PracticePrelimIa

PracticePrelimIa - Math 1920 Prelim 1 October 2 2008 7:30...

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Unformatted text preview: Math 1920, Prelim 1 October 2, 2008, 7:30 PM to 9:00 PM You are NOT allowed calculators, the text or any other book or notes. SHOW ALL WORK! Write your name and Lecture/Section number on each booklet you use. You may leave when you have finished, but if you have not handed in your exam booklet and left the room by 8:45, please remain in your seat so as not to disturb others who are still working. 1) a) (7 points) Calculate the distance from (1,1,1) to the plane with the equation a + 2y + 3z 2 0. b) (7 points) Find the equation of the plane through (0,0,0), (1, 1,0), and (1,2,3). 2) ( 14 points) The position vector of a particle moving in space is given by the vector function r(t) : (cos t)i + 2tj + (sin t)k, where t Z 0. For each t calculate the angle between the velocity vector and acceleration vector. 3) (14 points) The equation 3a2y + yz + zln(2a: — 1) = 0 defines at as a function of the two independent variables y and z. Find the partial derivative 833/82 at the point (1, ~1, ~3). ) 1 _ x2 + y2 — 1' a) (5 points) Determine the domain of f (a,y). b) (6 points) Determine and graph level curves in the x—y plane for f (:r, y) : ——2, 1, and 2. c) (5 points) Explain in complete sentences whether the level curve exists and what sets they form for : —1, —1/2, and 0. 5) Consider the curve r(t) : (4 sin t)i + (4 cos t)j + (3t)k. a) (8 points) Find the unit tangent vector of the curve. b) (8 points) Find the point on the curve such that the length along the curve from t 2 O is 1071' units in the direction of increasing t. 6) (14 points) Let f (x,y,z) : x2 + 2y2 + 3z2. Integrate f(a’,y,z) along the straight line segment from A to B where A : (1,0,0) and B : (0,1,1). 7) (12 points) The covalent bonds of the molecule methane CH; around the carbon atom can be described as three—dimensional unit vectors v1, v2, v3, v4, where V1+V2+V3+V4:0, and the angle 0 between any two of these vectors is same. Use vector operations to derive an expression for 0. Please leave your answer in terms of inverse trig functions. (Hint: Think about the vector dot product.) Math 1920, Prelim 1 October 2, 2008 Solutions 1) a) n = i + 2j + 3k is a normal of the plane and |n| : 12 + 22 + 32 = By looking at the equation of the plane then we see that P = (1,1, —1) is a point in the plane. If we set S 2 (1,1,1) then we have 1379—25 1 — 1,1 — 1,1 ~—(—1)>=< 0,0,2 > and the distance from (1,1,1) to the plane is d : |PS- = Iflififll 2 761—1. b) < 1—0,1~0,0—0 >=< 1,1,0 > and < 1—0,2—0,3—0 >=< 1,2,3 > are two vectors in the plane that are not parallel. A vector perpendicular to the plane is then given by the cross product between the two vectors. i j k <1,1,0>><<1,2,3>= 1 1 0 :(3—0)i—(3—0)j+(2—1)k:3i—3j+k 1 2 3 So the equation is given by 3st — 33,1 + z = 0, since the plane goes through the origin (0,0,0). 2) The velocity vector is v(t) = 3—: r: — sin(t)i + 2j + cos(t)k. The acceleration vector is a(t) : 2—: : — cos(t)i ~— sin(t)k. To calculate the angle between v(t) and a(t) consider the dot product between these two vectors: v(t) - a(t) : cos(t) sin(t) — cos(t) sin(t) : 0 The product is always 0. Hence, the velocity and acceleration vectors are always orthogonal to each other. 3) Differentiating both sides of equation w.r.t. z, we get 833 2 8x .2 ._.__ _ . .__._: 3y 3: az+y+ln(2x 1)+z 2m_1 8z 0 2z 85c —=— — 2 —1 (6xy+2$_1)az y ln(a: ) So at point (1,—1,—3) we get, <(6)(1)(—1)+ EKi) 8“: 2 -(—1) — ln(2(1) —— 1) and thus 2(1)—1 a 82: _ 1—0 1 52 _ —6+(——6) ~“1—2' 4) a) The domain is the set of all points (ac, y) in the plane that fulfill the equation x2+y2 7é 1. In other words it is the set of all points in the plane except for the unit circle. b) f(a:,y) = —2 or m2+;2_1 : ——2 gives 962 —|— y2 = g which is a circle centered at (0,0) with radius f (3:,y) : 1 or $2 +;2_1 : 1 gives 3:2 + 3/2 = 2 which is a circle centered at (0,0) with radius x/i any : 2 or '2—"2__ = 2 gives :32 + 3/2 = 3 which is a circle centered at 0,0 with radius a: +y 1 2 c) f(a:,y) : ——1 or fig—l : —-1 gives 322 + 3/2 = 0 so the level curve is the point (0,0). f (x, y) 2 —5 or zg+y2_1 : —% gives 3:2 + 3/2 = —1 which clearly can not hold for any point (at, Thus the level curve does not exist. f(ac,y) : 0 or 321:7: : 0 gives 1 = 0 and can thus clearly not hold for any point Hence the level curve does not exist. 5) a) V(t) : 4cos(t)i~4 sin(t)j+3k and |v(t)| : (4 cos(t))2 + (—4 sin(t))2 + 32 = \/ 42 + 32 = 5. The unit tangent vector is T : l—E% : %00s(t)i — gsin(t) j + gk, 13) Let t be the time when we are the point we want to find. Then we must have 1071' = f0t|v(7-)|d7 2 5t and thus 15 = 271'. Since r(27r) :< 0,4,67r > we see that the point is (0,4,677). 6) 1:3) 2 (0— 1)i+ (1—0)j+(1 —O)k 2 ~i+j+k is a vector parallel to the line through A and B. Using that vector and the point A z (1, 0, 0) then a parametrization of the line segment £rom A to B is r(t) : (1 — t)i—|—tj +tk where 0 S t S 1. Next we see that V(t) : —i —l—j +k and |v(t)| : 1/(—-1)2 + 12 + 12 : Finally we calculate the line integral, /Cf(w,y,z)ds = /01 ((1 — t)2 + 2t2 + 3H) x/idt : /1(1 — 2t + 6t2)\/§dt = x/fi[t—t2+2t3]3 :2\/§ 7) Since the angle between any different pair of the vectors v,- and vj is 9, and they are unit vectors we have vi - vj : cos 6, for 2' yé j. Then Ozvl-(V1+V2+v3+v4):vl-v1+v1-V2+v1-V3+v1-v4=1+3cos9. So 9 = cos—1(—1/3). ...
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This note was uploaded on 02/17/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

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PracticePrelimIa - Math 1920 Prelim 1 October 2 2008 7:30...

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