Math 251, Spring 2015-copyright Joe Kahlig Page 11. compca =a · c|c|=6√32. center: (−2, 4, 0) and radius =√453. |a × b| = ||a||b|sinθ| and will have a max when sin θ = 1.mas is |a||b| = 3 ∗ 4 = 124. Note: since we are identifying surfaces, this means we are graphing in 3 dimensions.cone x2− y2+ z2= 0elliptic Paraboloid 5x = 2y2+ 3z2circular cylinder 6z2+ 6y2= 8Hyperboloid of two sheet x2− 4y2− 2z2= 10ellipsoid 3x2+ 2y2+ 4z2= 95. Since the line is perpendicular to the plane, this means the direction vector of the line is parallelto the normal vector of the plane. so let the normal vector be some scalar multiple of thedirection vector.parametric equation of the line: x = 1 + 2t, y = −2 + 5t, z = 1 − 3tn = v = h2, 5, −3i.Answer: 2x + 5y − 3z = 46. r(t) =t, t2, t3and a(p) = h1 + 4p, 1 + 16p, 1 + 52pi.(a) x component; t = 1 + 4py component: t2= 1 + 16pnow solve: (1 + 4p)2= 1 + 16p to get 16p2−8p = 0 or 8p(2p −1) = 0 or p = 0 and p = 1/2.The points are a(0) = (1, 1, 1) and a(1/2) = (3, 9, 27)(b) For the point (2, 4, 8), t = 2 and r0(2) = h1, 4, 12ix − 2 =y −44=z −8127. either do a product rule or distribute the f(t) though the vector r(t).3t2et+ t3et, 5t4+ 3t2, 3t2sin(2t) + 2t3cos(2t)8. r0(t) =4t, t2, 0and |r0(t) =√16t2+ t4= t√16 + t2Length =3R0t√16 + t2dt =613. use u-sub with u = 16 + t29. (a) T (t) =1√4 + 4t2+ 9t4D2, 2t, 3t2E(b) r0(t) × r00(t) =6t2, −12t, 410. The angle between plane is the acute angle between the normal vectors .x + 2y + z = 4 has n1= h1, 2, 1i and |n1| =√63x + 6y + 2z = 12 has n2= h3, 6, 2i and |n2| = 7θ = cos−1177√6= 7.49 degreesMath 251, Spring 2015-copyright Joe Kahlig Page 211. (a) Domain: [−4, 1) ∪ (1, 4](b) This limit does not exist since limt→1sin tt2− 6t + 5= DNE. Note this limit is not a L’Hopitallimit.12. (a) The direction vectors for the lines are v1= h1, 4, 2i and v2= h2, 4, 0i. Since there is not anumber that can be multiplied by v1to get v2, the lines are not parallel.(b) first get the parametric equations for the second line. Be sure to use a variable differentfrom the variable for the first line. Note: If you use a t as the variable for both lines, thenyou are trying to find out if both lines intersect at the same point at the same time. Thisquestion doesn’t ask this, it only wants to know if the ”paths” intersect.Line 2: x = 1 + 2s, y = 5 + 4s, and z = 10z components equal gives 2t = 10 or t = 5.x components equal gives t + 2 = 1 + 2s or 7 = 1 + 2s or x = 3.Now check the y components for t = 5 and s = 3. Line 1 gives y = 21 and Line 2 givesy = 17. Since these are not the same, the lines are skew.13. There was an error in this problem. The given lines were actually skew(not parallel and notintersecting) and thus no plane will contain both lines.This problem was graded as if the lines intersected. Since this is what was intended.Steps that would have been performed if the lines intersected.step 1: find the direction vector of both lines.step 2: find cross product of these vectors. this is the normal vector of the desired plane.step 3: use any point on one of these lines and create the desired plane.14. Method 1: If there is a plane, P2, containing the line, L, where P2is parallel to the given plane,P , then the normal vector of P and the normal vector of P2are parallel. Since the line lies in P2then the dot product of the direction vector of the line and the normal vector of P2will be zero.In similar fashion the dot product of the direction vector of L and plane P will also be zero.The direction vector of the line is v = h3, 1, −5i and normal vector of plane P is n = h1, 2, 1i.v · n = 3 + 2 − 5 = 0.so the answer is yes there is a plane containing line L that will be parallel to plane P .Method 2: If there is a plane, P2, containing the line, L, where P2is parallel to the given plane,P , then the line will not intersect the plane. So check to see if there is a value of t will the linewill intersect the plane.x + 2y + z = 51 + 3t + 2(1 + t) + 1 − 5t = 55t + 4 − 5t = 54 = 5This is a contradiction so there is not any value of t where the line intersects the plane.Thus yes there is a plane containing line L that will be parallel to plane P
View Full Document
Unlocking...