Solutions to the Calculus and Linear Algebra problems on the Comprehensive Examination of February 1, 2008

Solutions to the Calculus and Linear Algebra problems on the Comprehensive Examination of February 1, 2008 Solutions to Problems 1-5 and 10 are omitted since they involve topics no longer covered on the Comprehensive Examination. 6. Evaluate the following integrals: Z 1Z 2 2 (a) e(y ) dy dx. 0 2x Solution: We need to switch the order of integration. The region of integration lies between y = 2x and y = 2 for 0 ≤ x ≤ 1. This is the same as the region between x = 0 and x = y/2 for 0 ≤ y ≤ 2. Thus Z 1 Z 2 e 0 (y 2 ) Z 2 Z y/2 Z0 2 2x 2 e(y ) dx dy dy dx = 0 1 xe dy = = 2 x=0 0 1 y2 2 1 4 = e = (e − 1) 4 4 0 y2 y/2 Z 2 2 yey dy 0 where we used the substitution u = y 2 , du = 2ydy. R (b) C (ex + y 3 )dx + (6y 2 x + x3 )dy, where C is the circle x2 +y 2 = 1, oriented counterclockwise. Solution: Apply Greens’s Theorem, letting D be the domain enclosed by C: ZZ ZZ Z x 3 2 3 2 2 2 (e + y )dx + (6y x + x )dy = (6y + 3x − 3y ) dA = 3 (y 2 + x2 ) dA C D D Now integrate on D using polar coordinates, easy enough since D is just a circle of radius 1 and the integrand turns into r2 :  4  Z 2π Z 1 r 1 3π 2 3 r · r dr dθ = 3(2π) = 4 r=0 2 0 0 7. Find the volume of the region that is inside the sphere x2 + y 2 + z 2 = 4 and above the p cone z = x2 + y 2 . Solution: We’ll be working with spherical coordinates here. The inside-the-sphere constraint is pretty obvious: it just means that r ≤ 2. The cone constraint is a bound on φ. To find the specific bound, it can be helpful to just consider the xz-plane. 1 Above the cone on that plane means that z ≥ x, which converted to spherical means tan(φ) ≤ 1, or φ ≤ π/4. Any θ satisfies the constraints, so integrating: ZZZ dV V = Z V 2π π/4 Z Z 2) = 0 = 2π = 0 r2 sin(φ) dr dφ dθ 0  r3 2  16π 3 3 − cos(φ) √ ! 2 1− 2 0 π/4  0 8. Consider the function ( f (x, y) = x3 x2 +y 2 0 if (x, y) 6= (0, 0), if (x, y) = (0, 0). (a) Find fx (0, 0) and fy (0, 0). Solution: Use the definition of partial derivative: fx (0, 0) = lim h3 h2 +02 h h→0 fy (0, 0) = lim 03 02 +h2 h→0 h h3 =1 h→0 h3 = lim 0 =0 h→0 h3 = lim (b) Use the definition of the directional p derivative p to find the directional derivative of f at (0, 0) in the direction ~u = ( (2)/2, (2)/2). Solution: From the definition of directional derivative, D~u = lim h→0 √ (h 2/2)3 √ √ (h 2/2)2 +(h 2/2)2 h √ √ h3 ( 2/2)3 2 = lim 3 √ = h→0 h 2( 2/2)2 4 (c) Is f differentiable at (0,0)? Justify your answer. Solution: Will not need to prove differentiability in the new comps. 9. Find the maximum value of the function f (x, y) = x2 y on the ellipse x2 + 2y 2 = 24. Solution: We use Lagrange multipliers here to find the max of f (x, y) = x2 y with the constraint g(x, y) = x2 + 2y 2 = 24. To find the points to test, we solve the system ∇f (x, y) = λ∇g(x, y) (so 2xy = λ(2x) and x2 = λ(4y)) and g(x, y) = 24. We’ll break 2 √ up the system into two cases here. If x = 0, then clearly y = ± 12. If x 6= 0, then from the first equation y = λ so substituting into the second equation, x2 = 4y 2 . We now substitute in for x in the last equation: 6y 2 = 24, which gives us y = ±2 and x = ±4 for four more critical points (±4, ±2). Now it is just a matter of plugging each of these 6 points into f , and we find that the maximum occurs at (±4, 2) for a value of 32. 11. Let   2 0 −1 A = 3 0 −5 . 1 0 0 (a) Find all eigenvalues of A. Solution: We want to find all possible λ ∈ R such that ∃~v 6= ~0 ∈ R3 such that A~v = λ~v , which means that the system (A − λI)~v = ~0 has a nontrivial solution. Thus, A − λI is non-invertible, or det(A − λI) = 0. We use the recursive formula for the determinant to simplify the calculation:       3 −5 2 − λ −1 2 − λ −1 det(A − λI) = −0 det + (−λ) det − 0 det 1 −λ 1 −λ 1 −λ = −λ(λ2 − 2λ + 1) = −λ(λ − 1)2 Thus the eigenvalues of A are 0 and 1. (b) Find an invertible matrix P such that P AP −1 is a diagonal matrix, or show that there is no such matrix P . Solution: The eigenvalue 1 has algebraic multiplicity 2, so let’s start the diagonalization process there: if we don’t get an eigenspace of dimension 2, then we know A is not diagonalizable so we don’t have to bother with the eigenvalue of 0. Without further ado, we find the nullspace of A − I:     1 0 −1 1 0 −1 3 −1 −5 −→ 0 1 2 1 0 −1 0 0 0 There is only one free variable, which means that the nullspace of (A − I) is only 1. Thus, the sum of the dimensions of the eigenspaces of A is 2 and not 3, so A is not diagonalizable (that is, there is no P such that P AP −1 is diagonal). 12. Suppose that u, v, and w are distinct vectors in a vector space V , and {u, v, w} is linearly independent. Prove that {u + 2v, v + 2w, u + 2w} is linearly independent. Solution: Given α1 ,α2 , and α3 in R such that α1 (u+2v)+α2 (v +2w)+α3 (u+2w) = 0, we can rearrange the terms to find that (α1 + α3 )u + (2α1 + α2 )v + (2α2 + 2α3 ) = 0. Since {u, v, w} is linearly independent, α1 + α3 = 0, 2α1 + α2 = 0, and 2α2 + 2α3 = 0. 3 We now find all possible solutions to that system matrix corresponding to the system:      1 0 1 1 0 1 1 2 1 0 −→ 0 1 −2 −→ 0 0 2 2 0 2 2 0 by considering the nullspace of the    0 1 1 0 0 1 −2 −→ 0 1 0 0 1 0 0 1 So the matrix has a trivial nullspace! Thus α1 = α2 = α3 = 0, so {u+2v, v+2w, u+2w} is linearly independent, as desired. QED 13. Suppose V , W , and Z are finite-dimensional vector spaces, T : V → W and U : W → Z are linear transformations, and T is onto. (a) Prove that range(U T ) = range(U ). Solution: ⊆: Given z ∈ range(U T ), by definition of range ∃v ∈ V such that U ◦ T (v) = z. But then U (T (v)) = z so z ∈ range(U ).X ⊇: Given z ∈ range(U ), by definition of range ∃w ∈ W such that U (w) = z. But since T is onto, ∃v ∈ V such that T (v) = w. Thus, U ◦ T (v) = z so z ∈ range(U T ).X Thus range(U T ) = range(U ) as desired. QED (b) Prove that nullity(U T ) = nullity(U )+nullity(T ). Solution: We use the rank-nullity theorem; there are a lot of functions and vector spaces flying around, so we start by defining some variables. Let a = nullity(U ), b = nullity(T ), c = nullity(U T ), α = dim(V ), β = dim(W ), q = rank(U ), r = rank(T ), and s = rank(U T ). Since T is onto, r = β and since we already showed that range(U T ) = range(U ), s = q. Applying the rank-nullity theorem to U , T , and U T respectively and using those two substitutions, we find: β =q+a α=β+b α=q+c Solving for c and substituting, we find: c = α − q = (β + b) − q = (q + a) + b − q = a + b as desired. QED 4