m2203fa10exam1ver2solutions

MATH 2203–Exam 1(Version 2)Solutions September 15,2010S.F.Ellermeyer NameInstructions.Your work on this exam will be graded according to two criteria:mathe-matical correctness and clarity of presentation .In other words,you must know what you are doing (mathematically)and you must also express yourself clearly.In particular,write answers to questions using correct notation and using complete sentences where appropriate.Also,you must supply su¢cient detail in your solutions (relevant calculations,written explanations of why you are doing these calculations,etc.).It is not su¢cient to just write down an “answer”with no explanation of how you arrived at that answer.As a rule of thumb,the harder that I have to work to interpret what you are trying to say,the less credit you will get.You may use your calculator but you may not use any books or notes.1.Show how to …nd the distance between the points P 0(4; 2;3)and P 1(2; 2;4).Solution:The distance between these points is !P 0P 1 =q (2 4)2+( 2 ( 2))2+(4 3)2=q ( 2)2+02+12=p .(a)Find the unit vector obtained by rotating the vector h 0;1i through an angle of120 counterclockwise about the origin.(b)Express the vector v =5i +15j as the product of its magnitude and direction.5i +15j =_____|{z }magnitude 0@__________|{z }direction1A .Solutions:For part a,notice that the vector h 0;1i is already a unit vector that points to the north (in the xy plane).If we rotate this vector 120 counterclockwise,we get a unit vector that points 30 south of west.Based on out knowledge of the unit circle,this vector is D p 32; 12E .For part b,we see that the magnitude of the given vector is j v j =s 52+ 15 2=p 6265Thus the direction of this vector is 1j v j v =5p 626 5i +15j =25p 626i +1p 626j .Therefore v =p 6265|{z }magnitude 0B B @25p 626i +1p 626j |{z }direction1C C A .2.Let u and v be the vectorsu = p 2i +p 3j v =p 2i +p 3j +2kand let be the angle between u and v .(a)Find u v ,j u j ,and j v j .(b)Find cos ( ).(c)Find the projection of u onto v (denoted by proj v u ).(d)Find the scalar component of u in the direction of v .Solution:u v = p 2i +p 3j p 2i +p 3j +2k = p 2 p 2 + p 3 p 3 +(0)(2)=1andj u j =r p 2 2+ p 3 2=p 5j v j =r p 2+ p 2+22=3.Thus cos ( )=u v j u j j v j =13p 5=p 515,proj v u =u v j v j 2v =19 p 2i +p 3j +2k ,and the scalar component of u in the direction of v isu v j v j =13.3.Which of the following statements are always true and which are not always true ?(Circle the correct choice.)(a)j v j =v v (always true,not always true )(b)u v = (v u )(always true,not always true )(c)(3u ) v =3(u v )(always true ,not always true)(d)u 0=0(always true ,not always true)(e)u ( u )=0(always true ,not always true)(f)u v= (v u)(always true,not always true)(g)u (v w)=(u v) w(always true,not always true)(h)(u v) v=j u v j j v j(always true,not always true)(i)j u v j=j u j j v j cos( )where is the angle between u and v.(always true,not always true)(j)u v=j u j j v j cos( )where is the angle between u and v.(always true,not always true)Grading of this question will be as follows:Number correct012345678910Points00.7 1.32 2.7 3.3481216204.Find either parametric equations of symmetric equations of the line that contains thepoint P0( 4;2; 4)and is perpendicular to the plane3x+3y+z= 1.Solution:The vector h3;3;1i is perpendicular to the given plane and is thus parallel to the line in question.Hence this is a direction vector for the line in question.A vector equation for the line ish x;y;z i=h 4;2; 4i+t h3;3;1i.Parametric equations arex= 4+3ty=2+3tz= 4+t1<t<1.5.The position vector of a particle moving in space is given byr(t)=(t+1)i+ t2 1 j+2t k.(a)Find the velocity vector,v(t).(b)Find the acceleration vector,a(t).(c)Find the speed,v(t).(d)Find parametric equations for the line that is tangent to the curve of motion atthe point on the curve corresponding to time t=1.Solution:v(t)=d rdt=i+2t j+2kanda(t)=d vdt=2jandv(t)=j v(t)j=q12+(2t)2+22=p4t2+5.At time t =1we have v (1)=i +2j +2k and this vector is tangent to the curve of motion at time t =1.Also the point on the curve of motion that corresponds to time t =1is (2;0;2).Thus the tangent line at this point is given by the parametric equationsx =2+ty =0+2tz =2+2t1<t <1.6.Please be as detailed as possible in answering this question (so that I can follow your reasoning easily):A projectile is …red with an initial speed of 500m =s at an angle of elevation of 45 .(a)Begin with the acceleration function a (t )= g j (where g =9:8m =s 2)andshow the steps involved in obtaining the velocity function,v (t ),and the position function,r (t ).(b)Show how to determine when and how far away (from where it is …red)theprojectile will strike the ground.Solution:Sincea (t )= g j ,thenv (t )= gt j +v (0).Alsov (0)=500cos (45 )i +500sin (45)j =250p 2i +250p 2j and thusv (t )=250p 2i + 250p 2 gt j .Therefore r (t )=250p 2t i + 250p 2t 12gt 2 j +r (0)and since r (0)=0we obtain r (t )=250p 2t i + 250p 2t 12gt 2 j .In parametric form,the x and y components of the path of motion arex (t )=250p y (t )=250p 2t 12gt 20 t T fwhere T f is the time that the projectile lands.To …nd T f ,we solve y (t )=0for t :Setting250p 2t 12gt 2=0,we obtain either t =0(obviously not T f )and 12gt =250p 2or t =T f =500p 2g72seconds.The range of the projectile isx (T f )=250p 2T f =250p 2 500p 2g!=5002g25;510meters(or about 25:51kilometers).。