1. Find and classify the extrema of h(x, y) = sin(x) sin(y) sin(x + y) on the square[0, π] [0, π]. (Keep in mind there is a boundary to check out).

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1 . Find and classify the extrema of hx, y sinx siny sinx + y on the square[, π] [, π]. Keep in mind there is a boundary to check out. Solution: h x cos x sin y sinx + y + sin x sin y cosx + y h y sin x cos y sinx + y + sin x sin y cosx + y Notice the second term in each equation is the same, so cos x sin y sinx + y sin x cos y sinx + y. Case I. sin y sin x does not occur on the interior of [, π] [, π], only on the boundary, which we ll deal with later. Case II. cos x cos y x y π/. In the original equations, this gives for and so it s no good. Case III. sinx + y implies x + y π in this domain. y π x sin y sin x, cos y cos x cos x sin x + sin x 4 sin x cos x + sin x 5 which together imply x or π, which is on the boundary. So, with sinx + y, we can divide to get cos x sin y sin x cos y sinx y x y in interior of square 7 Thus from and, i.e. x y 8 cos x sin x sin x + sin x sin x cos x We are assuming here that sin x the case sin x was dealt with in Case I, and so we can divide it out to get cos x sin x + sin x cos x i.e. sin x, so x π/ or π/. So our critical points are P π, π and P π, π. Plugging these points back into and, and remembering our -- triangle, we see they both work.

2 Then hp hp 8 MAX 8 MIN Also check the boundary: On the boundary, one of x or y is either or π, so h on the boundary. Therefore the max and min are at the points found above.. Find those points on the curve of intersection of the surfaces x xy + y z and x + y which are nearest the origin. Hint: what was Lagrange s first name? Solution: The function to be minimized here is the distance x + y + z, or equivalently by the usual trick the square of the distance, which we ll call f x + y + z. The two constraints are g x xy + y z and h x + y. Then the Lagrangian function is Lx, y, z, λ, µ f + λg + µh; looking for a critical point of L yields the five! equations Note that and imply x λx y + µx 9 y λ x + y + µy z λ z + µ x + y x xy + y z 4 xy z. 5 Now, implies λ OR z. Case I: z. Then 5 implies xy. But x + y, so possible points are P,,, P,,, P,,, P 4,, Check P in 9 and and get λ, λ + µ

3 so λ, µ will work. For P, we get λ, λ + µ which works as well. And P and P 4 check out as well. Therefore P, P, P, and P 4 are candidates, each at distance from the origin. Case II: λ. Then 9 and become 4 µx y 4 µy x, 7 and we also have xy z, x + y. If either x or y is zero, then or 7 implies that the other is zero, which contradicts them being on the circle. So assume neither one is zero. Then we can divide and 7 to get x y y x x y. Since x + y, we have x ± and y ±. Since z xy, we have four possibilities:,,,,,,,,,,,, each of which has distance from the origin. Thus P P 4 found before are the closest points. Note: The reference to Lagrange s first name is U of Alberta convention for hints. Lagrange s actual first name is not important.. Find the volume of the finite solid bounded by the surfaces az x + y, x + y + z a. Solution: The surfaces intersect in the curve x + y a ; z a, as can be seen by solving the two equations simultaneously. Thus the volume is V 4 a a y a x y dz x +y /a

4 the 4 comes from symmetry, and for the rest, draw a picture... which I can t duplicate easily here!. This is easier in cylindrical coordinates, where it can be written remembering that dv in cylindrical coordinates is r dr dz dθ V π π πa a a r r a a dθ [ r dz dr dθ a r r a a r / r4 a + / a a r dr 4a ] a 4. Set up the correct limits for both iterated integrals for fx, y da over D if D is:. The parallelogram with sides x, x 5, x y + 4, x y The triangle with sides y, y x, y 4 x.. The finite domain cut out by the curves y x, y 4 x. 4. The region bounded by the curve x 4 + y 9. Solution: For this question you really need to draw diagrams, which I don t have on the computer, so I ll just give the answers... a The intersection points are 5, 8, 5, 9,, 5, and,. i ii 5 5 b i ii +y x+4 x+ 4 y x y f + f f + f x +y 4+y f 4 f y f

5 c i ii y d i ii 4 4 x y x f + f 4 4 y f 4 y + y /9 f y /9 + x /4 f x /4 5. Transform the following to polar coordinates, and evaluate:. 4 x ln + x + y,. y arctan, x over the region defined by x + y 9, x y x. Solution: a We re integrating over the part of the disc of radius that lies in the first quadrant. 4 x ln + x + y, π π 4 π 4 dθ 5 u ln u u ] 5 5 ln 5 4 ln + r r dr dθ lnu du letting u + r π 4 5ln 5 + b Again, draw a diagram. The region call it R lies between the circles of radius and, and between the lines y x and y x, which 5

6 correspond to θ π and θ π, respectively. Thus y arctan θr dr dθ x π R [ θ π r] dθ π 4 π θ π π π π θ 9 dθ. Change to cylindrical or spherical coordinates and evaluate:. x x a z x + y dz,. The volume of the solid that lies above the cone φ π sphere with equation ρ 4 cosφ and below the Solution: a The fact that we have a x + y and a plain z appearing suggests that we use cylindrical coordinates. The z integral, therefore, is unchanged; what we need is to figure out the integral in x and y. The region in x and y over which we re integrating is x, y x x. To find the curve y x x, complete the square: so y x x x, and so x + y y x x This is the circle of radius, centred at,, which has polar equation r cos θ. Since we re taking the positive square root, the region of integration is the top half of this circle see figure below. If you don t remember that equation in polar coordinates, take the second line of the above set of equations and convert to r and θ: so r or r cos θ r sin θ r cos θ r cos θ r r cos θ

7 Thus the region is r cos θ, for θ π. If it s not clear that this is the proper range of θ, consider the following diagram: where some lines of constant θ have been marked. Therefore, the integral becomes remembering that dz r dz dr dθ x x a cos θ a 4a 4a a r] cos θ z x + y dz zr r dz dr dθ dθ 8a cos θ dθ sin θ cos θ dθ 4a sin θ sin θ ] π 4a cos θ a r dr dθ cos θ sin θ cos θ dθ 8a 9 b The sphere ρ 4 cos φ has radius and sits on top of the z-axis i.e. it has centre,,. The surface φ π is a cone. The volume element in spherical coordinates is ρ sin φ dρ dφ dθ, and so the volume is V 4 cos φ π ρ] 4 cos φ π π 4 4 ρ sin φ dρ dφ dθ dφ 4 cos φ sin φ dφ π π 5 π. 4 4 cos4 φ ] π 7

8 7. Prove that the function yx e xz + z dz satisfies the differential equation y x + y x. Solution: If yx sign, we have e xz dz, then differentiating under the integral + z Thus y x and y x y + y e xz x + z dz z e xz + z dz z e xz + z dz + z e xz + z dz e xz dz ] T lim T x e xz lim T x e xt x x as required. 8. Given a region D R in the plane and a function of two variables fx, y, let R be the region in space above D and below the graph of f. Then we have two expressions for the volume of R, namely fx, y da and dv Show that these two expressions are equal. D Solution: The region R can be described as {x, y D, z fx, y}. R 8

9 Therefore the triple integral over R can be written as an iterated integral: R dv as required. D D D fx,y z ] fx,y da fx, y da dz da where da is the area element in the xy plane 9. Let fx, y x 4 4x y + y. Show that on each line y mx, the function has a minimum at.. Show that, is not a minimum of f.. Make a sketch or use Maple showing those points x, y where fx, y > and fx, y <. Hint: If you have trouble with part b, perhaps try part c first. Solution: a If y mx, then fx, y fx, mx x 4 4mx + m x, and we can treat this by the usual methods for functions of one variable. Differentiate: f x mx + m x and since f at x, there is a critical point at the origin. By the one-variable Second Derivative Test, f x 4mx + m x m > and so the origin is a minimum along the line y mx. Note that the line m i.e. y also gives a minimum, since fx, x 4. Also note that the value of f here is f,. b & c Following the hint, we consider points where f > and f <. Note that f factors as fx, y x yx y, and so fx, y whenever y x or y x. Furthermore, f is negative if x < y < x and positive elsewhere. Put another way, the function f is zero on the two 9

10 curves y x and y x, negative between them, and positive everywhere else. Every neighbourhood of, contains points between the curves y x and y x, and thus contains points where f is negative. Therefore the origin is not a miniumum of f. Note: Many people said that the origin is not a minimum because the Hessian is zero. This is not correct. If the Hessian is zero, that means that the second derivative test gives us no information, which means just that: we have no information, i.e. we can t conclude anything. In particular, in this example we cannot conclude that, is not a minimum just because the Hessian is zero. We have to look more closely at the behaviour of the function. For example, the functions x 4 +y 4, x 4 y 4, and x 4 y 4 all have critical points at the origin where the Hessian is zero. However, they all have different behaviour there: minimum, saddle point, and maximum, respectively. Thus when the Hessian is zero, it means we need to do more analysis.