1. x = 2 t, y = 1 3t, z = 3 2t. 4. x = 2+t, y = 1 3t, z = 3+2t. 5. x = 1 2t, y = 3 t, z = 2 3t. x = 1+2t, y = 3 t, z = 2+3t


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1 Version 1 Homework 4 gri (11111) 1 This printout should have 1 questions. Multiplechoice questions ma continue on the net column or page find all choices before answering. CalC13e11b 1 1. points A line l passes through the point P(,,) is parallel to the vector 4, 3,. At what point Q does l intersect the plane? 1. Q(, 1, ). Q(, 1, ) 3. Q(,, 5) 4. Q(, 6, 1) 5. Q(6, 5, ) correct 6. Q(5, 6, ) Since the plane is given b =, we have to find an equation for l then set =. Now a line passing through a point P(a, b, c) having direction vector v is given parametricall b But for l, Thus r(t) = atv, a = a, b, c. a =,,, v = 4, 3,. r(t) = 4t, 3t, t, so = when t = 1. Consequentl, the line l intersects the plane at Q(6, 5, ). kewords: line, parametric equations, direction vector, point on line, intercept, coordinate plane CalC13e13a 1. points Find parametric equations for the line passing through the points P(1, 3, ) Q(3,, 5). 1. = t, = 1 3t, = 3 t. = t, = 13t, = 3t 3. = 1 t, = 3 t, = 3t correct 4. = t, = 1 3t, = 3t 5. = 1 t, = 3 t, = 3t 6. = 1t, = 3t, = 3t A line passing through a point P(a, b, c) having direction vector v is given parametricall b Now r(t) = atv, a = a, b, c. PQ =, 1, 3 is a direction vector for the given line, so Thus a = 1, 3,, v =, 1, 3. r(t) = 1t, 3 t, 3t. Consequentl, = 1t, = 3 t, = 3t are parametric equations for the line. CalC13e15a 3 1. points Find the point of intersection, P, of the lines 5 5 = 4 = 4 = 6 = 5 3,.
2 Version 1 Homework 4 gri (11111) 1. P(7, 6, 8). P( 7, 6, 8) 3. P(7, 6, 8) correct 4. P(7, 6, 5) 5. P(7, 6, 6) To determine where the lines intesect it is convenient first to convert the lines from equations in smmetric form to ones in parametric form: = 5t, = 4t, = 6t, = 5s, = 4s, = 53s. For then the lines intersect when the equations 5t = 5s, 4t = 4s, 6t = 53s, are satisfied simultaneousl. Solving the first two equations gives t = 1, s = 1 a check shows that these values then satisf the third equation. Consequentl, the lines intersect when s = t = 1, i.e., at the point P(7,6,8). CalC13e1b 4 1. points Determine where the plane passing through the points Q(, 1, ), S( 5, 1, 4), intersects the plane =. 6 4 = R( 3,, ), = correct = = = Since the points Q, R, S lie in the plane, the displacement vectors QR = 1, 3,, QS = 3,,, lie in the plane. Thus the cross product n = QR QS = is normal to the plane. i j k On the other h, if P(,, ) is an arbitrar point on the plane, then the displacement vector v = PQ =, 1, lies in the plane, so Now But then i j k v n =. = 6,, 9. v n = 6() ( 1) 9( ) = =. Consequentl, the plane passing through Q, R, S is =,
3 Version 1 Homework 4 gri (11111) 3 it will intersect the plane when =, i.e., when 6 4 =. CalC13e63s 5 1. points The distance d from P to the line l in It is convenient to take Q = ((), (), ()) = (4, 5, 1) R = ((1), (1), (1)) = (, 6, 3). With this choice of Q R, Q a b D R d P l while a =, 1,, b =,,, a b = = i j k 1 1 i j 1 k is given b d = a b a where Q, R are points on l a = QR, b = QP. Use this formula to find the distance from P(6, 7, 3) to theline l specified in parametric form b = 4 t, = 5t, = 1t. 1. distance =. distance = 14 correct Consequentl, = i8j 6k. d = i8j 6k, 1, = 14 kewords: distance, distance from line, cross product, vector product, vector, length, CalC13f3a 6 1. points Which one of the following equations has graph distance = 4. distance = 5. distance = 6. distance =
4 when the circular clinder has radius. Version 1 Homework 4 gri (11111) =. 4 = correct 3. 4 = 4. = 5. = kewords: quadric surface, graph of equation, clinder, 3D graph, circular clinder, trace CalC13f5c 7 1. points Which one of the following equations has graph 6. = The graph is a circular clinder whose ais of smmetr is parallel to the ais, so it willbethegraphofanequationcontainingno term. This alread eliminates the equations =, 4 =, =, 4 =. On the other h, the intersection of the graph with the plane, i.e. the = plane, is a circle centered on the ais passing through the origin as shown in = 1. = 1 3. = 1 4. = 1 5. = 1 correct But this circle has radius because the clinder has radius, so its equation is ( ) = 4 as a circle in the plane. Consequentl, after epansion we see that the clinder isthe graph of the equation 4 = = 1 The surface is a twosheeted hperboloid. As the graph shows, the trace on the = plane is a hperbola opening in the direction of the ais, so will be of the form a b = 1
5 Version 1 Homework 4 gri (11111) 5 for some choice of a, b. On the other h, for large m the trace on = m is a circle = r 4. where r depends on m. Consequentl, the graph is that of the quadric surface = 1. CalC13f5d 8 1. points Which one of the following quadric surfaces is the graph of the equation = 1? correct The graph of = 1 is a hperboloid. To decide which of the surfaces shown is the graph of this equation we look at traces on the coordinate planes.
6 Version 1 Homework 4 gri (11111) 6 Now the trace on the = plane is the hperbola = 1 which opens in the direction of the ais. On the other h, the trace on the = plane is the circle = 1 centered at the origin. Consequentl, the graph of the equation is = 1 5. (1) ellipsoid ( 1) ( ) = 1, Completing squares in all three variables gives (1) ( 1) ( ) = or = 1, a hper (1) boloid. ( 1) ( ) CalC14a5b 1 1. points A space curve is shown in black on the surface CalC13f36s 9 1. points Reduce the equation 4 = to stard form then classif the surface. 1. (1) ( 1) hperboloid correct. (1) hperboloid ( 1) ( ) ( ) = 1, = 1, Which one of the following vector functions has this space curve as its graph? 1. r(t) = sint, cost, cost 3. ( 1) ellipsoid ( 1) ( ) = 1,. r(t) = sint, cost, t 3. r(t) = cost, sint, sin4t correct 4. (1) hperboloid ( 1) ( ) = 1, 4. r(t) = cost, sint, cost 5. r(t) = cost, sint, t
7 Version 1 Homework 4 gri (11111) 7 6. r(t) = cost, sint, cos4t If we write then r(t) = (t), (t), (t), (t) (t) = 1 for all the given vector functions, showing that their graph will alwas lie on the clindrical clinder = 1 To determine which particular vector function has the given graph, we have to look more closel at the graph itself. Notice that the graphoscillateswithperiod4,so r(t)isoneof cost, sint, sin4t, cost, sint, cos4t. On the other h, it passes it through (1,, ) (, 1, ). Consequentl, the space curve is the graph of kewords: r(t) = cost, sint, sin4t. CalC14aa points 6. radius = 1, center = (, 1, ) Writing r(t) = (t)i(t)j(t)k, we see that (t) = 5 for all t, while (t) = 4cost, (t) = 14sint. Thusr(t)tracesoutacurveintheplane = 5 ((t) ) ((t) 1) = 16. Consequentl, r(t) traces out a circle with radius = 4, center = (, 1, 5). kewords: vector function, space curve, circle, plane, radius, center circle, CalC14a3b 1 1. points A circular clinder of radius 1 the hperbolic paraboloid intersect as shown in = The vector function r(t) = ( 4sint)i(14cost)j5k traces out a circle in 3space as t varies. Determine the radius center of this circle. 1. radius = 1, center = (5, 1, ). radius = 4, center = (, 1, 5)correct 3. radius = 4, center = (1, 5, ) 4. radius = 1, center = (5, 1, ) 5. radius = 4, center = (5, 1, )
8 Version 1 Homework 4 gri (11111) 8 Which vector function r(t) = (t)i(t)j(t)k has the curve of intersection as its graph if () = 1? 1. r(t) = (sint)i(cost)j(cost1)k. r(t) = (cost)i(sint)j(cost 1)k Using the trig identit cosθ = cos θ sin θ, we thus see that (t) = cost. Therefore, 3. r(t) = (sint)i(cost)j(cost)k 4. r(t) = (cost)i (sint)j (cost)k correct 5. r(t) = (cost)i(sint)j(cost1)k 6. r(t) = (sint)i(cost)j(cost 1)k The intersection shown in orange will be the graph of the vector function r(t) = (t)i(t)j(t)k when the components (t), (t) (t) satisf the equation for the clinder that for the hperbolic paraboloid simultaneousl. Now the circular clinder has radius 1 has the ais as a line of smmetr, so it is the graph of r(t) = (cost)i(sint)j(cost)k. kewords: space curve, vector function, curve of intersection, circular clinder, parabolic clinder, Intersection1d points The curve of intersection of the surfaces shown in in which case = 1, (t) (t) = 1. But all the given vector functions have this propert since cos θ sin θ = 1, On the other h, () = 1, so (t) = cost, (t) = sint. But the curve of intersection lies also on the hperbolic paraboloid, so (t) = (t) (t) = cos t sin t. is the graph of which of the following vector functions? 1. r(t) = sint, cost, 1 cost correct
9 Version 1 Homework 4 gri (11111) 9. r(t) = sint, cost, cost 3. r(t) = cost, sint, cost 1 4. r(t) = sint, cost, cost 1 5. r(t) = cost, sint, 1 cost 6. r(t) = cost, sint, cost If we write r(t) = (t), (t), (t), then each of the given vector functions has the propert that (t) (t) = 1. This is consistent with the fact that the curve of intersection lies on a circular clinder = 1 with the ais the line of smmetr as is shown in the figure. Thus we need to look more carefull at the vector functions to determine which lie on the parabolic clinder shown in the figure. Recall first that cosθ = 1 sin θ = cos θ 1. Now the crosssections of the parabolic clinder perpendicular to the ais are parabolas opening upwards with verte on the ais, so the parabolic clinder is the graph of = a, a >. As a result, the components of r(t) have to satisf an equation (t) = a(t), a >. Consequentl, the curve of intersection of the twosurfacesisthegraphofthevectorfunction r(t) = sint, cost, 1 cost. kewords: surface, space curve, parametric equation, 3D graph, circular clinder, paraboloid, VectorFunc4a points Determine as a linear relationin,, the plane given b the vector function when F(u, v) = aubvc a = i j 3k, b = i j k, c = i 3j k =. 5 = correct = 4. 5 = = 6. 5 = Set n = b c Then, if we see that r = ijk. r = F(u, v) = aubvc, r n = a nu(b n)v(c n) = a n, because b (b c) = = c (b c).
10 Version 1 Homework 4 gri (11111) 1 But b c = = Thus so i j k i 1 j 3 k. n = ij k, r n =, a n = 5. Consequentl, F describes the plane (5) =. CalC14ba points Determine the vector when r(t) = I = 1. I = ln, π, 4. I = 8, π, 8ln r(t) dt 8 t 8 1t, 1t, (1t). 3. I = π, ln, 4 correct 4. I = 8ln,, 8π 5. I = 8, ln, 8π 6. I = π,, 4ln For a vector function r(t) = f(t), g(t), h(t), the components of the vector are given b f(t)dt, I = r(t) dt g(t) dt, h(t)dt, respectivel. But when 8 t 8 r(t) = 1t, 1t, (1t), we see that while Consequentl, f(t)dt = 8 1t dt [ ] 1 = 8tan 1 t g(t)dt = = t 1t dt [ ] 1 ln(1t ) = ln, 8 h(t)dt = (1t) dt [ = 8 ] 1 1t = 4. I = π, ln, 4. CalC14b3a points Find an equation in parametric form for the tangent line to the graph of at the point (1, 1, 1). r(s) = s 5, s 6, s 3 1. (t) = 4t 1, (t) = 5t 1, t 1
11 Version 1 Homework 4 gri (11111) 11. (t) = 14t, (t) = 15t, (t) = 1t 3. (t) = 15t, (t) = 16t, (t) = 13t correct 4. (t) = 1 5t, (t) = 16t, (t) = 1 3t 5. (t) = 5t 4, (t) = 6t 5, (t) = 3t 6. (t) = 5t 1, (t) = 6t 1, (t) = 3t 1 The graph of r(s) has tangent vector r (s) = 5s 4, 6s 5, 3s. On the other h, (1, 1, 1) = r(1). Thus, at (1, 1, 1) the equation of the tangent line in vector form is L(t) = r(1)tr (1) = 15t, 16t, 13t, which in parametric form becomes (t) = 15t, (t) = 16t, (t) = 13t. kewords: tangent line, tangent vector, vector function, derivative, vector form, parametric form, CalC14b31s points The curves given parametricall b (t) = t, (t) = t 5, (t) = t 6, (t) = sint, (t) = sint, (t) = t intersect at the origin. Find the cosine of their angle, θ, of intersection. 3. cosθ = 7 4. cosθ = 6 5. cosθ = 1 6 correct 6. cosθ = 1 5 The angle of intersection of the two curves is the angle between the tangent vectors at the point of intersection. Now the tangent vector at the origin to the first curve is r 1 () = 1, 5t 4, 6t 5 = 1,,, t= while the tangent vector at the origin to the second curve is r () = cost, cost, 1 = 1,, 1. t= But then cosθ = r 1 () r () r 1 () r () = Consequentl, cosθ = 1 6. CalC14b37s points Evaluate the integral when I = a(t) b(t)dt 1 (4) 1/. a(t) = e t i4j 3ln(1t)k 1. cosθ = 5 b(t) = 3ti j k.. cosθ = I = 3e51ln
12 Version 1 Homework 4 gri (11111) 1. I = 6e56ln 3. I = 3e 5 6ln 4. I = 3e 51ln correct 5. I = 6e5 1ln 6. I = 6e 5 6ln When we see that Thus a(t) = e t i4j 3ln(1t)k b(t) = 3ti j k a b = 6te t 46ln(1t) points Which one of the points P(1, 3, 3), Q(, 6, 4), R(, 4, ) in 3space is closest to the plane? 1. Q(, 6, 4). P(1, 3, 3) 3. R(, 4, ) correct The distance of a point (a, b, c) in 3space from the plane is given b c. Consequentl, of the three points P(1, 3, 3), Q(, 6, 4), R(, 4, ) the one closest to the plane is R(, 4, ). I = {6te t 46ln(1t)}dt. kewords: plane, distance in 3space, Now, b substitution, while 6te t dt = 4dt = 6ln(1t)dt [3e t ] 1 = 3(e 1), [ ] 1 4t = 4, [ ] 1 = 6 (1t)ln(1t) (1t) = 6(ln 1), using integration b parts. Consequentl, I = 3e 51ln. CalC13aa CalC13a5b 1. points Determine the distance of the point Q(4, 5, 1) from the coordinate plane. 1. distance = 41. distance = 4 3. distance = 6 4. distance = 4 5. distance = distance = 5 correct 7. distance = 1
13 Version 1 Homework 4 gri (11111) 13 Since the distance of a point P(,, ) from the , , coordinate planes is given respectivel b,, the point Q(4, 5, 1) has from the plane. distance = 5 kewords: coordinate plane, projection, point, distance, 3space CalC13d4s 1 1. points When P is a point not on the plane passing through the points Q, R, S, then the distance, d, from P to that plane is given b the formula where a = QR, d = a (b c) a b b = QS, c = QP. Use this formula to determine the distance from P(1, 5, 4) to the plane through the points Q(1, 1, ), R(, 1, 3), S(, 1, ). 1. distance = 4. distance = 3 For the given points, while In this case a b = a = QR = ij, b = QS = ik, c = QP = 4jk. in which case i j k = i j k, a b = 1 = 3. On the other h, 1 a (b c) = = 8. Consequentl, P is at a distance = 8 3 = 8 3 from the plane through Q,R S. kewords: distance from plane, distance, plane, scalar triple product, cross product, 3. distance = 9 4. distance = 8 3 correct 5. distance = distance = 3
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