Review for the First Midterm Exam

Transcription

1 Review for the First Midterm Exam Thomas Morrell 5 pm, Sunday, 4 April 9 B9 Van Vleck Hall For the purpose of creating questions for this review session, I did not make an effort to make any of the numbers work out nicely. As such, there may be some truly horrible numbers which appear here which, if tomorrow you find them on your exam, will indicate with high certainty that you have made a mistake. Note also that many of the questions presented here are too difficult for the exam I am trying to fit nearly two dozen topics into only a few questions, so every question covers multiple ideas. The exam will not do this some of the topics covered in this review will not appear on the exam. Likewise, it is very possible that I miss a topic or two, so be sure to study all the homework assignments and quizzes.. For each of the following functions, determine whether the limit as (x, y) (, ) exists, and if so, find it: (a) f(x, y) = x y x + y, (b) g(x, y) = sin(x y ) x + y. Solution. (a) Since the numerator and denominator are both of degree two, it seems unlikely that the limit exists, so let s try to find two paths with different limits as (x, y) (, ). First, let s try the path x =. Parameterizing, r(t) =, t. Then t lim f(x(t), y(t)) = lim t t + t = lim =. t Next, let s try the path y = x. Parameterizing, r(t) = t, t. Then t t lim f(t, t) = lim t t t + t = lim =. t Since, we have found two paths with different limits as (x, y) (, ), so the overall limit does not exist. (b) This limit will exist, but if you didn t know that, you could always try picking a few easy paths, each of which you would find had limit g, so we expect that to be the limit.

2 In order to actually show this, however, we need to find functions which bound g and which have easy to find limits of as (x, y) (, ). There are probably a lot of ways to go about doing this, but the one I notice first is that a sin a a, so x y x + y g(x, y) + x y x + y. Now note that x y x + y (by the triangle inequality) and x, y x + y, so x x y y x y x + y g(x, y) + x y x + y + x x + y y. The expression on the left is x y, while the expression on the right is + x + y. By the sandwich theorem, = lim ( x y ) lim g(x, y) lim ( + x + y ) = implies that as desired. lim g(x, y) =,. Let f(x, y) = e x 4 cos(y ). (a) Find the plane tangent to z = f(x, y) at the point (, ). (b) Find the linear approximation of f and use it to estimate f(.9,.5). (c) Find the tangent line to the level set of f at the point (,, ). Solution. (a) Note that f(, ) =, f x = e x 4 cos(y ) = f x (, ) =, f y = e x 4 sin(y ) =, so the tangent plane is z = + (x ) + (y ). (b) The linear approximation is the same as the tangent plane, so f(x, y) + (x ). Therefore, f(.9,.5) + (.9 ) =. =.8. (c) To get the tangent to the level set, all we have to do is take the tangent plane and set z equal to whatever level set we are on (in this case, z = ), so = + (x ) + (y ) = x =.. Compute dz dt at time t = if h(x, y, z) = xy z sin x + yz = and (x(t), y(t)) = r(t) = t t, e t. Solution. The problem will require both the chain rule and implicit differentiation. Note that z is defined implicitly as a function of x and y from the condition that h(x, y, z) =. Call this

3 function z = f(x, y). Then the curve r(t) gives x and y as functions of time, so z = f(x(t), y(t)). By the chain rule, dz dt = f(r()) r (). By the implicit function theorem, f x = z = hx x h z r(). and f y = z y = hy h z, each evaluated at the point Now we explicitly compute these quantities for the given problem. Note first that r() =, and r (t) = t, e t, so r () =,. Meanwhile, h(,, z) = z = = z = and y z cos x h = x + z sin x + z y = h(,, ) =. Now f x (, ) = =, f y(, ) =, and ( ) ( ) dz dt = = t= /. 4. Find and classify the critical points of f(x, y) = x + y xy. Solution. First compute the gradient and set it equal to the zero vector: ( ) ( ) x y f = =. y x By the first equation, y = x, and by the second equation, x = y. Substituting, y = x = (y ) = y 4, so either y = or y = = y =. If y =, then x = y =, while if y =, then x = y =, so the only two critical points are (, ) and (, ). Now we compute the second derivatives in order to classify the critical points: f xx = 6x, f xy =, and f yy = 6y, so D = f xx f yy f xy = 6x 6y ( ) = 6xy 9. At the point (, ), D = 9 <, so the origin is a saddle point. At the point (, ), D = 7 > and f xx = 6 >. Therefore, the point (, ) is a local minimum. 5. Find the points on the spheroid x + 4y + 9z = 6 which are closest and furthest from the point (,, ). Solution. First, we pick our f and our g. For the constraint g, we see that we only care about points on the spheroid. Therefore, g(x, y, z) = x + 4y + 9z = 6. Then f is the quantity that we want to minimize or maximize. For this problem, that is the distance to the point (,, ), which is given by the equation d = (x ) + y + z. If you don t know this already, I really hate square roots, so let s get rid of this one by taking f(x, y, z) = d = (x ) + y + z. Note that this is okay because d is always positive, so squaring it does not change which points are a minimum or a maximum just the values at those points. Now that we have selected f and g, the method of Lagrange multipliers tells us that we have to solve a system of equations: f = λ g, g = 6, or the exceptional case, g =, g = 6.

4 This latter case is harder to remember to do (and it is quicker), so let s do it first. We compute g = x, 8y, 8z =,, when (x, y, z) = (,, ). But g(,, ) = 6, so this point is not on the constraint, and therefore we can ignore it. (If we had found that g = 6, then this point would have been a critical point, so we would have to add it to our list of critical points, which we are creating to plug into f and determine the global minimum and maximum.) Now let us consider the main case. f = x, y, z, so x x x( λ) =, y = λ 8y = λ = /4 or y =, z 8z λ = /9 or z =. The condition for the first equation looks complicated, so let s ignore it for now and start by looking at the conditions for the second and third equations. We get four cases: CASE I: λ = /4 and λ = /9. This case is stupid λ can t possibly take on two different values, so we get no critical points. CASE II: λ = /4 and z =. This information tells us nothing about y, but we know that x = λ = /4 = 4. In order to solve for y, we will need to use the constraint g = 6: 6 = x + 4y + 9z = y + 9 = y = 9 = y = ±4. CASE III: y = and λ = /9. This information tells us nothing about z, but we know that x = /9 = 9 8. In order to solve for z, we will need to use the constraint g = 6: 6 = x + 4y + 9z = z = z = = z = ± 4. CASE IV: y = and z =. This information is insufficient to determine anything about x, so we use the constraint: 6 = x + 4y + 9z = x = x = x = ±4. Now we take each of the six critical points we found and plug them into f: f(4/, 4 /, ) =, f(4/, 4 /, ) =, f(9/8,, 94/4) = 9 7, f(9/8,, 94/4) = 9, f(4,, ) = 9, f( 4,, ) = 5. 7 Of these, the point ( 4,, ) is furthest away from (,, ) (at a distance of 5 away), while the points (9/8,, ± 94/4) are tied for closest (at a distance of 9/7 away). 4

5 6. Compute the following double integral: y e x dx dy. Solution. Note that e x does not have an elementary integral, so we are stuck. Therefore, the only thing to do is to switch the order of integration. This is probably best achieved by drawing a picture, but for these notes, we will switch them algebraically. Note that y and y x from the given bounds. These can be combined into a single string y x, so x and y x. Therefore, y e x dx dy = x e x dy dx = xe x dx. Now let u = x (where u ranges from = to = ), so that the remaining integral becomes u= eu du = eu u= = (e ). 7. Find the center of mass of the circular plate x + y 4 if the density is given by the function ρ = 4 + x y. Solution. Note that although we have a disk, so polar coordinates are advisable, the center of mass is always the point ( ) xρ da yρ da ( x, ȳ) =,. ρ da ρ da These integrals a best performed in polar coordinates, where the bounds are r and θ π, while x = r cos θ, y = r sin θ, and ρ = 4 + r cos θ r sin θ. Also, don t forget that da = r dr dθ. Then the mass is π ρ da = (4 + r cos θ r sin θ)r dr dθ = π, while and xρ da = yρ da = π π r cos θ(4 + r cos θ r sin θ)r dr dθ = r sin θ(4 + r cos θ r sin θ)r dr dθ = 4π. Note that in order to compute these integrals, you may need the double angle formulas sin θ = cos θ and cos θ = + cos θ, as well as the trick that cos θ = ( sin θ) cos θ (along with u-substitution), but the details are not included here. Therefore, the center of mass is ( xρ da ( x, ȳ) =, ρ da ) ( yρ da = ρ da π, 4π ) ( =, ). π 5 5

6 8. Consider the region bounded by the cylinder x + y = 4 and the cone z = (x + y ) contained within the first octant. (a) Express the volume of this region as an integral in cylindrical coordinates. (b) Express the volume of this region as an integral in spherical coordinates. (c) Use either of the integrals you set up to determine the volume. Solution. (a) First, we must write each of the equations in cylindrical coordinates. The cylinder becomes r = 4 = r = (note that we only consider positive r when writing bounds, so we don t have to worry about the case r = ). Meanwhile, the cone becomes z = r, or z = ±r. Since we are in the first octant, z, and we know that r, so we need only consider the positive case z = r. Note that the bounds corresponding to being in the first octant are θ π and z. From the picture (not drawn here), we see that the bounds on z are z r, while r (assuming we integrate r after z) and θ π. Therefore, the volume is π/ r r dz dr dθ. Although you only had to pick one order to integrate, it is worth writing down the integral in the case where you decided to integrate r first: π/ z/ r dr dz dθ. (b) To find the volume as an integral in spherical coordinates, we must first write the bounds in spherical coordinates. The cylinder is x + y = ρ sin φ = 4, so ρ = 4 csc φ, or ρ = csc φ (again, technically ±, but csc φ since φ π, even before we start looking at restrictions from the bounds, and of course we always assume ρ ). The cone becomes ρ cos φ = z = (x + y ) = ρ sin φ = tan φ =, or tan φ = ± /. This has solutions φ = π, 5π. Since we are in the first octant, φ π, so only the former of these values 6 6 matters, and θ π. Then the volume becomes π/ π/ csc φ π/6 ρ sin φ dρ dφ dθ. (c) The integral in spherical coordinates turns out to be quite easy to do, so long as you remember the integral of csc φ, but I think that most of us would agree that the integral in cylindrical coordinates looks simplest to do, so let s pick that one. Then the volume is π/ r r dz dr dθ = π/ r dr dθ = π/ 8 dθ = 4π. 6

7 9. Consider the region in the first quadrant bounded by the curves xy =, xy =, y = x, and y = x. Find the integral of the function f(x, y) = y inside this region via a change of coordinates. Solution. We choose our u and v so that the boundary curves are of the form u = c or v = c (as much as possible anyway). Note that each of the bounds is of the form xy = c or y/x = c, so let u = xy and v = y/x. Then the bounds are u =, u =, v =, and v =, respectively. Now we need to find the Jacobian and write f in terms of u and v, each of which involve solving for x and y as functions of u and v. Note that uv = y, so y = ± uv. In the region, y is positive, so y = uv. Similarly, x = u/v. Therefore, the Jacobian is J = det ( uv v/u since u, v > due to x, y >. Also, f = uv. Then the integral is f(x, y) da = = = [ ) u/v = u/v 4v + 4v = v uv du dv = v ] u / u= ( dv = v / u= [ 8 v/ 4 v/ ] v= v= = 8 6 u v du dv ) 4 v / v / 4 dv