Midterm Exam I, Calculus III, Sample A

1 Midterm Exam I, Calculus III, Sample A1. (10 points) Show that the 4 pointsP1= (0,0,0),P2= (2,3,0),P3= (1, 1,1),P4= (1,4, 1)are coplanar (they lie on the same plane), and find the equation of the plane that : u= P1P2= 2,3,0 ,v= P1P3= 1, 1,1 ,w= 1,4, 1 , and the scalar tripleproduct is equal tou (v w) = 2301 1114 1 = 2 114 1 3 111 1 + 0 1 114 = 0,so the volume of the parallelepiped determined byu,v, andwis equal to 0. This means thatthese vectors are on the same plane. So,P1,P2,P3, andP4are (10 points) Find the equation of the plane that is equidistant from the pointsA= (3,2,1)andB= ( 3, 2, 1) (that is, every point on the plane has the same distance from the twogiven points).Solution:The midpoint of the pointsAandBis the pointC=12(3,2,1)+( 3, 2, 1)]=(0,0,0). A normal vector to the plane is given by AB= 3,2,1 3, 2, 1 = 6,4,2.

2 So, the equation of the plane is 6(x 0) + 4(y 0) + 2(z 0) = 0, that is, 3x+ 2y+z= (6 points) Find the vector projection ofbontoaifa= 4,2,0 andb= 1,1,1 .Solution:Since|a|2= 42+ 22= 20, the vector projection ofbontoais is equal toprojab=a b|a|2a= 4,2,0 1,1,1 20=620 4,2,0 =35 2,1,0 .4. (12 points) Consider the curver(t) = 2 costi+ sintj+ sintk.(a) (8 points) Find the unit tangent vector functionT(t) and the unit normal vector functionN(t).(b) (4 points) Compute the curvature .Solution:(a)r (t) = 2 sinti+ costj+ costkand|r (t)|= 2 sin2t+ cos2t+ cos2t= 2. So, the unit tangent vectorT(t) is is equal toT(t) =r (t)|r (t)|= sinti+ 22costj+ (t) = costi 22sintj 22sintkand|T (t)|= cos2t+12sin2t+12sin2t= 1,the normal vector is equal toN(t) = costi 22sintj (b) One can use the formula (t) =|T (t)||r (t)|and|r (t)|= 2 form part (a) and|T (t)|= 1form part (b) to get (t) =1 2= can also use the formula (t) =|r (t) r (t)||r (t)|3.

3 Sincer (t) = 2 costi sintj sintk,|r (t)|= 2, andr (t) r (t) = ijk 2 sintcostcost 2 cost sint sint = 2j+ 2k= 0, 2, 2 .Then|r (t) r (t)|= 2 and (t) =2( 2)3= (10 points) Find the length of the curve with parametric equation:r(t) = et,etsint,etcost ,between the points (1,0,1) and (e2 ,0,e2 ).Solution:r (t) = et,etsint+etcost,etcost etsint and|r (t)|= e2t+e2t(sint+ cost)2+e2t(sint cost)2=et thatr(0) = 1,0,1 andr(2 ) = e2 ,0,e2 . So, the length of the curve is equal tolength = 2 0|r (t)|dt= 2 0 3etdt= 3(e2 1).6. (12 points) A spaceship is traveling with accelerationa(t) = et,t,sin 2t .Att= 0, the spaceship was at the origin,r(0) = 0,0,0 , and had initial velocityv(0) = 1,0,0 .Find the position of the spaceship att= .Solution:The velocity is equal tov(t) =v(0) + t0a(s)ds= 1,0,0 + t0 es,s,sin 2s ds= 1,0,0 + t0esds, t0s ds, t0sin 2s ds = 1,0,0 + et 1,t22,12(1 cos 2s) = et,t22,12(1 cos 2t).

4 Sincer(t) =r(0) + t0v(s)ds,r(t) = t0 es,s22,12(1 cos 2s) ds= t0esds, t0s22ds,12 t0(1 cos 2s)ds = et 1,t36,t2 14sin 2t .So,r( ) = e 1, 36, 2 .27. (10 points) Write the equation of the tangent line to the curve with parametric equationr(t) = t,1,t4 ,at the point (1,1,1).Solution:r (t) = 12 t,0,4t3 . At (1,1,1),t= 1 andr (1) = 1/2,0,4 . Thus theparametric equations of the tangent line arex=t/2 + 1, y= 1, z= 4t+ (12 points) Using cylindrical coordinates, find the parametric equations of the curve that isthe intersection of the cylinderx2+y2= 4 and the conez= x2+y2. (This problem refersto the material not covered before Midterm 1.)9. (6 points) Letf(x,y) = sin(x2+y2) + arcsin(y2). Calculate: 2f x y.(This problem refers to the material not covered before Midterm 1.)10. (12 points) (This problem refers to the material not covered before Midterm 1.)

5 Show thatthe following limit does not exist:lim(x,y) (0,0)7x2y(x y)x4+y4 Justify your answer. (This problem refers to the material not covered before Midterm 1.)3 Midterm Exam I, Calculus III, Sample B1. (6 Points) Find the center and radius of the following spherex2+y2+z2 6x+ 4z 3 = the squares:0 =x2+y2+z2 6x+ 4z 3 = (x2 6x+ 9) +y2+ (z2+ 4z+ 4) 3 9 4= (x 3)2+y2+ (z+ 2)2 , the equation of the sphere is (x 2)2+y2+ (z ( 2))2= 42, the center is (3,0, 2) andradius (6 Points) Write each combination of vectors as a single vector(a) AB+ BC(b) AC BC(c) AD+ DB+ :(a) AB+ BC= AC,(b) AC BC= AB,(c) AD+ DB+ BA= (6 Points) Find the cosine of the angle between the vectorsa= 1,2,3 andb= 2,0, 1 .Solution:If is the angle betweenaandb, thencos =a b|a||b|= 1,2,3 2,0, 1 1,2,3 || 2,0, 1 |= 1 14 5= 1 (6 Points) Find the vector projection ofvontouifu= 2i kandv= 2i+ :The vector projection ofvontouis equal toprojuv=u v|u|2u=(2i k) (2i+ 3j)|2i k|2[2i k]=45[2i k].

6 45. (7 Points) Find the area the triangle with verticesP= (2,1,7),Q= (1,1,5),R= (2, 1,1).Solution:Since PQ= 1,0, 2 and PR= 0, 2, 6 , and PQ PR= ijk 10 20 2 6 = 4i 6j+ 2k,the area of the triangle4 PQRis equal to12| PQ PR|=12 ( 4)2+ ( 6)2+ 22= 562= (5 Points) Show that the linex= 3 +t, y= 1 + 2t, z= 1 2tis parallel to the plane2x+ 3y+ 4z= :A line is parallel to the plane if it is perpendicular to a normal vector to theplane. A normal vector to the plane is given by 2,3,4 and the direction of the line is givenby the vector 1,2, 2 . Compute the dot product of these vectors: 1,2, 2 2,3,4 = 2 + 6 8 = , the line is parallel to the (6 Points) Find a vector parallel to the line of intersection for the two planesx+ 2y+ 3z= 0 andx 3y+ 2z= :A vector which gives the direction of the line of intersection of these planes isperpendicular to normal vectors to the planes.

7 A norma vector to the first plane 1,2,3 anda normal vector to the second is is given by 1, 3,2 . Then the vector 1,2,3 1, 3,2 = ij k12 31 3 2 = 13,1, 5 is parallel to the line of intersection of these (6 Points) Find cosine of the angle between intersection planes2x+y+z= 0and3x y+ 2z= :The angle between planes is equal to the angle between normal vectors of thetwo planes. Since 2,1,1 is a normal vector to the plane and 3, 1,2 is a normal vector tothe second plane,cos = 2,1,1 3, 1,2 | 2,1,1 || 3, 1,2 =7 6 14= (6 Points) Match the following equations with their graphs.(a)z=x2 y2.(b)x2+z2 1 = 0.(c)x2+y2+z24= :(a) and III, (b) and VI, (c) and (6 Points) Change the point (2, 2,2 2) in rectangular coordinates to spherical coordinates.(This problem refers to the material not covered before Midterm 1.)

8 11. (6 Points) Change the equationr z= 1 in cylindrical coordinates into rectangular coordi-nates. (This problem refers to the material not covered before Midterm 1.)12. (8 Points) A particle moves with position functionr(t) = t,t2,3t . Find the tangentialcomponent of :TheaT(t) is equal toaT(t) =r (t) r (t)|r (t)|.Sincer (t) = 1,2t,3 andr (t) = 0,2,0 and|r (t)|= 10 + 4t2, one hasaT= 1,2t,3 0,2,0 10 + 4t2=4t 10 + (8 Points) Consider the curver(t) = 3 sinti+ 4tj+ 3 unit tangent vectorT(t) =35costi+45j 35sintkis given. Find the : (t) =|T (t)||r (t)|. Sincer (t) = 3 costi+4j 3 sintk,|r (t)|= 32cos2t+ 16 + ( 3)2sin2t=5,T (t) = 35sinti 35costk, and|T (t)|=35, the curvature is equal to (t) = (8 Points) Find the length of the curver(t) = t2,2t,lnt from the point (1,2,0) to the point(e2,2e,1).

9 Solution:At the point (1,2,0),t= 1 and at the point (e2,2e,1),t=e. Sincer (t) = 2t,2,1/t and|r (t)|= (2t)2+ 4 + (1/t)2= (2t+ 1/t)2= (2t+ 1/t) =2t2+ 1t,length = e12t2+ 1tdt= 2 e1t dt+ e11/t dt=e2 1 + 1 = (8 Points) Consider a particle whose acceleration is given bya(t) = t,t2,cos 2t with initial velocityv(0) = 1,0,1 . Find the velocity of the particle as a function of :v(t) =v(0) + t0a(s)ds= 1,0,1 + t0 s,s2,cos 2s ds= 1,0,1 + t0s ds, t0s2ds, t0cos 2s ds = 1,0,1 + t2/2,t3/3,sin 2t/2 = 1 +t2/2,t3/3,1 + sin 2t/2 .7 Midterm Exam I, Calculus III, Sample C1. (10 Points) Find an equation of the sphere that has center at (1, 2, 5) and that passesthrough the :The radius of the sphere is equal tor= 12+ ( 2)2+ ( 5)2= 30. So, theequation of the sphere is given by(x 1)2+ (y+ 2)2+ (y+ 5)2= (6 Points) Find a vectoruin the opposite direction asv= 5,3,7 , and has length : u= 6v|v|.

10 Since|v|= ( 5)2+ 32+ 72= 83,u= 30 83, 18 83, 42 83 .3. LetA= (1,0,0),B= (1,2,2),C= (3,0,1).(a) (6 Points) Find the area of triangleABC(b) (6 Points) Find the equation of the plane passing throughA,B, andC(Write theanswer in the formax+by+cz=d).Solution:(a) The area of the triangle4 ABCis equal to12| AB AC|. Since AB= 0,2,2 and AC= 2,0,1 , AB AC= i j k0 2 22 0 1 = 2i+ 4j 4k= 2,4, 4 and| AB AC|= 22+ 42+ ( 4)2= 6. So, the area of the4 ABCis equal to 3.(b) A normal vector to the plane is AB AC= 2,4, 4 and the plane is passing throughthe pointA= (1,0,0). So, the equation of the plane is given by 2(x 1) + 4y 4z= 0, ,x+ 2y 2z= (a) (6 Points) Find parametric equations of the line which passes through the point (1,1,1)and is perpendicular to the plane 2x+y+z= 0.