Solutions to the Calculus and Linear Algebra problems on the Comprehensive Examination of January 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