f(x, y) = 1 2 x y2 xy 3

Transcription

1 Problem. Find the critical points of the function and determine their nature. We have We find the critical points We calculate f(x, y) = 2 x y2 xy 3 f x = x y 3 = 0 = x = y 3 f y = 3y 3xy 2 = 0 = y = xy 2 = y = y 5 = y = 0, y =, or y =. (0, 0), (, ) (, ). f xx =, f xy = 3y 2, f yy = 3 6xy. At the critical point (0, 0) the Hessian is [ ] 0 H f = 0 3 which is positive definite, hence (0, 0) is a local minimum. At (x, y) = ±(, ), the Hessian is [ ] 3 H f = 3 3 which has negative determinant, hence it is definite, so ±(, ) are both saddle points.

2 Problem 2. Consider the function f(x, y) = y2 x 4. (i) Draw the level curves of f. (ii) At the point P (, ), find the unit direction of steepest increase for f. (iiii) For the vector u = 4 i+3 j 5, find the directional derivative of f at P in the direction u, and use it to estimate f((, ) +.0 u). (i) The function f is undefined at x = 0. The level curves of f are f(x, y) = c y 2 = cx 4 y = ± cx 2. When c > 0, we get two arches of a parabola i.e. the parabola minus the point (0, 0). When c = 0, we get the axis y = 0 minus the point (0, 0). (ii) We calculate the gradient f = ( 4y2 x 5, 2y ) x 4 which at P equals f(p ) = ( 4, 2). The unit direction of steepest increase is (iii) We have We obtain u = f ( 4, 2) ( 2, ) = =. f 20 5 ( 4 u f = f u = ( 4, 2) 5, 3 ) = 2. 5 f((, ) +.0 u)) f(, ) +.0 u f =.02 =.98.

3 Problem 3. Show that the function f : R 2 R 2 given by f(x, y) = (e x + e 2y, e 3y + e 2x ) is locally invertible at every point. We calculate the Jacobian [ ] e x 2e J f = 2y which has determinant det J f = 3e x+3y 4e 2x+2y 0. Thus J f is invertible at every point, so by the inverse function theorem, f is a local isomorphism. 2e 2x 3e 3y

4 Problem 4. Consider the region R given by x 2 + 4y Find the global minimum and global maximum of the function f(x, y) = 2 + 8y x 2 y 2 over the region R. We find the critical point of f in the interior of R. We have f x = 2x = 0 = x = 0 f y = 8 2y = 0 = y = 9. The point (0, 9) is however not in the interior of R. We use Lagrange multipliers to optimize f on the boundary of R which is g(x, y) = x 2 + 4y 2 = 00. We have f = λ g ( 2x, 2y + 8) = λ(2x, 8y) or else g = (2x, 8y) = 0 = x = y = 0 which however does not lie on the boundary. We must have x = λx = x = 0 or λ =, y + 9 = 4λy. If x = 0, then y = ±5 since x 2 + 4y 2 = 00, and f(0, 5) = 67 while f(0, 5) = 3. If λ = we obtain y + 9 = 4y hence y = 3. We must have x = ±8 and f(±8, 3) = 25. The global minimum occurs at (±8, 3) while the global maximum is at (0, 5).

5 Problem 5. [7 points.] Consider the functions Calculate at (x, y, z) = ( 2,, ). f(x, y, z) = (x 2 + 3xy, z cos x, z ln y) g(a, b, c) = ab(c + ) sin c. (g f) x and We calculate 2x + 3y 3x 0 Df = z sin x 0 cos x ln y 0 z y Dg = (b(c + ) sin c, a(c + ) sin c, ab(sin c + (c + ) cos c)). For (x, y, z) = ( 2,, ) we have a = 2, b = cos( 2), c = 0. Thus Dg( 2, cos 2, 0) = (0, 0, 2 cos 2) and 6 0 Dg( 2, cos 2, 0) Df( 2,, ) = (0, 0, 2 cos 2) sin 2 0 cos The first entry in the product is 0 hence (g f) = 0. x

6 Problem 6. Calculate the limits below or explain why they do not exist (i) lim x,y,z 0 x 2 y 2 z 2 x 4 +y 4 +z 4. (ii) lim x,y 0 xy 2 x 2 +4y 4. (i) When y z, we have y 2 z 2 z 4 x 4 + y 4 + z 4, hence y 2 z 2 x 4 + y 4 + z 4 = 0 x 2 y 2 z 2 x 4 + y 4 + z 4 x2 0. The limit is 0 by the squeeze theorem. When y > z, the argument is similar. (ii) If we let x, y 0 along the parabola x = my 2, the fraction becomes xy 2 x 2 + 4y 4 = my 4 m 2 y 4 + 4y 4 = m m This does depend on m, hence the limit does not exist.

7 Problem 7. Find the closest point to the origin lying on both planes x y 2z = 5, 2x y + 2z = 8. We use Lagrange multipliers to minimize the function f(x, y, z) = x 2 + y 2 + z 2. We have f = λ(,, 2) + µ(2,, 2) which gives We have We have (2x, 2y, 2z) = (λ + 2µ, λ µ, 2λ + 2µ). x = λ 2 + µ, y = λ 2 µ, z = λ + µ. 2 x y 2z = 5 = 3λ µ 2 = 5 2x y + 2z = 6 = λ 2 + 9µ 2 = 8. We solve λ = 2 and µ = 2 hence x = 3, y = 2, z = 0. The point is (3, 2, 0).

8 Problem 8. Consider the surface S R 3 with equation xy 2 z 3 = 0 in the first octant. (i) Show that γ : R 2 R 3 given by is a parametrization of S for u, v > 0. γ(u, v) = (u 3, v 3, uv 2 ) (ii) Using this parametrization, find vectors spanning the tangent plane to S at (,, ). (i) Clearly, the image of γ is in S. We show γ is bijective. Indeed, the inverse is given by u = x 3, y = v 3 = x = u 3, y = v 3 = z 3 = u 3 v 6 = z = uv 2. We show that Dγ(u, v) is injective. Indeed, we calculate 3u 2 0 Dγ(u, v) = 0 3v 2. v 2 2uv Clearly, the columns of Dγ are linearly independent since otherwise u = v = 0 which is not allowed. (ii) The tangent plane is the column space of Dγ(, ) which is spanned by the columns of Dγ(, ). The 3 0 basis is given by 0, 3. 2

9 Problem 9. Prove that the function f(x, y, z) = { x 7 x 6 +y 6 +z 6 if (x, y, z) (0, 0, 0) 0 if (x, y, z) = (0, 0, 0) admits all directional derivatives, but it is not differentiable. We fix a unit vector u = (u, u 2, u 3 ) and we use the definition of directional derivative to calculate f u = lim f(hu) f(0) f(hu) h 7 u 7 = lim = lim h 0 h h 0 h h 0 h 6 (u 6 + u6 2 + u6 3 ) h = u 7 u 6 + u u6 3 Thus directional all directional derivatives exist. Note that for the vectors u = i, we have f x = while for u = j and v = k, we obtain f y = f z = 0. Thus, if f is differentiable, its derivative equals the Jacobian which is Df(0, 0, 0) = (, 0, 0). But then f u = Df u = (, 0, 0) (u, u 2, u 3 ) = u which is not the answer we found above. Therefore, f is not differentiable at the origin.

10 Problem 0. Consider the surface S R 4 given by the equations x 2 + y 2 xz + w 2 = 2 xyz xw + z 2 + w 2 = 2. (i) Show that S is a smooth manifold near (, 0, 0, ). (ii) Show that near (, 0, 0, ) we can solve for z, w in terms of x and y: (z, w) = g(x, y). (iii) Find the derivative of g at (, 0). (iv) Find vectors spanning the tangent plane to S at (, 0, 0, ). (i) We calculate the derivative of the two equations to obtain the matrix [ ] 2x z 2y x 2w. yz w xz 2z xy x + 2w At (, 0, 0, ) this matrix becomes [ 2 0 ] This is clearly surjective since the columns span R 2. The result follows from the implicit function theorem. (ii) In fact, since the second block g of class C. (iii) We have Dg(, 0) = [ ] 2 is invertible, we can solve (z, w) = g(x, y) for some function 0 [ ] [ ] = [ ] [ ] = [ ] (iv) The tangent plane is the null space of the matrix [ ] We can either row reduce, or we can observe that the vectors 0 4, 0 are in the nullspace. 0

11 Problem. Consider U the set of n n matrices A such that 2I A 2 is invertible. (i) Prove that U is an open set in Mat n n. (ii) Find the derivative of the function f(a) = (2I A 2 ) at A = I. [ + (iii) Using linear approximation, estimate the inverse of 2I A 2 where A = (i) We let F (A) = det(2i A 2 ). The function F is a composition of two continuous functions, the determinant and the function A 2I A 2. Thus F is continuous. We have that U is the preimage of the open set R \ {0} under the continuous function F, hence U is open. (ii) We have f = g h where g(b) = B and h(a) = 2I A 2. We calculated in class the derivative ]. Dg(B)(H) = B HB = Dg(I)(H) = H and We have Dh(A)(H) = AH HA = Dh(I)(H) = 2H. Df(I) = Dg(h(I)) Dh(I) = Dg(I) Dh(I) = Df(I)(H) = 2H. (iii) By linear approximation, we have f(i + H) f(i) + Df(I)(H) = I + 2H = (2I A 2 ) I + 2 [ ] [ + = ]