Solutions Exam 1 Math 221 For each problem, show your work and state clearly the answer you obtain. There are ten problems. The first 9 are worth 12 points each. The last is worth 20 points. 1. Find vectors p and q such that p is parallel to 2, 1, 2, q is perpendicular to 2, 1, 2, and p + q = 4, 3, 1. This task is exactly what the vector projection of u onto v is designed to do: it gives p. If you think back to the discussion of vector projection, it was introduced not as a formula, but as the thing that fit a certain description. The picture (figure 1) shows two black vectors, one parallel to the green vector and one perpendicular to it, adding to the blue vector. The first black vector is the vector projection of the blue vector onto the green one. The other is simply the result of subtracting the first from the blue vector. It was from this characterization of the vector projection that we arrived at the formula for it. Here, v = 2, 1, 2, u = 4, 3, 1, and the formula spits out p = 4, 3, 1 2, 1, 2 2, 1, 2 = 26/9, 13/9, 26/9. 2, 1, 2 2, 1, 2 Now q = 4, 3, 1 p = 10/9, 14/9, 17/9. If, given the task of finding the vector projection, without the task being labeled with the words vector projection, it didn t occur to you to use the vector projection, all was not lost. You would have had an algebra problem. What does it mean for two vectors to be parallel? It means that each is a scalar multiple of the other. (Maybe you d say antiparallel if the scalar were negative.) What does it mean for two vectors to be perpendicular? It means that neither is zero, but their dot product is zero. So here, the question translates to this: find a number t (one number, the same number every time the letter t appears), so that if p = 2, 1, 2 and p + q = 4, 3, 1 (so that q = ( 4, 3, 1 t 2, 1, 2 )), then ( 4, 3, 1 t 2, 1, 2 ) 2, 1, 2 ) = 0. Grinding out the arithmetic translates this into 13 9t = 0. So p = (13/9) 2, 1, 2 and the rest goes just like it did in the other solution. 2. Find a vector p such that the angle between p and 2, 1, 3 is equal to the angle between p and 1, 2, 2. The zero vector doesn t make an angle with any vector and thus cannot serve as an answer. 1
Can they both be right angles? Yes. So you take the cross product and get 4, 1, 3. There are other answers. The next simplest would be to take u = 1 14 2, 1, 3 and v = 1 3 1, 2, 2, set w = u + v, and give w as your answer. Nobody said simplify, so you d be done. The reason it works is that if you have unit vectors, the midpoint of the line joining the endpoints, coming from the origin, is also the angle bisector because it s an isosceles triangle. There s a purely algebraic way to do this as well. You want a vector w = a, b, c say, with the property that the cosines of the angles between w and the two given vectors are equal to each other. That translates into 2a + b + 3c 14 a2 + b 2 + c = a + 2b + 2c 2 3 a 2 + b 2 + c. 2 Canceling the common factors and simplifying, you want (6 14)a + (3 2 14)b + (9 2 14)c = 0. Now here s the thing: if you detach yourself just a little from the task at hand and think about what sort of task this is, it amounts to this: Find a point on a given plane! The variables are a, b, and c rather than x, y, and z, but no matter. You have only to pick any old values for a and b and then solve for c. That gets you out of the difficulty of trying to find a specific tree when there s a whole forest blocking your view. Here, taking a = b = 0 doesn t work because then c = 0 and that wasn t what you wanted; you wanted a nonzero list. But taking a = 1 and b = 0, say, leads to this equation: (6 14) + (9 2 14)c = 0. So c = (6 14)/(9 2 14) and you have your vector. Finally. It would have been better to take the cross product, but if inspiration doesn t strike, algebra will grind it out. All other answers are combinations sw + t 4, 1, 3 with at least one of s and t nonzero. 3. Find the point on the line x = t, y = 2t, z = 2t nearest (7, 8, 9). If you re comfortable with vector projections, you d just project 7, 8, 9 onto 1, 2, 2 ; the answer is (41/9) 1, 2, 2. If not, you can use the distance formula. The distance is the square root of (7 t) 2 + (8 2t) 2 + (9 2t) 2. The minimum distance occurs at the same point at which the square of the distance is minimal. The minimum occurs at the value of t where the derivative (of the square of the distance, to keep things simple) is zero. Thus you want ( 1)(7 t) 2(8 2t) 2(9 2t) = 0, or equivalently, 41 9t = 0 and again t = 41/9. 4. Consider the surface H with equation z = x 2 y 2. (a) Which sort of quadric surface is H? (It s not a sphere.) It s a hyperbolic paraboloid. 2
Figure 1: Vector Projection (b) The intersection of H with the plane z = 1 is a curve. Give the most appropriate single-word description of this intersection, and sketch the intersection. When z = 1, the curve is 1 = x 2 y 2 which is a hyperbola. The asymptotes are the lines x = ±y in the plane z = 1. (c) Same question but with the plane now being y = 2. Now the equation is z = x 2 4 which is a parabola that opens up, in the plane y = 2, with its vertex at (0, 2, 4). 5. (a) Find 1, 2, 3 2, 1, 1. The answer is 1, 5, 3. (b) Find the volume of the set of all points inside the solid with corners (0, 0, 0), (1, 2, 3), (3, 3, 4), (2, 1, 1) and the result of adding (1, 4, 2) to each of the first four. Another characterization of the solid whose volume is required is that it is the set of all points of the form s 1 (1, 2, 3) + s 2 (2, 1, 1) + s 3 (1, 4, 2) with 0 s 1 1, 0 s 2 1, and 0 s 3 1. This is a parallelopiped, and you get the volume by the (absolute value of) the determinant of three vectors from any chosen vertex along an edge to three adjacent vertices. This is also equal to the (absolute value again) of the dot product of one of them with the cross product of the other two. Since you presumably already have the cross product of the first two, you just dot that with 1, 4, 2 and the volume is 13. Almost any choice of three vectors between vertices of this solid would work; you just have to avoid taking them all in the same plane. Thus, the triple product formula using just the vectors from the origin to the first three points listed would yield 0 because the corners form a parallelogram with the (3, 3, 4) = (1, 2, 3) + (2, 1, 1). 3
6. Let f(x, y) = xy/(x 2 + y 2 ). (a) State the exact meaning, using ϵ and δ, of the statement lim f(x, y) = 1 (x,y) (0,0) 2. The statement unpacks (unzips, might we say?) to this: For all ϵ > 0 there exists δ > 0 such that if the distance from (x, y) to the origin is positive but less than δ, then the distance from f(x, y) to 1/2 is less than ϵ. (That s what the statement says, but for this function and that declared limit, it s not true.) (b) Now let ϵ = 1/4. For any δ > 0, let (x, y) = (δ/2, δ/2). Alice claims that this choice of ϵ, and this strategy for choosing (x, y), is the key to a proof that the limit of f at (0, 0), if it exists at all, is at any rate not equal to 1/2. Alice has made a poor choice of (x, y). Patch her choice and fill in the missing details and thus prove that the statement in part (a) is false. Alice had the bad luck to choose a formula for (x, y) that actually always gives 1/2. Almost any tweak fixes that. For instance, taking (x, y) = (δ/2, 0) would yield f(x, y) = 0, which has a distance from 1/2 of 1/2. As that is more than the leeway 1/4 at issue, this proves that the limit is (if it exists at all, which it does not) is at any rate not 1/2. 7. Find D x,y e xy 2x 3y. (Equivalently, find 2 x y e xy 2x 3y.) Any second derivative, whether in calculus 1 or now, is got by taking the derivative of the original function f, and then taking the derivative of what you just got. The only thing new is that first you differentiate with respect to y, and then differentiate the result with respect to x. (The derivative can be calculated either by differentiating first with respect to x, and then the result of that with respect to y, or in the reverse order. This is because all the partial derivatives are continuous, being formulas in x and y with no risk factors, so to speak.) So taking the x derivative first gives (y + 2)e xy 2x 3y, and then differentiating that with respect to y gives [ 1 (x+3)( (y +2))]e xy 2x 3y = (xy +2x+3y +5)e xy 2x 3y. 8. Find the directional derivative of 8x + x 3 + 2x 2 y + 3xy 2 in the direction parallel to 3, 4 at (5, 2). Recall that the directional derivative is the limit of the change in f divided by the distance displaced. Given a function f and a displacement vector hu, the change in f, when hu is short, is approximately given by f hu. The distance displaced is h u. In the limit, that approximation is exact, and the directional derivative is ( f hu)/(h u ). The h s cancel. 4
The gradient of f here is 8+3x 2 +4xy+3y 2, 2x 2 +6xy, if we don t specify x and y. But we should, because the problem says at (5, 2). Plugging in x 5, y = 2 gives f = 135, 110. Now since we want to divide by the distance, that means we need to keep in mind that the length of u is 5, not 1. The unit vector in the chosen direction is (1/5) 3, 4. Dot f = 135, 110 with 3/5, 4/5 to get the answer, 169. Remark: what if the question had been T/F, asking which was greater, the directional derivative corresponding to 3, 4, or that corresponding to 4, 3? Would that have been more work, or less? Less, because it s clear that 4, 3 is more nearly parallel to the gradient than 3, 4, because of having the larger number, 4, in the same slot as the larger 135 in the gradient. So the answer would have been that the greater directional derivative was that in the direction of 4, 3. 9. Find a vector normal to the surface x 2 y 2 3z 2 = 9 at (4, 2, 1). The gradient of x 2 y 2 3z 2 is 2x, 2y, 6z which evaluates at (4, 2, 1) to 8, 4, 6 and that is the answer. 10. Let f(x, y) = xy(1 x y)e (x2 +y 2). The picture shows a contour plot of f on the domain D = [ 2, 2] [ 2, 2]. (a) Judging from the contour plot, how many critical points does f have inside D? (The graphic displayed on the overhead gave a better view, because color is preserved for one thing, and because higher resolution was possible. This solution also includes a higher resolution plot, but it has to be viewed at magnification to sidestep Moiré effects.) First note that whenever contours cross we have a critical point because the function is remaining flat in two directions so the tangent plane is flat there. The iconic example is the function f(x, y) = xy, which has contours along the x and y axis and a saddle point where they cross. Here, the function is zero along three lines, x = 0, y = 0, and x + y = 1. These three lines intersect in pairs at three points: (0, 0), (1, 0, and (0, 1). This shows up in the picture as a crisscross of lines forming a triangle. That gives us three saddle points. The lines divide the plane into seven regions. In each, there will be a local maximum somewhere, or a local minimum, depending on whether the function is positive or negative in that region. The factor e x2 y 2 ensures that the limit as (x, y) goes to is zero, so the maximum or minimum in each zone will not be on the receding boundary once the box is made big enough. Going by the contour plot, the box is already big enough and these max or min occur inside the plot rather than further out. There are seven summits or hollows visible (one for each region.) Three saddle points and seven locals makes for ten critical points. (b) Judging from the contour plot, it seems that there is a critical point at or near (1, 1). Determine, with paper calculations, whether or 5
2 1 0-1 -2-2 -1 0 1 Figure 2: Countour Plot 6 2
not there really is a critical point exactly at (1, 1). (There really is. Taking the gradient and setting x = y = 1 yields 0. (c) Whether it s exactly at (1, 1) or not, what sort of critical point is it? Local minimum, local maximum, or neither? Explain. (An explanation need not be a proof, for this item, though a proof would be a good explanation. You could also give reasons related to the computer plot or to the formula for f.) Judging from all those contour loops around it, it s not a saddle. Now f itself is negative at (1, 1), while at neighboring points such as (1, 0), it s zero. Thus f seems to have a local minimum at (1, 1). 7