Vector Calculus PDF

Vector Calculus By Dr. Ehab Said  Objectives: 1) Define scalars and vectors 2) Represent the vector symbolically 3) Introduce the notion of unit vectors 4) Understand Scalar product 5) Understand Vector product  SCALARS AND VECTORS...

Vector Calculus By Dr. Ehab Said  Objectives: 1) Define scalars and vectors 2) Represent the vector symbolically 3) Introduce the notion of unit vectors 4) Understand Scalar product 5) Understand Vector product  SCALARS AND VECTORS I. Scalars  A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities: F P= A  Distance  Time  Mass  Pressure  Temperature  Density  II. Vectors  A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quantities:  Displacement  Velocity  Acceleration  Force  Vector Diagrams  Vector diagrams are shown using an arrow in 2D and 3D z  The length of the arrow represents its k magnitude y  The direction of the arrow shows its j i direction x a = a1 i + a2 j =  a1 , a2 , a = a12 + a2 2 r = x i + y j + z k =  x, y, z , r = x 2 + y 2 + z 2  Example: Practical Applications  Here we see a table being pulled by a force of 50 N at a 30º angle to the horizontal y=25 N 3 30º Fx = 50 Cos 30 = 50 =43.3 N 2 x=43.3 N 50 Fy = 50Sin 30 = =25 N 2  We can see that it would  When resolved we see that this is the be more efficient to pull the table with a horizontal same as pulling the table up with a force force of 50 N of 25 N and pulling it horizontally with a force of 43.3 N F = 43.3 i + 25 j  Calculating the Magnitude of the Perpendicular Components  If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are:  x = v Cos θ y=v Sin θ  y = v Sin θ y θ x x=v Cos θ  Proof: x y Cos = Sin = v v x = vCos y = vSin Example: Calculating the magnitude of perpendicular components A force of 15 N acts on a box as shown. What is the horizontal component of the force? Solution: N Vertical 12.99 Horizontal Component = x = 15Cos 60 = 7.5 N Component Vertical Component = y = 15Sin60 = 12.99 N 60º Then F = 7.5 i + 12.99 j Horizontal 7.5 N Component  Unit Vectors  A unit vector is a vector of length 1.  They are used to specify a direction.  By convention, we usually use i, j and k to represent the unit vectors in the x, y and z directions, respectively (in 3 dimensions).  i= points along the positive x-axis z  j= points along the positive y-axis k  k= points along the positive z-axis y  Unit vectors for various coordinate systems: j i  Cartesian: i, j, and k x  Cartesian: we may choose a different set of  unit vectors, e.g. we can rotate i, j, and k  Unit Vectors To find a unit vector, â , in an arbitrary direction, for example, in the direction of vector a, where a=, divide the vector by its magnitude (this process is called normalization). a a  a1 , a2  aˆ = = = a a +a 2 1 2 2 a12 + a22 Example: If a=, then is a unit vector in the same direction as a. Position Vectors 1) Find the magnitude of v = i + 4j. 2) Find the magnitude of the vector A = 2i – 5j + 4k, using vector algebra. 3) A ball is thrown with an initial velocity of 70 feet per second, at an angle of 35° with the horizontal. Find the vertical and horizontal components of the velocity.  Scalar Product of Two Vectors in 3D  If A & B are vectors, their Scalar Product is defined as: A B ≡ AB cosθ  In terms of vector components & unit vectors i, j, k are along the x, y ,z axes: A = Axi + Ayj + Azk B = Bxi + Byj + Bzk  Using i i = j j = k k = 1, i j= i k = j k = 0 gives A B = AxBx + AyBy + AzBz  Dot Product clearly a SCALAR.  The condition of perpendicularity  Given two nonzero vectors a = a1, a2, a3 and b = b1, b2, b3, we can say that the vector a perpendicular to b if and only if A B ≡ AB cos(π/2)=AB (0) b ab=0 a a1b1 + a2b2 + a3b3 = 0 ˆi  ˆj = ˆi  kˆ = ˆj  kˆ = 0  Note that: ˆi  ˆi = ˆj  ˆj = kˆ  kˆ = 1 1) Find the scalar product of vector A= 2i + 5j +3k and vector B= 3i + j +2k. 2) A particle covers a displacement from position 2i + j + 2k to 3i + 2j +5k due to uniform force of ( 7i + 5j +2k) N. If the displacement is in mitres calculate the work done. 3) Find the value of m such that vectors A = 2 i + 3j + k and B = 3 i + 2 j + mk may be perpendicular.  Cross or Vector Product + − + A  B = A B sin  nˆ ˆi ˆj kˆ → Ay Az A Ay C = A  B = Ax Ay Az == ˆi − Ax Az ˆj + x kˆ By Bz Bx Bz Bx By Bx By Bz = ( Ay Bz − Az By ) ˆi − ( Ax Bz − Az Bx ) ˆj + ( Ax By − Ay Bx ) kˆ y  The cross product of two vectors says something about how i perpendicular they are: j i   x  Magnitude: → C = A  B = AB sin  j k z k   is smaller angle between the vectors  Cross product of any parallel vectors = zero  Cross products of Cartesian unit vectors: n̂ iˆ  ˆj = kˆ; iˆ  kˆ = − ˆj; ˆj  kˆ = iˆ iˆ  iˆ = 0; ˆj  ˆj = 0; kˆ  kˆ = 0  The condition of parallelism  Since a vector is completely determined by its magnitude and direction, we can now say that a  b is the vector that is perpendicular to both a and b, whose orientation is determined by the right-hand rule, and whose length is | a | | b |sin (θ)= | a | | b |sin (0)= | a | | b |(0)  The two nonzero vectors a and b are parallel if and only if ab=0 Note that: ˆi  ˆi = ˆj  ˆj = kˆ  kˆ = 0 ˆi  ˆj = kˆ , ˆj  kˆ = ˆi , kˆ  ˆi = ˆj  Example  Given A = 2ˆi + 3ˆj; B = − ˆi + 2ˆj  Find A B  Solution: ˆi ˆj kˆ A B = 2 3 0 = ˆi(0 − 0) − ˆj(0 − 0) + kˆ (4 − ( −3)) −1 2 0 = 7 kˆ Example  If a = 1, 3, 4 and b = 2, 7, –5, then = (–15 – 28)i – (–5 – 8)j + (7 – 6)k = –43i + 13j + k Problems 1) Prove that vectors U = 2i + 3j + k and V = 4i – 2j + 2k are perpendicular to each other. 2) Two vectors are given by, a = 2i + j + k and b = i + j + k. Find the dot product of these two vectors. 3) Two vectors are given by, a = 2i + j + k and b = i + j + k. Find the cross product of these two vectors. 1) Two vectors are given by, a = i + j + k and b = i – 2j + 3k. Find the dot product of these two vectors. 2) Two vectors are given by, a = 4i +2 j +2 k and b = 2i + 2j + 2k. Find the cross product of these two vectors. 1) Two vectors are given by, a = 2i + j + k and b = i + j + k. Find the cross product of these two vectors. 2) A and B are two perpendicular vectors given by A =5i+10j−3k and B =2i+4j−ck. The value of c is: 1) The angle between two vectors P =3i+2j​+3k and Q ​=2i−j​+3k is: 2) Let’s say A = 4i + 3j and B = 5i + 4j. Find the resultant vector from the addition of these two vectors. 3) Let’s say A = 5i + 5j and B = 3i + 3j. Find the resultant vector from the addition of these two vectors.