Math 323 Eam 2 - Practice Problem Solutions 1. Given the vectors a = 2,, 1, b = 3, 2,4, and c = 1, 4,, compute the following: (a) A unit vector in the direction of c. u = c c = 1, 4, 1 4 =,, 1+16+ 17 17 (b) The angle between a and b (to the nearest tenth of a degree). cosθ = a b a b = (2)(3)+()( 2)+( 1)(4) 2 =. 4++1 9+4+16 29 ( ) 2 Then θ = arccos.4 29 (c) a b a i j k b = 2 1 3 2 4 = i( 2) j(+3)+ k( 4+) = 2 i 11 j 4 k (d) The area of the triangle determined b a and b. A = 1 2 a b = 1 141 4+121+16 =.937 square units. 2 2 (e) comp b a. comp b a = a b b = 2 29 (f) A non-ero vector that is perpendicular to c. We want 1, 4, a,b,c =, or a 4b =. One wa to do this (there are infinitel man choices) is to take a = 4. Then b = 1 and c can be whatever we choose, so v = 4,1,3 is a non-ero vector that is perpendicular to c (check that the dot product works out). 2. Given the vectors a = 1,2,, b = 1,,2, and c =,1,1, compute the following: (a) 2 a+3 b = 2,4, + 3,,6 = 1,4,6 (b) a b = (1)( 1)+(2)()+()(2) = 1. (c) A unit vector in the direction of a. u = a a = 1,2, = 1 2,, 1+4 (d) The component of c along a. comp a c = c a a = (1)()+(2)(1)+()(1) = 2 (e) The Volume of the parallelepiped determined b a, b and c. ( V = a ) 1 2 b c = 1 2 = 1( 2) 2( 1 )+( ) = 2+2+ = (what does this tell ou about 1 1 these vectors?...) (f) The angle between b and c cosθ = b c b c = ()+()+(2) = 2. 1++4 +1+1 2 ( ) 2 Then θ = arccos. 1
3. Decide whether each of the following are true or false. If true, eplain wh. If false, give a countereample. (a) If a b = a c, then b = c. False. If we take a =, then a b = for an vector b. (in fact, it is still false when we assume all vectors are non-ero. Can ou show this?) (b) If b = c then a b = a c. True. Let a = a 1,a 2,a 3, b = b 1,b 2,b 3, and c = c 1,c 2,c 3 If b = c, then b 1 = c 1, b 2 = c 2, and b 3 = c 3. Then, a b = a 1 b 1 +a 2 b 2 +a 3 b 3 = a 1 c 1 +a 2 c 2 +a 3 c 3 = a c. (c) a a = a False. Again let a = a 1,a 2,a 3. Then a a = a 2 1 +a 2 2 +a 2 3 = a 2 (d) If a > b, then a c > b c False. Let a =,2, b = 1,, and c = 2,. Then a = 2 > 1 = b, but a c = while b c = 2 (e) If a = b, then a = b False. If a = 1, and b =,1, then a = b = 1, but clearl a b. 4. Let P=(2,1,2) and Q=(3,1,1) be points. (a) Find parametric equations for a line through these points. PQ = 1,, 1, so the line (using P) is given b: = 2+t l : = 1 t R = 2 t (b) Find the equation of a plane containing P,Q, and the origin. Let R = (,,). First, we find a normal vector to the plane: PQ i j k PR = 1,, 1 2, 1, 2 = 1 1 2 1 2 = i( 1) j( 2 2)+ k( 1 ) = 1 i+4 j 1 k Then the plane (using P) has equation: 1( 2)+4( 1) 1( 2) =, or +2+4 4 +2 =, which simplifies to give 4 + =. Given the lines l 1 : = 3; = 6 2t; = 3t+1 l 2 : = 1+2s; = 3+s; = 2+2s: (a) Show that the lines intersect b finding the coordinates of a point of intersection. Equating the first pair of coordinate equations for these lines, 3 = 1+2s, or 2 = 2s, so s = 1 Then, looking at the third pair of equations with s = 1, 3t+1 = 2+2s = 4, so 3t = 3 or t = 1. Checking this in the middle pair of equations, 6 2(1) = 4 and 3+1 = 4. Hence this pair of lines intersect at the point (3,4,4). (b) Find a vector orthogonal to both lines. The vectors associated with this pair of lines are v =, 2,3 and v = 2,1,2 respectivel. i j k To find a vector that is orthogonal (perpendicular) to both, we compute v w = 2 3 2 1 2 = i( 4 3) j( 6)+ k(+4) = 7 i+6 j +4 k (c) Find the equation of a plane containing both lines. We need onl find the equation for a plane with normal vector n = 7 i+6 j+4 k and containing the point P(3,4,4). This is given b: 7( 3)+6( 4)+4( 4) =, or 7+6 +4 19 = 6. Given the two planes 2+ = 4 and 3 + = 6 (a) Find normal vectors to each plane. n 1 = 2,1, 1 and n 2 = 3, 1,1
(b) Find parametric equations for the line of intersection of the two planes. Solving each plane s equation for, we have = 2+ 4 and = 3+ +6. Then 2+ 4 = 3+ +6, so = 1, or = 2. Substituting this into our previous equation, = 4+ 4, so =. = 2 Thus, if we take = t, we have the line of intersection: l : = t t R = t 7. Find both the parametric and smmetric equations for the line through the points P(3,,7) and Q( 6,2,1). To find parametric equations, We find PQ = 9, 3, 6. A slightl nicer vector in the opposite direction is: 3,1,2. = 3+3t From this, using P, l : = +t t R = 7+2t To put this in smmetric form, we solve each equation for t, ielding: 3 3 = = 7 2. Find an equation of the plane through the point (1,4, ) and parallel to the plane defined b 2 +7 = 12. Since the planes are parallel, we use the same normal vector: n = 2,,7. Therefore, using P(1,4, ), we have a plane with equation: 2( 1)+ ( 4)+7( +) =. 9. Sketch at least 3 traces, then sketch and identif the surface given b each of the following equations: Note: In the interest of getting these posted in a timel fashion, I am not including the graphs of the traces, although ou should include these on our eam solutions. (a) 2 + 2 = If =, we have 2 + 2 =, a point. If = k, we have 2 + 2 = k, which is a circle if k >, and is empt if k <. If =, we have 2 =, or = 2, a parabola. If =, we have 2 = or = 2, a parabola. Therefore, the figure is a paraboloid: -7. -4. -2.. 3.. -2.. -7. -4.. 3.
(b) 2 + 2 = 1 This is a circular clinder sitting over the unit circle in the -plane: (c) 4 2 +4 2 2 2 = 4. If =, we have 4 2 2 2 = 4, or 2 2 2 = 1 a hperbola. If =, we have 4 2 2 2 = 4, or 2 2 2 = 1, a hperbola. If =, we have 4 2 +4 2 = 4 or 2 + 2 = 1, a circle. Therefore, the figure is a hperboloid of 1 sheet: 3-2 -7-1 - 1 1 - -1 (d) 4 2 + 2 2 = If =, we have 2 2 =, or 2 = 2, so we have = ±, a pair of lines. If =, we have 4 2 2 =, or 4 2 = 2, so we have = ±2, a pair of lines. If =, we have 4 2 + 2 =, a point. If = k, we have 4 2 + 2 = k 2, an ellipse. Therefore, the figure is a cone: 1-2 -12-22 1 - -1 (e) 2 4 = 4 2 + 2 If =, we have 2 4 = 2, or 2 2 =, a hperbola. If =, we have 4 2 + 2 = 4, which is empt. If = k, we have 2 4 = 4 2 + 2, which is a point if = 2 and is an ellipse if > 2 If =, we have 2 4 = 4 2, or 2 4 2 = 4, a hperbola.
Therefore, the figure is a hperboloid of 2 sheets: 1-2 -12-22 1 - -1 - -1 (f) 2 = 2 If =, we have = 2, or = 2, a parabola opening downward. If =, we have 2 =, or = 2, a parabola opening upward. If =, we have 2 = 2, or = ±, a pair of lines. If = k, we have 2 2 = k, which are hperbolas whose vertices depend on the sign of k. Therefore, the figure is a hperbolic paraboloid (or saddle ): -1-1 1 - -1 1 - -1 1. Let r(t) = t 2,4+3t,4 3t be a vector valued function. Calculate the following: (a) r (t) = 2t,3, 3 (b) r (t) = 2,, (c) r(t)dt = t 2 dt, 4+3t dt, 4 3t d = 1 3 t3,4t+ 3 2 t2,4t 3 2 t 1 = 1 3, 11 2, 2 (d) the arc length of r(t) for t 2. (Just set up the integral, ou do not need to evaluate it). 2 L = [f (t)] 2 +[g (t)] 2 +[h (t)] 2 dt = 2 (2t)2 +3 2 +3 2 dt = 2 4t2 +1 dt (e) Find the values of t for which r(t) and r (t) are perpendicular. We need r(t) r (t) =. That is, t 2,4 + 3t,4 3t 2t,3, 3 = (t 2 )(2t) + (4 + 3t)(3) + (4 3t)( 3) = 2t 3 +12+9t 12+9t = 2t 3 +1t = Then 2t(t 2 +9) =, so either t =, or t 2 = 9, which is impossible. Therefore, t = is the onl solution. 11. Let r(t) = 3, 4cos(t), 4sin(t). (a) Sketch the curve traced out b the vector-valued function r(t). Indicate the orientation of the curve. What is geometric shape of this curve? Using algebra, 4 = cost and 4 = 3. (See graph) 2 = sint, so 16 + 2 = 1. Therefore, the curve is a circle or radius 4 in the plane 16
(b) Find an equation in terms of t for s(t), the arc length of the curve traced out b r(t) as a function of t for t. t t t t t s(t) = 2 +16cos 2 +16sin 2 d = 16(cos 2 +sin 2 ) d = 16 d = 4 d = 4 = 4t (c) Find r (t). r (t) =, 4sint,4cost (d) Find an epression for the angle between the vectors r(t) and r (t). Recall that cosθ = r(t) r (t) r(t) r (t) = 16sintcost+16costsint 9+16cos2 t16sin 2 t +16sin 2 +16cos 2 t = +4 = Therefore, θ = arccos() = π 2 or 9 (hence the are perpendicular at all times). (e) Find the force acting on a kilogram object moving along the path given b r(t), in units of meters and seconds (assume that no other forces are acting on this mass). Recall that F = m a. Also, a(t) = r (t) =, 4cost, 4sint. Thus F(t) =, 4cost, 4sint =, 2cost, 2sint (in Newtons). (f) Find and draw the position and tangent vectors when t = π. Find 2 different vectors that are normal to r(π) and r (π). Draw these vectors (carefull labeled) on the same aes (as the position and tangent vectors). r(π) = 3, 4, and r (π) =,, 4 To find vectors normal to both of these, we take the cross product: i j k r(π) r (π) = 3 4 4 = i( 16+) j(12+)+ k(+) = 16 i 12 j + k Another vector normal to both is: 16 i+12 j + k (the dotted line vectors represent the pair of vectors normal to both r(π) and r (π)) r ( π) r( π) 12. Let v(t) = t 2 2t,4t 3,t 3. (a) Find a(t) a(t) = 2t 2,4,3t 2
(b) If r() = 4, 1,, find r(t). r(t) = r(t) dt = 1 3 t3 t 2,2t 2 3t, 1 4 t4 + C Since r() = 4, 1,, C = 4, 1, and hence r(t) = 1 3 t3 t 2 +4,2t 2 3t 1, 1 4 t4 (c) Find the force acting on an object of mass kg with position function r(t) (in units of meters per second). F(t) = m a(t) = 2t 2,4,3t 2 = 1t 1,2,1t 2 (in Newtons). (d) Find the speed of the object at time t = 2. Since v(2) = 4 4, 3, =,,, v(2) = +2+64 = 9. 13. Suppose that a projectile is launched with initial velocit v = 1 ft/s from a height of feet and at an angle of θ = π 6. (a) Assuming that the onl force acting on the object is gravit, find the maimum altitude, horiontal range, and speed at impact of this projectile. Recall that the downward force of gravit is 32ft/s 2. Since the onl forces acting on this projectile are gravit and its initial velocit, we can model this situation in 2 dimensions. We take j to be the unit vector in the upward direction. Then a(t) = 32 j, so, integrating: v(t) = i 32 j + C. Now, using trigonometr, v() = 1cos π 6 i+1sin π 6 j = 3 i+ j. Thus v(t) = 3 i+( 32t) j. Since r() =, r(t) = 3t i+(t 16t 2 ) j. The maimum altitude occurs when the vertical component of v(t) is ero, that is, when 32t =, or when t = 32 = 2 16. Therefore, the maimum altitude is found b looking at the vertical position at that time: ( ) ( 2 16 16 2 2 16) = 39.62 feet. The horiontal range is found b first finding the time when the projectile hits the ground, and then finding the horiontal component of position at that time: t 16t 2 = when t = or 16t =. That is, when t = ) 27.633 feet. 16 = 2. Then the range is: 3 ( 2 The speed at impact is found b looking at the magnitude of the velocit vector at the time of impact: v ( 2 ) = ( 3 ) 2 + ( 32 ( 2 )) 2 = 7+2 = 1feet/sec. (b) Find the landing point of this projectile if it weighs 1 pound, is launched due east, and there is a southerl wind force of 4 pounds. Because of the wind, we must move to a 3D coordinate sstem (let the positive -ais be East and positive be Up). I worded this a bit clumsil, so just to clarif, m intention is that the wind is gusting as that its force is 4 times that of gravit (I know, this is not ver realistic...) w = 4(32) j and g = 1(32) k, so a(t) = i 12 j 32 k. v() = 1cos π 6 i+ j +1sin π 6 k = 3 i+ j + k. Then v(t) = ( i 12t j 32t k)+( 3 i+ j + k) = 3 i 12t j + 32t k As before, the projectile hits the ground after t = 2 seconds. Notice that we set up our coordinate sstem so that r() =. Therefore, r(t) = 3t i 64t 2 j +(t 16t 2 ) k. Thus the final position of the projectile is: r( 2 ) = 3 ( ) ( 2, 64 2 ) 2, ( 2 ) ( 16 2 ) 2 27.63, 62, where the coordinates are in feet.