Solutions to te Multivariable Calculus and Linear Algebra problems on te Compreensive Examination of January 3, 24 Tere are 9 problems ( points eac, totaling 9 points) on tis portion of te examination. Sow all of your work.. Find te critical points of te function f(x, y) = x 4 4xy + 2y 2 and classify as a local maximum, local minimum, or a saddle point. Solution: Since f is a polynomial, it is differentiable on R 2. Te critical points occur wen f x (x, y) = 4x 3 4y = and f y (x, y) = 4x + 4y =. Te second equation gives x = y, and substituting it into te first gives 4x 3 4x =, or x(x + )(x ) =. Tus x = or x = ±. Terefore te critical points of f are (, ), (, ), and (, ). To classify of te critical points, we use te second derivative test. First let us compute te second derivatives: f xx (x, y) = 2x 2 f xy (x, y) = 4 f yy (x, y) = 4 D(x, y) = f xx (x, y)f yy (x, y) (f xy (x, y)) 2 = 48x 2 6 D(, ) = 6 <, so (, ) is a saddle point; D(, ) = 32 > and f xx (, ) = 2 >, so (, ) is a local minimum; and D(, ) = 32 >, f xx (, ) = 2 >, so (, ) is also a local minimum. 2. Suppose te plane z = 2x y is tangent to te grap of z = f(x, y) at P = (5, 3). (a) Determine f(5, 3), x (5, 3) and (5, 3). x Solution: We know te graps of z = 2x y and z = f(x, y) intersect at P (5, 3, f(5, 3)), so f(5, 3) = 2(5) 3 = 6. Furtermore, recall tat an equation of te tangent plane to z = f(x, y) at te point P (x, y, z ) is So we ave z z = x (x, y )(x x ) + y (x, y )(y y ) z = 2x y = 6 + (5, 3)(x 5) + (5, 3)(y 3). x y Comparing coefficients of x and y, we obtain (5, 3) = 2 x (5, 3) =. y (b) Estimate f(5.2, 2.9). Solution: We use te linear approximation of f at (5, 3): f(5.2, 2.9) f(x, y ) + x (x, y )(x x ) + y (x, y )(y y )
f(5, 3) + (5, 3)(5.2 5) + (5, 3)(2.9 3) x y 6 + 2 (.2) + ( ) (.) = 6.5. Here is anoter way to do tis. Near (5, 3), te grap z = f(x, y) is approximated by te tangent plane at (5, 3), wic is given as z = 2x y. Tus f(5.2, 2.9) 2(5.2) (2.9) = 6.5. 3. Calculate te volume of te region inside spere x 2 + y 2 + z 2 = a 2 and outside te cylinder x 2 + y 2 = b 2, were a > b, by using an appropriate double integral. Solution: We are removing a vertical cylinder of radius b from a spere of radius a. Wen we tink of tis as double integral, te region in te plane is R = {(x, y) b x 2 + y 2 a} = {(r, θ) θ 2π, b r a}. Te top of te figure is te top alf of te spere, given by z = a 2 r 2, and te bottom is te bottom alf of te spere, given by z = a 2 r 2. Here, we are using cylindrical coordinates. Ten te double integral giving te volume is V = a2 r 2 ( a 2 r 2 ) da R 2π a = 2r a 2 r 2 dr dθ b [ = 2π 2 ( a 2 r 2) a 3/2] 3 = 4π 3 ( (a 2 a 2 ) 3/2 (a 2 b 2 ) 3/2) b = 4π 3 ( a 2 b 2) 3/2. 4. Suppose tat r(t) = (3 2t, e 3t, e 3t ) describes te position of an object at time t. (a) Calculate te acceleration of te object at time t. Solution: Te velocity and acceleration of te object at time t are v(t) = r (t) = (3 2, 3e 3t, 3e 3t ) a(t) = r (t) = (, 9e 3t, 9e 3t ). (b) Calculate te speed of te object at time t. Simplify by factoring te expression under te square root. Solution: Te speed of te object at time t is v(t) = (3 2) 2 + ( 3e 3t ) 2 + (3e 3t ) 2 = 8 + 9e 6t + 9e 6t = 9(e 6t + 2 + e 6t ) = 3 (e 3t + e 3t ) 2 = 3(e 3t + e 3t ). 2
(c) Calculate te distance traveled by te object between times t = and f =. Solution: Te total distance traveled by te object between t = and t = is D = 5. Consider te function v(t) dt = 3e 3t + 3e 3t dt = e 3t e 3t = e 3 e. 3 3y 3 if (x, y) (, ), f(x, y) = x 2 + y 2 if (x, y) = (, ). (a) Sow tat f is continuous at (, ). Solution: Recall tat f is continuous at (, ) iff lim (x,y) (,) f(x, y) = f(, ). 3y 3 Hence we must sow tat lim =. In polar coordinates, x = r cos θ, (x,y) (,) x 2 + y2 y = r sin θ, and (x, y) becomes r. Ten te limit is lim (x,y) (,) 3y 3 x 2 + y 3r 3 sin 3 θ 2 r r 2 r 3r sin 3 θ =, were te last equality follows since 3r and sin 3 θ is bounded. (b) Find f x (, ) and f y (, ). Solution: We will directly apply te definition of partial derivatives. and f x (, ) f(, ) f(, ) f y (, ) f(, ) f(, ) 2 + 3 3 + 2 = 3 3 = 3. 6. Suppose T : V V is a linear transformation, B = {b, b 2, b 3 } is a basis for V, and te matrix representation of T wit respect to B is 2 3 5 7 3. 9 Determine T (2b + 4b 3 ) as a linear combination of b, b 2, and b 3. Solution: Te coordinate vector of 2b + 4b 3 relative to B is a = coordinate vector of T (2b + 4b 3 ) relative to B is 2 3 5 2 24 b = Aa = 7 3 = 2. 9 4 2 Hence T (2b + 4b 3 ) = 24b + 2b 2 2b 3. 3 2. Ten te 4
2 7. Let A = 2 6 2. 3 (a) Compute te eigenvalue(s) of A. Solution: Te eigenvalues of A are te roots of its caracteristic polynomial 2 2 λ det(a λi 3 ) = det 2 6 2 λ = det 2 6 λ 2 3 3 λ [ ] 6 λ 2 = (2 λ) det = (2 λ)(6 λ)( λ). λ Tus A as tree eigenvalues: λ =, λ = 2, and λ = 6. (b) Find an invertible matrix C suc tat C AC is diagonal. Solution: We must first compute bases of te eigenspaces corresponding to te eigenvalues of A. λ = : Te eigenspace is te solution space of (A I) x =. Being omogeneous, we need only row reduce te coefficient matrix: 2 5 2 2/5, 3 so we ave x = x 2 + (2/5)x 3 = wit free variable x 3. Te solutions are x x 2 = ( 2/5)x 3 = x 3 2/5. x 3 x 3 Tus te eigenspace of A associated to λ = as basis 2/5. λ = 2: Te eigenspace is te solution space of (A 2I) x =. Being omogeneous, we row reduce te coefficient matrix: /3 2 4 2 2/3, 3 so we ave x (/3)x 3 = x 2 + (2/3)x 3 = wit free variable x 3. Te solutions are x (/3)x 3 /3 x 2 = (2/3)x 3 = x 3 2/3. x 3 x 3 4
/3 Tus te eigenspace of A associated to λ = 2 as basis 2/3. λ = 6: Te eigenspace is te solution space of (A 6I) x =. As above, we row reduce te coefficient matrix: 4 2 2, 3 5 so we ave x = x 3 = wit free variable x 2. Te solutions are x x 2 = x 2 = x 2. x 3 Tus te eigenspace of A associated to λ = 6 as basis. Finally, because A as tree linearly independent eigenvectors, it is diagonalizable (te problem implies tis anyway). Its eigenvectors form te columns of te desired matrix /3 C = 2/5 2/3. (Note: One can ceck tat C AC is diagonal, but doing so is not recommended because te calculations are extremely time-consuming.) 3 8. Let A = 2, were α is a real number. 2 2 4 α (a) For wat values of α does Ax = b ave at least one solution for all b R 3? Solution: Let b = b b 2 b 3. We can row reduce to obtain 3 b 3 b 2 b 2 2 b 2 b + b 2 b + b 2 2 2 4 α b 3 2 α 6 b 3 2b α 6 b 3 4b 2b 2 To finis te row reduction, tere are two cases, depending on α 8: α 6 We can divide te last row by α 6 to obtain a leading in te 4t column. Hence we ave at least one solution, no matter ow we coose b R 3. α 6 = If we pick b R 3 so tat b 3 4b 2b 2, ten te last row will ave a leading in te last column of te augmented matrix, wic implies tat te system as no solution. So b R 3 can be cosen so tat tere is no solution. Conclusion Ax = b as at least one solution for all b R 3 if and only if α 6. 5
(b) For te remainder of te problem set α =. Find te general solution of Ax =. Solution: Set α = in te partial row reduction of part (a). If we focus on te coefficient matrix (Ax = is omogeneous), we get te row reduction 3 2 2, 2 2 4 8 wic gives te equations x x 2 = x 3 = x 4 = wit free variable x 2. Tus te general solution is x x 2 x 2 x 3 = x 2 = x 2. x 4 9. Suppose {u, v} is a basis for a vector space V. Prove tat {u + 2v, 3u v} is also a basis for V. Solution: Our ypotesis implies tat V as dimension 2, and we are asked to prove tat {u + 2v, 3u v} is a basis for V Recall tat n vectors in an n-dimensional vector basis form a basis tey span tey are linearly independent. Here, it is easier to sow tat u+2v, 3u v are linearly independent. Suppose a (u+2v)+b (3u v) =, were a, b R. Ten au + 2av + 3bu bv = (a + 3b)u + (2a b)v =. Since {u, v} is a basis for V, we know tat u and v are linearly independent. Terefore te last equation implies tat a + 3b = and 2a b =. It is easy to sow tat te only solution to te above system of equations is a = b =, wic in turn implies tat u + 2v and 3u v are linearly independent. As noted above, V as dimension 2, so tat any set of 2 linearly independent vectors in V is a basis. Terefore, {u + 2v, 3u v} is a basis for V. 6