Math 211, Section 01,05 Solutions to Practice Problems 1

Math 211, Section 01,05, Fall 2018 Solutions to Practice Problems 1 1. Find an equation for the plane that passes throught the point (1, 2, 2) and contains the line ~r(t) = 2t~i + (3 − t)~j + (1 + 3t)~k. Answer. By its equation, the line (and hence the plane) is parallel to the vector ~v = h2, −1, 3i. Plugging in t = 0, the line also contains the point (0, 3, 1). The plane therefore also parallel to the vector from (0, 3, 1) to (1, 2, 2), which is w ~ = h1, −1, 1i. So any normal vector to the plane is orthogonal to both ~v and w, ~ so we can take ~i ~j ~k ~n = ~v × w ~ = 2 −1 3 = (−1 + 3)~i + (3 − 2)~j + (−2 + 1)~k = h2, 1, −1i. 1 −1 1 With that normal vector, and since the point (1, 2, 2) is on the plane, the plane has equation 2(x − 1) + (y − 2) − (z − 2) = 0. That is, 2x + y − z = 2. Check (if time permits): 2(2t) + (3 − t) − (1 + 3t) = 4t − t − 3t + 3 − 2 = 2, so the line is indeed in the plane, and 2(1) + 2 − 2 = 2, so the point (1, 2, 2) is also. [Note: there are other ways to do this. For example, one could find two points on the line by plugging in two different values of t, and subtract both resulting points from (1, 2, 2) to get a different pair of vectors parallel to the plane. Then take their cross product, etc.] 2.Find an equation for the plane containing the three points (1, −2, 4), (2, 1, 1), and (−4, 0, −1). Answer. The vector from [say] the first point to the second is ~a = h2−1, 1−(−2), 1−4i = h1, 3, −3i, and the vector from the first point to the third is ~b = h−4 − 1, 0 − (−2), −1 − 4i = h−5, 2, −5i. Their cross product is ~i ~ ~n = ~a × b = 1 −5 ~j ~k 3 −3 = (−15 + 6)~i + (15 + 5)~j + (2 + 15)~k = h−9, 20, 17i, 2 −5 which is normal to the plane, since ~a and ~b both lie in the plane. Using the fact that [say] the first point lies in the plane, we get the following equation for the plane: −9(x − 1) + 20(y + 2) + 17(z − 4) = 0, i.e., − 9x + 20y + 17z = 19. Check (if time permits): −9(1)+20(−2)+17(4) = −9−40+68 = 19, and −9(2)+20(1)+17(1) = −18 + 20 + 17 = 19, and −9(−4) + 20(0) + 17(−1) = 36 − 17 = 19. 3(a). Find an equation for the line of intersection of the planes x − 2y + z = 1 and 2x − y − z = 5. Answer. The two normal vectors to the planes are ~n1 = h1, −2, 1i and ~n2 = h2, −1, −1i. Let ~k ~i ~j ~v = ~n1 × ~n2 = 1 −2 1 = h3, 3, 3i. 2 −1 −1 Since ~v is perpendicular to both normal vectors, it must lie in both planes, and hence on the line. And since any scalar multiple of ~v is parallel to ~v , it follows that w ~ = h1, 1, 1i is parallel to the line. To find a point on the line, we simultaneously solve the equations for the planes, arbitrarily setting [say] z = 0, giving x − 2y = 1 and 2x − y = 5, which gives the point (3, 1, 0). Using this point and the vector w, ~ we get ~r(t) = ht + 3, t + 1, ti as an equation for the line. Check (if time permits): Plane #1: (t + 3) − 2(t + 1) + t = 1. Plane #2: 2(t + 3) − (t + 1) − t = 5. [There are LOTS of other ways to do this. On the one hand, we could use the above method but find a different point on the line. Alternately, we could simply simultaneously solve the two equations in general, rather than to pick a single point, as follows. Adding the two equations [to cancel the z term] gives 3x − 3y = 6, i.e., x = y + 2. Plugging this back into the first equation gives −y + z + 2 = 1, so z = y − 1. Using y = t, then we get ~r(t) = ht + 2, t, t − 1i as an equation for the line.] 3(b). Find the cosine of the angle between these two planes. Answer. The cosine of the angle between the planes is the cosine between their normal vectors ~a = h1, −2, 1i and ~b = h2, −1, −1i. So the desired cosine is ~a · ~b 1 2+2−1 3 √ =√ =√ √ = . 2 1+4+1 4+1+1 6 6 k~akk~bk 4. Find an equation for the plane that contains the line in problem 3(a) above and also contains the point (1, 0, −2). Answer. Plugging t = 0 into the equation for the line gives (3, 1, 0), so the vector from that to (1, 0, −2), namely ~v = h2, 1, 2i, is parallel to the plane. Taking the cross product of ~v and w ~ = h1, 1, 1i (which is parallel to the line) gives ~i ~j ~k ~n = ~v × w ~ = 2 1 2 h−1, 0, 1i. 1 1 1 So the plane is −x + z = d for some d; using the point (1, 0, −2) gives d = −3, so the plane is −x + z = −3. [You can also write x − z = 3, of course.] Check (if time permits): The point is in the plane because −1 − 2 = −3, and the line is in the plane because −(t + 2) + (t − 1) = −3. [Of course, if we had messed up the equation for the line, this check wouldn’t help us much.] [There are other ways to do this, like finding two points that both lie on both of the original planes, and then using those two points and (1, 0, −2) as three points on the plane, to get two vectors and take their cross product, as we did in problem 2.] 5. Let ~a = 5~i − 3~j + 2~k and ~b = ~i − 3~k. Find the (vector) projection of ~a onto ~b, and find the (vector) projection of ~b onto ~a. ~a · ~b~ 5 − 0 − 6 1 Answer. First, proj~b~a = b= h1, 0, −3i = h−1, 0, 3i. ~b · ~b 1+9 10 1 3 1~ 3~ [You can write that as h− 10 , 0, 10 i or − 10 i + 10 k if you prefer.] ~b · ~a 5−0−6 1 Second, proj~a~b = ~a = h5, −3, 2i = h−5, 3, −2i. ~a · ~a 25 + 9 + 4 38 6(a). Confirm that the four points P1 = (1, 1, −1), P2 = (3, −2, 0), P3 = (2, −3, 0), and P4 = (4, −6, 1) form the vertices of a parallelogram. −−−→ Answer. The vector ~v1 = P1 P2 is h3 − 1, −2 − 1, 0 − (−1)i = h2, −3, 1i. −−−→ We also have P3 P4 = h4 − 2, −6 − (−3), 1 − 0i = h2, −3, 1i = ~v1 , so those two line segments are parallel. −−−→ −−−→ Meanwhile, the vector ~v2 = P1 P3 is h2 − 1, −3 − 1, 0 − (−1)i = h1, −4, 1i, while the vector P2 P4 is h4 − 3, −6 − (−2), 1 − 0i = h1, −4, 1i = ~v2 , so those two line segments are parallel. Thus, the four points are the vertices of a parallelogram. 6(b). Find the area of the parallelogram with vertices P1 , P2 , P3 , P4 . Answer. The parallelogram is given by the vectors ~v1 and ~v2 from part (a), so its area is k~v1 ×~v2 k. We compute ~i ~j ~k ~v1 × ~v2 = 2 −3 1 = h−3 − (−4), 1 − 2, −8 − (−3)i = h1, −1, −5i. 1 −4 1 So the area is kh1, −1, −5ik = √ √ √ 1 + 1 + 25 = 27 = 3 3. 7. For each of the following pairs of lines, decide whether they cross, are parallel, or are skew. If they cross, find their point of intersection and the cosine of the angle between them. 7(a). ~r1 (t) = h1 + 3t, 3 − t, ti and ~r2 (t) = ht + 4, 2t − 5, 5 − ti. Answer. The vectors ~a = h3, −1, 1i and ~b = h1, 2, −1i are parallel to the two lines; but these vectors are not parallel to each other (since they are not scalar multiples), so the lines are not parallel. To see whether they intersect, we must try to solve h1 + 3s, 3 − s, si = ht + 4, 2t − 5, 5 − ti for s and t. [Note: we’re using different parametric variables for the two lines.] The x-coordinate gives t = 3s − 3, so that the y-coordinate gives 3 − s = 6s − 11, i.e. 7s = 14. Thus, s = 2 and so t = 3. That also satisfies the z-coordinate s = 5 − t; so the lines do intersect, at the point ~r1 (2) = ~r2 (3) = h7, 1, 2i. The angle between the lines is the angle between the parallel vectors ~a and ~b, so its cosine is ~a · ~b 3−2−1 √ = 0. = √ 11 · 6 k~akk~bk 7(b). ~r1 (t) = h1 − 2t, 3t + 1, t − 2i and ~r2 (t) = h1 + t, −2t, 5t + 2i. Answer. The parallel vectors are h−2, 3, 1i and h1, −2, 5i, which again are not scalar multiples, so the lines are not parallel. To look for an intersection point, we try to solve h1 − 2s, 3s + 1, s − 2i = h1 + t, −2t, 5t + 2i for s and t. The x-coordinate gives t = −2s, and therefore the y-coordinate says 3s + 1 = 4s. Thus, s = 1, and so t = −2. But in the last equation, we must then have −1 = −8, which is false. So the two lines are skew. 7(c). ~r1 (t) = h3 + 2t, 4t − 7, 5 − 6ti and ~r2 (t) = h2 − 3t, 1 − 6t, 9t + 4i. Answer. The parallel vectors are ~v1 = h2, 4, −6i and ~v2 = h−3, −6, 9i. Clearly, ~v1 = − 32 ~v2 , so the vectors, and hence the lines, are parallel to each other. [Actually, even though I didn’t ask, it’s worth noting that they are not actually the same line as each other. To see this, we need to show that they never meet. Solving ~r1 (s) = ~r2 (t) gives 3 + 2s = 2 − 3t in the first coordinate, 4s − 7 = 1 − 6t in the second coordinate, and 5 − 6s = 9t + 4 in the third. Subtracting twice the first equation from the second gives −13 = −3, which is false; so the lines don’t meet. Thus, the two lines are actually different, and parallel.] 7(d). ~r1 (t) = h5 − t, 4 − 3t, 2t − 1i and ~r2 (t) = ht + 2, 3t − 5, 5 − 2ti. Answer. The parallel vectors are ~v1 = h−1, −3, 2i and ~v2 = h1, 3, −2i. Then ~v1 = −~v2 , meaning that the vectors, and hence the lines, are parallel. [Note: closer observation shows that ~r1 (0) = h5, 4, −1i = ~r2 (3). So the lines are parallel and pass through the same point, meaning that they are actually the same line! We could also have discovered this by trying to solve ~r1 (s) = ~r2 (t) as in the comment in part (c) above, only this time we would have found a solution, namely any time t = s + 3, since it turns out that ~r1 (s) = ~r2 (s + 3).] 7(e). ~r1 (t) = h5 − t, 4 − 3t, 2t − 1i and ~r2 (t) = h4t − 1, t − 5, 5 − ti. Answer. The parallel vectors are ~v1 = h−1, −3, 2i and ~v2 = h4, 1, −1i, which are not scalar multiples (e.g., looking at first coordinates, we’d have to multiply ~v1 by −4, but that doesn’t work for the second coordinates), so the lines are not parallel to each other. Solving ~r1 (s) = ~r2 (t) gives 5 − s = 4t − 1, 4 − 3s = t − 5, and 2s − 1 = 5 − t. Adding the last two equations gives 3 − s = 0, so s = 3, and then t − 5 = −5, so t = 0. However, in that case the first equation says 2 = −1, which is false. Thus, the two lines do not cross; that is, they are skew. 8. Sketch graphs of the following surfaces. Answers. I can’t easily sketch the graphs here, but I will describe them and how to figure them out. See me for the graphs. [I’ll use technical names like ellipsoid below, but again, you do not need to know such terms for the exam.] 8(a). 4x2 + y 2 + 25z 2 = 100 The intersection with the xy-plane (i.e., z = 0) is 4x2 + y 2 = 100, i.e., x2 /52 + y 2 /102 = 1, which is an ellipse hitting the x-axis at ±5 and the y-axis at ±10. The intersection with the xz-plane (i.e., y = 0) is 4x2 + 25z 2 = 100, i.e., x2 /52 + z 2 /22 = 1, which is an ellipse hitting the x-axis at ±5 and the z-axis at ±2. The intersection with the yz-plane (i.e., x = 0) is y 2 + 25z 2 = 100, i.e., y 2 /102 + z 2 /22 = 1, which is an ellipse hitting the y-axis at ±10 and the z-axis at ±2. [And slices by other planes like x = c give nothing at all if |c| is too big (e.g. x = 10 gives y 2 + 25z 2 = −300, which has no solutions), but smaller ellipses if |c| is not too big.] So this graph is an ellipsoid, which can be sketched by sketching the above ellipses and tying it all together. Make sure you label the points (5, 0, 0), (0, 10, 0), and (0, 0, 2), where it hits the x-, y-, and z-axes, respectively. 8(b). x2 + 4y 2 − z 2 = 4 The intersection with the xy-plane (i.e., z = 0) is x2 + 4y 2 = 4, i.e., x2 /22 + y 2 = 1, which is an ellipse hitting the x-axis at ±2 and the y-axis at ±1. The intersection with the xz-plane (i.e., y = 0) is x2 − z 2 = 4, which is a hyperbola that crosses the x-axis at x = ±2 and doesn’t cross the z-axis. The intersection with the yz-plane (i.e., x = 0) is 4y 2 − z 2 = 4, which is a hyperbola that crosses the y-axis at y = ±1 and doesn’t cross the z-axis. [Slices by planes z = c give bigger ellipses x2 + 4y 2 = 4 + c2 . Slices by planes y = c give x2 − z 2 = 4 − 4c2 , which will still be hyperbolas (except for c = ±2, which gives two crossing lines x2 = z 2 ); when |c| is small they don’t cross x = 0, and when |c| is big they don’t cross z = 0. Slices by x = c give 4y 2 − z 2 = 4 − c2 , which are hyperbolas like the y = c cases.] After sketching the intersections with the coordinate planes, it is not too difficult to see that the surface is a hyperboloid of one sheet, running vertically. Your graph should clearly show that there is an ellipse at each end, and that it narrows in the middle and widens at the ends (i.e., that it’s not a cylinder). It would be nice to specifically mark in the ellipse x2 + 4y 2 = 4 where it intersects the xy-plane. 8(c). z = 4x2 + y 2 The intersection with the xy-plane (i.e., z = 0) is 4x2 + y 2 = 0, which is just the point x = y = 0, i.e., the origin. So it may be more useful to notice that any other horizontal slice z = c gives nothing if c < 0, but it gives an ellipse 3x2 + y 2 = c if c > 0. The intersection with the xz-plane (i.e., y = 0) is z = 4x2 , which is a parabola opening upward. The intersection with the yz-plane (i.e., x = 0) is z = y 2 , which is a parabola opening upward. Drawing the intersections with the xz- and yz-planes along with a slice z = c for some c > 0, it becomes clear that the surface is an elliptic paraboloid (i.e., a bowl) opening upward. 8(d). 3x2 − z 2 = 9 The variable y is not involved, so for any plane y = c parallel to the xz-plane, the √ trace is the curve 3x2 − z 2 = 9. This trace curve is a hyperbola; it intersects the x-axis at x = ± 3, and it does not intersect the z-axis at all. On the xz-plane in the usual sketch of R3 , that means the hyperbola has one arc opening towards us and the other arc opening away from us (as opposed to one arc opening up and the other one opening down). Thus, the surface is a hyperbolic cylinder, running parallel to the y-axis (i.e., left to right). To sketch it, draw one copy of the hyperbola over on the right side of your picture, and another identical copy on the left side. (Both are drawn in perspective, of course, since they are translates of the hyperbola in the xz-plane. Then connect the endpoints of the two hyperbolas by horizontal lines (i.e., lines parallel to the y-axis). The final picture should have one hyperbolic sheet opening towards us, and the other one opening away from us, with both lying “on their sides”. 8(e). z = 4x2 − y 2 The intersection with the xy-plane (i.e., z = 0) is 4x2 − y 2 = 0, which is 4x2 = y 2 , or in other words y = ±2x; that is, it is two straight lines crossing in an X at the origin. The horizontal slice z = c for c > 0 is a hyperbola intersecting the x-axis but not the y-axis; for c < 0, it is a hyperbola intersecting the y-axis but not the x-axis. That is, the horizontal slices above the xy-plane are hyperbolas opening toward us and away from us; the slices below the xy-plane are hyperbolas opening to our left and to our right. Setting x = 0 gives z = −y 2 , which is a parabola opening downwards in the yz-plane; setting y = 0 gives z = 4x2 , which is a parabola opening upwards in the xz-plane. The resulting surface is a hyperbolic paraboloid. To sketch it, draw the two parabolas— the upward one on the xz-plan in perspective, and the downward one face-on in the yz-plane. (FYI: as you’ll see if you practice the sketch once first, portions of the back half of the upward parabola will ultimately be hidden from view and hence should be drawn as a dotted line in your final sketch; but the rest of the upward parabola, and all of the downward parabola, should be visible.) Then, draw the two arcs of a hyperbola opening toward and away from you coming off the endpoints of the top parabola. Similarly, draw the two arcs of a hyperbola opening to your left and right coming off the endpoints of the bottom parabola. (The back halves of those last two hyperbolic arcs will ultimately be mostly hidden from view and so should be mostly dotted lines. As before, you won’t know this until you do a practice sketch first.) Make these hyperbolic arcs fairly wide, to get ready for the next step. To finish the surface, draw some approximately (but not quite) vertical lines, to connect each of the four endpoints of the bottom hyperbola to the endpoint of the top hyperbola that is approximately above it. In the end, it should look something like Figure 8 on page 853, but rotated 90◦ . Take particular note of the way Stewart drew the parabolas, hyperbolas, and vertical-ish lines in that figure. 8(f). z 2 = 4x2 − y 2 Setting z = 0 gives 4x2 = y 2 , which is the two crossing lines y = ±2x in the xy-plane. [Similarly, setting y = 0 gives z 2 = 4x2 , which is the two crossing lines z = ±2x in the xz-plane. Meanwhile, setting x = 0 gives y 2 + z 2 = 0, which is just the single point at the origin, so that’s probably not worth considering.] For any c 6= 0, the plane x = c (parallel to the yz-plane) gives the circle y 2 + z 2 = 4c2 . [It’s also true that the planes where we set either y = c or z = c give hyperbolas, but those trace curves are not particularly useful at this point.] Thus, the surface is a double circular cone, lying along (i.e., centered along) the x-axis, with one cone opening towards us and the other opening away from us. So it looks like Figure I on page 857, except that it runs along the x-axis instead of along the y-axis. To sketch it, draw a circle in back and a circle in front, and connect the points of those circles that lie in the xy-plane to each other along the pair of crossing lines in the xy-plane. 8(g). y = 4z 2 + x2 Setting x = 0 gives the parabola y = 4z 2 opening along the positive y-axis in the yz-plane, and setting z = 0 gives the parabola y = x2 opening along the positive y-axis in the xy-plane. [Setting y = 0 gives 4z 2 + x2 = 0, which is simply the point at the origin.] Setting y = c gives 4z 2 + x2 = c, which is nothing if c < 0, just the point at the origin if c = 0, and an ellipse if c > 0. Moreover, the ellipse is twice as wide in the x-direction as in the z-direction, and it gets larger as c increases. Combining all this information, we see that the surface is an elliptic paraboloid opening along the positive y-axis. To sketch it, draw the ellipse 4z 2 + x2 = c in perspective on the plane y = c parallel to the xz-plane, trying to show that this ellipse is wider than it is tall. Also draw the parabolas in the xy- and yz-planes, ending at the ellipse you just drew. 8(h). y = 4z 2 − x2 Setting y = 0 gives 4z 2 = x2 , which is two crossing lines. Setting y = c gives 4z 2 − x2 = c gives a hyperbola— if c > 0, it opens along both directions of the z-axis, but if c < 0, it opens along both directions of the x-axis. Setting x = 0 gives y = 4z 2 , which is a parabola in the yz-plane, opening to the right; setting z = 0 gives y = −x2 , which is a parabola in the xy-plane, opening to the left. Thus, we have a hyperbolic paraboloid. As in part (e), sketch it by first sketching the two parabolas. Then, from the endpoints of the parabolas, draw the hyperbolas. Finally connect the endpoints of the hyberbolas by (this time) roughly horizontal lines. See Figure V on page 857 for an idea of what I mean, except that our hyperbolic paraboloid is a mirror image (across the xz-plane) of the one in Figure V. 8(i). x2 − 4y 2 + z 2 = −4 Rearranging, the equation is −x2 + 4y 2 − z 2 = 4. Setting x = 0 gives the hyperbola 4y 2 − z 2 = 4 in the yz-plane, opening along both directions of the y-axis. Setting z = 0 gives the hyperbola 4y 2 − x2 = 4 in the xy-plane, again opening along both directions of the y-axis. Setting y = 0 gives −x2 − z 2 = 4, which has no solutions, but setting y = c for |c| > 1 gives the circle x2 + z 2 = 4(c2 − 1). Thus, the surface is a hyperboloid of two sheets, opening along both directions of the y-axis. To sketch it, draw circles (in perspective) at both ends of the y-axis, on planes parallel to the xz-plane. Then draw the hyperbolic arcs on both sides, lying in the xy- and xz-planes. Each hyperbolic arc should start and end on the same circle; the ones on the left should pass through the point (0, −1, 0), and the ones on the right should pass through (0, 1, 0), since these are the two points on the y-axis at the “bases” of the two sheets. 9(a). Find a vector-valued function tracing out the curve of intersection of the cylinder x2 + z 2 = 9 and the plane x + y + z = 1. Answer. The circle x2 + z 2 = 9 in the xz-plane can be parametrized by x = 3 cos t, z = 3 sin t. So to be in the plane x + y + z = 9, we have y = 1 − x − z = 1 − 3 cos t − 3 sin t. So the curve in question can be parametrized by ~r(t) = h3 cos t, 1 − 3 cos t − 3 sin t, 3 sin ti. Check (if time permits): The curve satisfies x2 + z 2 = 9, since (3 cos t)2 + (3 sin t)2 = 9(cos2 t + sin2 t) = 9; and it satisfies x + y + z = 1, since 3 cos t + (1 − 3 cos t − 3 sin t) + 3 sin t = 1. 9(b). Find the tangent line to the curve in part (a) at the point (3, −2, 0). Answer. The point (3, −2, 0) is ~r(0). Meanwhile, ~r 0 (t) = h−3 sin t, 3 sin t − 3 cos t, 3 cos ti. So the tangent vector at (3, −2, 0) is ~r 0 (0) = h0, −3, 3i. So the tangent line is h3, −2 − 3t, 3ti. [We could also use the vector h0, 1, −1i, which is parallel to h0, −3, 3i, which would give the tangent line as h3, t − 2, ti. That’s the same line, but a different parametrization of it.] 1 10. Let ~r(t) = h1 − 2t, , 2 ln ti. t 10(a). Find the domain of ~r, and compute the derivative ~r 0 (t) and speed k~r 0 (t)k (for any t in that domain). Answer. 1 − 2t is defined for all t ∈ R; 1/t is defined for t 6= 0, and 2 ln t is defined for t > 0. So the domain of ~r is where all three are simultaneously defined, i.e.,√(0, ∞). The derivative is √ 0 −2 −1 0 −4 ~r (t) = h−2, −t , 2t i, and the speed is k~r (t)k = 4 + t + 4t−2 = t−4 + 4t−2 + 4 = t−2 + 2. 10(b). Compute the arclength of the curve from t = 1 to t = 3. Answer. Since we already computed, k~r 0 (t)k = t−2 + 2, the arclength from t = 1 to t = 3 is Z 3 Z 3 14 1 3 0 k~r (t)k dt = t−2 + 2 dt = −t−1 + 2t 1 = − + 6 + 1 − 2 = . 3 3 1 1 11. Sketch the curve traced out by each of the following vector-valued functions, and indicate with an arrow the direction in which the curve is traced. Provide some brief extra description of the curve, either by labelling points, giving a brief but precise verbal description of it, or by an alternate mathematical equation for the curve, or some combination of such descriptions. Answers: I can’t easily sketch the graphs here, so I’ll merely describe them. See me if you’d like me to draw the graphs for you. 11(a). ~r(t) = het , e2t − 1i, for t ∈ R. Answer. Note that the x- and y-coordinates are related by y = x2 − 1. Thus, the curve traced out is (part of) the parabola y = x2 − 1. However, since et outputs values in the interval (0, ∞), only the right half of the parabola, not including the vertex point (0, −1), is actually on the curve. Since x = et increases as t increases, this right half of the parabola is traced from left to right. 11(b). ~r(t) = h2t, 4 sin(5t), cos(5t)i, for t ∈ R. Answer. Note that the y- and z-coordinates are related by y 2 + 16z 2 = (4 sin 5t)2 + 16(cos 5t)2 = 16(sin2 5t + cos2 5t) = 16. Thus, if we ignore the x-coordinate, ~r(t) moves clockwise (because y is the sine and z is the cosine) around the ellipse y 2 + 16z 2 = 16 in the yz-plane. (This ellipse is four times as wide as it is tall— it hits the y-axis at y = ±4, and it hits the z-axis at z = ±1.) However, when we do take the x-coordinate into account, we see that as the y- and z-coordinates rotate around that ellipse, the point is moving towards us along the x-axis, since x = 2t steadily increases as t increases. Thus, the curve is an elliptical helix wrapped around the x-axis. The particle moves towards us while orbiting clockwise around the x-axis, all the while staying on the elliptical cylinder y 2 + 16z 2 = 16. 11(c). ~r(t) = ht cos t, t sin t, t2 i, for t ≥ 0. Answer. Note that if we take x2 + y 2 (as suggested by the cos t and sin t that appear in the xand y-coordinates), we get x2 + y 2 = t2 cos2 t + t2 sin2 t = t2 = z. Thus, the curve lies in the elliptic paraboloid z = x2 +y 2 , opening along the positive z-axis. Viewed from above, the x- and y-coordinates (t cos t, t sin t) rotate counterclockwise around the z-axis in ever larger loops. Meanwhile, the z-coordinate z = t2 is increasing faster and faster as t ≥ 0 increases. Thus, starting at the origin ~r(0) = h0, 0, 0i, the particle moves counterclockwise and up the elliptic paraboloid z = x2 + y 2 . 12. For each of the following vector-valued functions, write down, but do not evaluate, a definite integral giving the arclength of the curve parametrized by the function. 12(a). ~r1 (t) = ht2 , e3t , 2t + 3i for 0 ≤ t ≤ 4. Answer. ~r1 0 (t) = h2t, 3e3t , 2i, Z so the arclength is 4p 4t2 + 9e6t + 4 dt 0 12(b). ~r2 (t) = ht sin t, t cos t, 2ti for −2 ≤ t ≤ 3. Answer. ~r2 0 (t) = hsin t + t cos t, cos t − t sin t, 2i, which has k~r2 0 (t)k2 = (sin t + t cos t)2 + (cos t − t sin t)2 + 4 = sin2 t + cos2 t + t2 (cos2 t + sin2 t) − 2t sin t cos t + 2t sin t cos t + 4 = 5 + t2 . Z 3p 5 + t2 dt So the arclength is −2 12(c). ~r3 (t) = h3 cos t, 2 sin ti. Answer. We are not told an interval for ~r3 (t) = h3 cos t, 2 sin ti, but this function clearly retraces itself every 2π, so we can use t ranging in any interval of length 2π, like [0, 2π]. [FYI: in fact, the curve is an ellipse, but you don’t need to know that to do the problem.] Z 2π p Z 2π p 2 0 2 Since ~r3 (t) = h−3 sin t, 2 cos ti, the arclength is 9 sin t + 4 cos t dt = 4 + 5 sin2 t dt. 0 0