique solution with x ∈ R5 ? Solution: No. There are 3 equations in 5 variables so there are at least 2 free variables, which means infinitely many solutions. (c) Assume that the span of the columns of C is all of R3 . Can you determine the nullity (= dimension of the null space or kernel) of C? Solution: Let n = nullity and r = rank of C. Since the columns of C span all R3 , r = 3. Thus by the rank-nullity theorem, n = 5 − r = 2. 10. [8 points] Consider the matrix   1 3 A= . 4 2 (a) Determine the eigenvalues and eigenvectors of A. Solution: We want to find all possible nontrivial values v ∈ R2 and λ ∈ R such that Av = λv, which means that the system (A − λI)v = 0 has a nontrivial solution. Thus, A − λI is non-invertible, or det(A − λI) = 0. This gives us the characteristic polynomial (1 − λ)(2 − λ) − 12 = 0. The solutions to this quadratic are λ = 5 and λ = −2. Now we solve for the nullspace of A − λI for each λ; these are our eigenvectors (plus zero):     −4 3 4 −3 λ=5: −→ . 4 −3 0 0 3     3 3 The nullspace of this matrix is clearly c , c ∈ R (that is, span ). 4 4     3 3 1 1 λ = −2 : −→ . 4 4 0 0     −1 −1 The nullspace of this matrix is clearly c , c ∈ R (that is, span ). 1 1   3 So to conclude, A has an eigenvalue of 5 with corresponding eigenvectors c , 4   −1 c 6= 0, and an eigenvalue of −2 with corresponding eigenvectors c , c 6= 0. 1 (b) Find an invertible matrix P and a diagonal matrix D such that A = P DP −1 . Solution: Following the diagonalization procedure,  we let D be the matrix with −2 0 A’s eigenvalues in the diagonal entries: D = . P is the matrix consisting 0 5 of each eigenvectors, taking care to keep the order  eigenvalue’s   the same as in D: −1 3 −4/7 3/7 P = . This is easily checkable; we find that P −1 = , and a 1 4 1/7 1/7 simple (if boring) exercise in matrix multiplication shows that A = P DP −1 . 11. [10 points] Let T : V → V be a linear transformation on a finite dimensional vector space V . Suppose T is one-to-one (injective). Prove that if {v1 , . . . , vn } is a basis for V , then {T (v1 ), . . . , T (vn )} is also a basis for V . Solution: First, we show that {T (v1 ), . . . , T (vn )} is linearly independent. Given scalars α1 , . . . , αn such that α1 T (v1 ) + · · · + αn T (vn ) = 0: Since T is linear, T (α1 v1 + · · · + αn vn ) = α1 T (v1 ) + · · · + αn T (vn ) = 0. Note that T is linear, so T (0) also equals 0. Since T is one-to-one, this means that α1 v1 +· · ·+αn vn = 0. We know that {v1 , . . . , vn } is a basis, so it is linearly independent, which means αi = 0 ∀i ∈ {1, . . . , n}. But that’s what we were trying to prove! So {T (v1 ), . . . , T (vn )} is linearly independent , as desired. Now note that if {v1 , . . . , vn } is a basis for V , then V has dimension n. Thus since {T (v1 ), . . . , T (vn )} is a set of n linearly independent vectors, it is a basis for V (in an n-dimensional vector space, n vectors span ⇔ they are linearly independent ⇔ they form a basis). QED 12. [15 points] Let T : P2 → R2 , where P2 = {a + bt + ct2 : a, b, c ∈ R}, be defined by   p(1) T (p) = . p(2) You may assume that T is linear. 4 (a) Find bases of the null space (kernel) and range of T . Solution: One way to find the kernel is to note that T (p) = 0 ⇒ p(1) = 0, p(2) = 0. The only quadratics with zeros at 1 and 2 are of the form c(x − 1)(x − 2) = c(t2 − 3t + 2), c ∈ R. Thus t2 − 3t + 2 is a basis of the kernel of T . Alternatively, we could rewrite T using the basis {1, t, t2 }:       a a+b+c 1 1 1   b . T (a + bt + ct2 ) = = a + 2b + 4c 1 2 4 c Then row reduce:       1 1 1 1 1 1 1 0 −2 −→ −→ 0 1 3 1 2 4 0 1 3    2   −3, so that 2 · 1 − 3 · t + 1 · t2 Hence the nullspace of the matrix is span 1 spans the kernel of T . Since T maps to R2 , the range of T is the column space of its matrix representation. From the matrix, we can readily see that the range is all of R2 , since the first two columns span R2 . (b) Find the matrix of this transformation with respect to the bases  representation    1 −1 {1, t, t2 } and , . 0 1 Solution: The procedure here is to find what T does to each member of the basis for P2 , and put those results in the given basis of R2 . So:       −1 1 1 T (1) = =2 +1 1 0 1       1 1 −1 T (t) = =3 +2 2 0 1       1 1 −1 T (t2 ) = =5 +4 4 0 1 The representations of T (1), T (t), and T (t2 ) were found by inspection. The more systematic method would be (for T (1), to pick an example) to solve the system of equations:       1 1 −1 =x +y 1 0 1   2 3 5 At any rate, the matrix representation is . 1 2 4 5