Physics 21APBS101 Vectors PDF

Physics 21APBS101 Vectors Vectors are quantities in physics and science characterized by both magnitude and direction such as displacement, velocity, force, etc,. It is denoted by a directed line segment 𝐴𝐵 from point 𝐴 to point 𝐵, where 𝐴 is the initial point and 𝐵 is the terminal or...

Physics 21APBS101 Vectors Vectors are quantities in physics and science characterized by both magnitude and direction such as displacement, velocity, force, etc,. It is denoted by a directed line segment 𝐴𝐵 from point 𝐴 to point 𝐵, where 𝐴 is the initial point and 𝐵 is the terminal or end point. The vector 𝑃𝑄 may be denoted by 𝑨 or 𝐴. The magnitude or the length of the vector is denoted by |𝑃𝑄| , 𝐴 , 𝑨 , or simply 𝐴. Vector algebra Vector addition Consider two vectors 𝑨 and 𝑩 as shown in the figure. The resultant or the sum of the vector 𝐶 is expressed as 𝑪=𝑨+𝑩 If a vector 𝑨 = 𝑩 then the term 𝑨 − 𝑩 is defined as the null or zero vector, represented by the symbol 0 or 𝟎. 𝑪 Vector algebra Scalar multiplication A vector 𝑨 multiplied by a scalar 𝑚 produces a vector 𝑚𝑨 having a magnitude |𝑚| times the magnitude of 𝑨. The direction of 𝑚𝑨 is in the same or opposite direction of 𝑨 defined by whether 𝑚 is negative or positive. If 𝑚 = 0, then 𝑚𝑨 = 0, a null vector 𝑪 Proposition Vector addition Scalar multiplication 1. 𝑨 + 𝑩 + 𝑪 = 𝑨 + 𝑩 + 𝑪 5. 𝑚 𝑨 + 𝑩 = 𝑚𝑨 + 𝑚𝑩 associative 6. (𝑚 + 𝑛)𝑨 = 𝑚𝑨 + 𝑛𝑨 2. 𝑨 + 𝟎 = 𝟎 + 𝑨 = 𝑨 7. 𝑚 𝑛𝑨 = 𝑚𝑛 𝑨 3. 𝑨 + −𝑨 = −𝑨 + 𝑨 = 𝟎 8. 1 𝑨 = 𝑨 4. 𝑨 + 𝑩 = 𝑩 + 𝑨 commutative The above are 8 axiom that defines the abstract nature of vector space Unit vectors Definition Vectors having unit length are known as unit vectors. 𝑨 If 𝑨 is a vector having length |𝑨| > 0, then is the unit |𝑨| vector denoted by 𝒂, which has the same direction as that of 𝑨. (𝒂) 𝑪 Example Unit vectors 1) If 𝑨 = 7, then 𝑨 𝒂= |𝐴| 𝑨 𝒂= is the unit vector in the direction of 𝑨 7 2) 𝑨 = 3𝑖 + 7𝑗 − 𝑘, 𝑩 = 2𝑖 + 7𝑘, then 𝑨 + 𝑩 = 3 + 2 𝒊 + 7 + 0 𝒋 + −1 + 7 𝒌 𝑨 + 𝑩 = 5𝒊 + 7𝒋 + 6𝒌 Unit vectors Rectangular unit vectors Rectangular unit vectors In three dimensional rectangular co-ordinate system, the unit vectors are denoted by 𝒊, 𝒋, and 𝒌 for the respective positive 𝑥, 𝑦, and 𝑧 axis If a nonzero vectors 𝑨, 𝑩, 𝑪 have the same initial point and are not coplanar. Then 𝑨, 𝑩, 𝑪 are said to form a right-handed system or dextral system if a right-threaded screw rotated through an angle less than 180° from 𝑨 to 𝑩 will advance in the direction 𝑪 Vector Components of vector A vector 𝑨 in three dimension can be represented with an initial point at the origin 𝑂 = (0,0,0) and its end points at some points say (𝐴1 , 𝐴2 , 𝐴3 ). Then the vectors 𝐴1 𝒊, 𝐴2 𝒋, 𝐴3 𝒌 are called component vectors of 𝑨 in the 𝑥, 𝑦, and 𝑧 direction, and the scalars 𝐴1 , 𝐴2 , 𝐴3 are called the components of 𝑨. 𝑨 = 𝐴1 𝒊 + 𝐴2 𝒋 + 𝐴3 𝒌 𝑨 = 𝐴12 + 𝐴22 + 𝐴23 Vector Position vector If we a point 𝑃(𝑥, 𝑦, 𝑧) in space, the the vector 𝒓 drawn from the origin 𝑂 to point 𝑃 is called the position vector or the radius vector represented by 𝒓 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒓 = 𝑥2 + 𝑦2 + 𝑧2 Vector Vector field Suppose for each point (𝑥, 𝑦, 𝑧) or a region 𝑨 in space there corresponds a vector 𝑽(𝑥, 𝑦, 𝑧). Then 𝑽 is called vector function of position and this is the vector field defined on 𝑨. We can say that vector field is the assignment of a vector 𝑽 to each point in a subset of space A vector field 𝑽 which is independent of time is called a stationary or steady-state vector field. Example: The function 𝐕 𝑥, 𝑦, 𝑧 = 18 𝑖 + 9 𝑗 + 𝑘 defines a vector field. Example An automobile travels 3 miles due north, then 5 miles northeast. Represent these displacements graphically and determine the resultant displacement 𝐴𝑥 = 𝐴 cos 𝜃 𝐴𝑥 = 3 cos 90 = 3 0 = 0 𝐴𝑦 = 𝐴 𝑠𝑖𝑛𝜃 𝐴𝑦 = 3 sin 90 = 3 𝐴 = 𝐴𝑥 + 𝐴𝑦 𝐴 = 0 𝑖 + 3𝑗 = 3𝑗 Example An automobile travels 3 miles due north, then 5 miles northeast. 𝐵𝑥 = 𝐵 cos𝜃 𝐵𝑥 = 5 cos 45 = 3.53 𝐵𝑦 = 𝐵 𝑠𝑖𝑛𝜃 𝐵𝑦 = 5 sin 45 = 3.53 𝐵 = 𝐵𝑥 + 𝐵𝑦 𝐵 = 3.53 𝑖 + 3.53 𝑗 Example An automobile travels 3 miles due north, then 5 miles northeast. Since 𝐴 + 𝐵 = 𝐶 Resultant 𝐶 = 3 𝑗 + 3.53 𝑖 + 3.53 𝑗 𝐶 = 3.53 𝑖 + 6.53 𝑗 Magnitude of C is 𝐶 = 3.53 2 + 6.53 2 𝐶 = 12.46 + 42.64 𝐶 = 7.42 Vector Dot or Scalar product The dot product two vectors 𝑨 and 𝑩 is defined as the product of the magnitude of 𝑨 and 𝑩 and the cosine of the angle between them 𝑨. 𝑩 = 𝑨 𝑩 cos 𝜃 0≤𝜃≤𝜋 Here 𝑨. 𝑩 is a Scalar Proposition 1. 𝑨. 𝑩 = 𝑩. 𝑨 commutative 2. 𝑨. 𝑩 + 𝑪 = 𝑨. 𝑩 + 𝑨. 𝑪 distributive 3. 𝒎 𝑨. 𝑩 = 𝒎𝑨. 𝑩 = 𝑨. 𝒎𝑩 = 𝑨. 𝑩 𝒎 4. 𝒊. 𝒊 = 𝒋. 𝒋 = 𝒌. 𝒌 = 𝟏, 5. 𝒊. 𝒋 = 𝒋. 𝒌 = 𝒌. 𝒊 = 𝟎 6. If 𝑨. 𝑩 = 𝟎, then 𝐴 and 𝐵 are perpendicular If 𝑨 = 𝐴1 𝒊 + 𝐴2 𝒋 + 𝐴3 𝒌 and 𝑩 = 𝐵1 𝒊 + 𝐵2 𝒋 + 𝐵3 𝒌, then 𝑨. 𝑩 = 𝐴1 𝐵1 + 𝐴2 𝐵2 + 𝐴3 𝐵3 Example Example 1: 𝑨 = 3𝒊 + 5𝒋 − 𝒌 and 𝑩 = 2𝒊 − 3𝒋 + 6𝒌, then 𝑨. 𝑩 = 3 2 + 5 −3 + (−1)(6) 𝑨. 𝑩 = 6 − 15 − 6 𝑨. 𝑩 = −15 Example Example 2: Find the angle between 𝑨 = 2𝒊 + 3𝒋 − 2𝒌 and 𝑩 = −𝒊 + 𝒋 + 2𝒌 𝑨 = 2 2 + 3 2 + −2 2 𝐴 = 17 𝑩 = −1 2 + 1 2 + 2 2 𝐵 = 6 𝑨 𝑩 = 17 × 6 𝑨 𝑩 = 10.09 Example 𝑨. 𝑩 = 2 −1 + 3 1 + (−2)(2) 𝑨. 𝑩 = −2 + 3 − 4 𝑨. 𝑩 = −3 We know that 𝑨. 𝑩 = 𝑨 𝑩 𝑐𝑜𝑠𝜃 𝑨. 𝑩 𝑐𝑜𝑠𝜃 = = −0.297 𝑨 𝑩 𝜃 = cos−1 −0.297 𝜃 = 107.2° Cross product The cross product of vector A and B is equal to the product of the magnitude of A and B and the sine of the angle between them. 𝑨 × 𝑩 = 𝑨 𝑩 𝑠𝑖𝑛𝜃𝒖 0≤𝜃≤𝜋 In the above 𝒖 is the unit vector indicating the direction of 𝑨×𝑩 If 𝑨 is parallel to 𝑩 or if 𝑨 = 𝑩, then 𝑠𝑖𝑛𝜃 = 0, then cross product 𝑨×𝑩=0 Proposition 𝑨 × 𝑩 = −(𝑩 × 𝑨) Commutative 𝑨× 𝑩+𝑪 =𝑨×𝑩+𝑨×𝑪 Distributive 𝑚 𝑨 × 𝑩 = 𝑚𝑨 × 𝑩 = 𝑨 × 𝑚𝑩 = 𝑨 × 𝑩 𝑚 𝒊 × 𝒊 = 𝒋 × 𝒋 = 𝒌 × 𝒌 = 0, 𝒊 × 𝒋 = 𝒌, 𝒋 × 𝒌 = 𝒊, 𝒌 × 𝒊 = 𝒋 𝒊 × 𝒌 = −𝒋, 𝒋 × 𝒊 = −𝒌, 𝒌 × 𝒋 = −𝒊 If 𝑨 × 𝑩 = 0, then 𝑨 and 𝑩 are parallel The magnitude of 𝑨 × 𝑩 is same as the area of parallelogram having side 𝑨 and 𝑩 Proposition 𝒊 × 𝒊 = 𝒋 × 𝒋 = 𝒌 × 𝒌 = 0, 𝒊 × 𝒋 = 𝒌, 𝒋 × 𝒌 = 𝒊, 𝒌 × 𝒊 = 𝒋 𝒊 × 𝒌 = −𝒋, 𝒋 × 𝒊 = −𝒌, 𝒌 × 𝒋 = −𝒊 Cross product If 𝑨 = 𝐴1 𝒊 + 𝐴2 𝒋 + 𝐴3 𝒌 and 𝑩 = 𝐵1 𝒊 + 𝐵2 𝒋 + 𝐵3 𝒌 then 𝒊 𝒋 𝒌 𝑨 × 𝑩 = 𝐴1 𝐴2 𝐴3 𝐵1 𝐵2 𝐵3 𝐴2 𝐴3 𝐴1 𝐴3 𝐴1 𝐴2 𝑨×𝑩= 𝒊− 𝒋+ 𝒌 𝐵2 𝐵3 𝐵1 𝐵3 𝐵1 𝐵2 Example Example : Evaluate 𝑖 2𝑗 × 4𝑘 𝑖𝑖 𝑗 × −4𝑘 𝑖𝑖𝑖 − 2𝑖 × − 7𝑘 𝑖𝑣 2𝑗 × 3𝑖 − 2𝑘 1. 2𝑗 × 4𝑘 = 8 𝑗 × 𝑘 = 8𝑖 2. 𝑗 × −4𝑘 = −4 𝑗 × 𝑘 = −4𝑖 3. −2𝑖 × −7𝑘 = 14 𝑖 × 𝑘 = −14𝑗 4. 2𝑗 × 3𝑖 − 2𝑘 = 6 𝑗 × 𝑖 − 2𝑘 = −6𝑘 − 2𝑘 = −8𝑘 Example Example : Evaluate 𝑖 2𝑗 × 4𝑘 𝑖𝑖 𝑗 × −4𝑘 𝑖𝑖𝑖 − 2𝑖 × − 7𝑘 𝑖𝑣 2𝑗 × 3𝑖 − 2𝑘 1. 2𝑗 × 4𝑘 = 8 𝑗 × 𝑘 = 8𝑖 2. 𝑗 × −4𝑘 = −4 𝑗 × 𝑘 = −4𝑖 3. −2𝑖 × −7𝑘 = 14 𝑖 × 𝑘 = −14𝑗 4. 2𝑗 × 3𝑖 − 2𝑘 = 6 𝑗 × 𝑖 − 2𝑘 = −6𝑘 − 2𝑘 = −8𝑘 Example Example: If 𝐴 = 𝑗 + 2𝑘 and 𝐵 = 𝑖 + 2𝑗 + 3𝑘, evaluate 𝐴 × 𝐵, and 𝐵 × 𝐴 𝑖 𝑗 𝑘 𝐴 × 𝐵 = 𝑗 + 2𝑘 × 𝑖 + 2𝑗 + 3𝑘 = 0 1 2 1 2 3 1 2 0 2 0 1 𝐴×𝐵 =𝑖 −𝑗 +𝑘 2 3 1 3 1 2 𝐴×𝐵 =𝑖 3−4 −𝑗 0−2 +𝑘 0−1 𝐴 × 𝐵 = −𝑖 + 2𝑗 − 𝑘 Example Example: If 𝐴 = 𝑗 + 2𝑘 and 𝐵 = 𝑖 + 2𝑗 + 3𝑘, evaluate 𝐴 × 𝐵, and 𝐵 × 𝐴 𝑖 𝑗 𝑘 𝐵 × 𝐴 = 𝑖 + 2𝑗 + 3𝑘 × 𝑗 + 2𝑘 = 1 2 3 0 1 2 2 3 1 3 1 2 𝐵×𝐴=𝑖 −𝑗 +𝑘 1 2 0 2 0 1 𝐵 × 𝐴 = 𝑖 4 − 3 − 𝑗 2 − 0 + 𝑘(1 − 0) 𝐵 × 𝐴 = 𝑖 − 2𝑗 + 𝑘 The results shows that 𝑨 × 𝑩 = −𝑩 × 𝑨 Dot and Cross Vector Triple product The dot and cross multiplication of three vectors 𝑨, 𝑩, and 𝑪 produce products known as triple products 𝑨. 𝑩 𝑪, 𝑨. 𝑩 × 𝑪 , and 𝑨 × (𝑩 × 𝑪) Proposition 𝑨. 𝑩 𝑪 ≠ 𝑨 𝑩. 𝑪 𝑨× 𝑩×𝑪 ≠ 𝑨×𝑩 ×𝑪 𝑨. 𝑩 × 𝑪 = 𝑩. 𝑪 × 𝑨 = 𝑪. (𝑨 × 𝑩) 𝑨 × 𝑩 × 𝑪 = 𝑨. 𝑪 𝑩 − 𝑨. 𝑩 𝑪 𝑨 × 𝑩 × 𝑪 = 𝑨. 𝑪 𝑩 − 𝑩. 𝑪 𝑨 If 𝑨 = 𝐴1 𝒊 + 𝐴2 𝒋 + 𝐴3 𝒌, 𝐵 = 𝐵1 𝒊 + 𝐵2 𝒋 + 𝐵3 𝒌 then 𝐴1 𝐴2 𝐴3 𝑨. 𝑩 × 𝑪 = 𝐵1 𝐵2 𝐵3 𝐶1 𝐶2 𝐶3 Reciprocal set The sets 𝒂, 𝒃, 𝒄 and 𝒂′ , 𝒃′ , 𝒄′ are called reciprocal sets of vectors if 𝒂. 𝒂′ = 𝒃. 𝒃′ = 𝒄. 𝒄′ = 1 𝒂′. 𝒃 = 𝒂′. 𝒄 = 𝒃′. 𝒂 = 𝒃′. 𝒄 = 𝒄′. 𝒂 = 𝒄′. 𝒃 = 0 Proposition The sets 𝒂, 𝒃, 𝒄 and 𝒂’, 𝒃’, 𝒄’ are reciprocal sets of vectors if and only if ′ 𝒃×𝒄 ′ 𝒄×𝒂 ′ 𝒂×𝒃 𝒂 = 𝒃 = 𝒄 = 𝒂. 𝒃 × 𝒄 𝒂. 𝒃 × 𝒄 𝒂. 𝒃 × 𝒄 Where 𝒂. 𝒃 × 𝒄 ≠ 0 Example Find the projection of the vector 𝐴 = 𝑖 − 2𝑗 + 3𝑘 on the vector 𝐵 = 𝑖 + 2𝑗 + 2𝑘. Solution: The unite vector of 𝐵 can be expressed as 𝐵 𝑏= |𝐵| 𝑖 + 2𝑗 + 2𝑘 𝑏= 1 2 + 2 2 + 2 2 𝑖 + 2𝑗 + 2𝑘 1 2 2 𝑏= = 𝑖+ 𝑗+ 𝑘 3 3 3 3 Type equation here.The projection of 𝐴 on vector 𝐵 is 1 2 2 𝐴. 𝑏 = 𝑖 − 2𝑗 + 3𝑘 𝑖+ 𝑗+ 𝑘 3 3 3 1 4 6 𝐴. 𝑏 = + − + 3 3 3 𝐴. 𝑏 = 1 Co-ordinate system Cartesian system For two dimension the Cartesian co-ordinate is also known as the rectangular co-ordinate. The point 𝑃(𝑥, 𝑦) in space in analogous of the point plotted in the 𝑥 and 𝑦 axis. Co-ordinate system Cartesian system For three dimensional system the point 𝑃(𝑥, 𝑦, 𝑧) is represented as shown in the figure. 𝒓 is the radius vector or the position vector, where 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 𝑟 = 𝑥2 + 𝑦2 + 𝑧2 Co-ordinate system Polar system In polar co-ordinate system the point in a plane is represented by 𝑟, 𝜃. The conversion from Cartesian to Polar co-ordinate is 𝑥 = 𝑟𝑐𝑜𝑠𝜃 𝑦 = 𝑟𝑠𝑖𝑛𝜃 𝑟2 = 𝑥2 + 𝑦2 𝑡𝑎𝑛𝜃 = 𝑦/𝑥 Co-ordinate system Cylindrical system For the cylindrical system, the point 𝑃 in space is represented by (𝜌, 𝜙, 𝑧). The transformation equation is 𝑥 = 𝜌𝑐𝑜𝑠𝜙 𝑦 = 𝜌𝑠𝑖𝑛𝜙 𝑧=𝑧 𝜌2 = 𝑥 2 + 𝑦 2 𝑡𝑎𝑛𝜙 = 𝑦/𝑥 Co-ordinate system Spherical system Here, the point 𝑃 in space is specified by (𝑟, 𝜃, 𝜙) Conversion from Cartesian to spherical 𝑥 = 𝑟 𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜙 𝑦 = 𝑟 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜙 𝑧 = 𝑟𝑐𝑜𝑠𝜃 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2 𝑥2 + 𝑦2 𝑡𝑎𝑛𝜃 = 𝑧 𝑡𝑎𝑛𝜙 = 𝑦/𝑥 Derivative of vector Suppose 𝑹(𝑢) is vector depending on a scalar variable 𝑢, then Δ𝑹 𝑹 𝑢 + Δ𝑢 − 𝑹(𝑢) = Δ𝑢 Δ𝑢 limit exists, the ordinary derivative of vector 𝑅(𝑢) with respect to the scalar 𝑢 is as follows 𝑑𝑹 𝛥𝑹 𝑹 𝑢 + Δ𝑢 − 𝑹(𝑢) = 𝑙𝑖𝑚 = 𝑙𝑖𝑚 𝑑𝑢 𝛥𝑢→0 𝛥𝑢 𝛥𝑛→0 𝛥𝑢 𝑑𝑹 In the above, is the derivative of 𝑅 with respect to 𝑢. 𝑑𝑢 Space curve If we consider a point 𝑃(𝑥, 𝑦, 𝑧) in space, the position vector 𝑟(𝑢) joining the origin 𝑂 and any point (𝑥, 𝑦, 𝑧), then 𝑟 𝑢 =𝑥 𝑢 𝑖+𝑦 𝑢 𝑗+𝑧 𝑢 𝑘 With the change in 𝑢, the terminal points of 𝑟 describes a space curve having parametric equations 𝑥 = 𝑥 𝑢 , 𝑦 = 𝑦 𝑢 , 𝑧 = 𝑧(𝑢) The following is a vector in the direction of Δ𝑟 if Δ𝑢 > 0 and in direction of −Δ𝑟 if Δ𝑢 < 0 Δ𝑟 𝑟 𝑢 + Δ𝑢 − 𝑟(𝑢) = Δ𝑢 Δ𝑢 𝛥𝒓 𝑑𝑟 If limit exist i.e 𝑙𝑖𝑚 = 𝛥𝑢→0 𝛥𝑢 𝑑𝑢 Then the limit will be a vector in the direction of tangent to the space curve at (𝑥, 𝑦, 𝑧), given by 𝑑𝑟 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑖+ 𝑗+ 𝑘 𝑑𝑢 𝑑𝑢 𝑑𝑢 𝑑𝑢 Derivative of vector Velocity and acceleration If a particle 𝑃 moves along a space curve C, whose parametric equations are 𝑥 = 𝑥 𝑡 , 𝑦 = 𝑦 𝑡 , 𝑧 = 𝑧(𝑡), where 𝑡 represents time. Then the position vector of 𝑃 along the curve is 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗+𝑧 𝑡 𝑘 In such case, the velocity and acceleration of the particle 𝑃 is given by 𝑑𝑟 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑣=𝑣 𝑡 = = 𝑖+ 𝑗+ 𝑘 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑 2 𝑟 𝑑𝑣 𝑑 2 𝑥 𝑑2 𝑦 𝑑2 𝑧 𝑎=𝑎 𝑡 = 2= = 2𝑖+ 2𝑗+ 2𝑘 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡  Proposition 𝑑 𝑑𝑨 𝑑𝑩 1. 𝑑𝑢 𝑨×𝑩 = 𝑑𝑢 + 𝑑𝑢 𝑑 𝑑𝑩 𝑑𝑨 2. 𝑑𝑢 𝑨. 𝑩 = 𝑨. + 𝑩. 𝑑𝑢 𝑑𝑡 𝑑 𝑑𝑩 𝑑𝑨 3. 𝑑𝑢 𝑨×𝑩 =𝑨× + ×𝑩 𝑢 𝑑𝑢 𝑑 𝑑𝑨 𝑑𝜙 4. 𝑑𝑢 𝜙𝑨 = 𝜙 + 𝑨 𝑑𝑢 𝑑𝑢 𝑑 𝑑𝑪 𝑑𝐵 𝑑𝐴 5. 𝑑𝑢 𝑨. 𝑩 × 𝑪 = 𝑨. 𝑩 × + 𝐴. 𝑑𝑢 𝑑𝑢 ×𝐶+ 𝑑𝑢.𝐵 × 𝐶 𝑑 𝑑𝑪 𝑑𝑩 𝑑𝑨 6. 𝑑𝑢 𝑨× 𝑩×𝑪 =𝑨× 𝑩× 𝑑𝑢 +𝑨× 𝑑𝑢 × 𝑪 + 𝑑𝑢 × 𝑩×𝑪  Example Example 1: If 𝑨 = 5𝑢2 𝒊 + 𝑢 𝒋 + 𝑢3 𝒌, 𝑩 = 𝑠𝑖𝑛𝑢 𝒊 − 𝑐𝑜𝑠𝑢 𝒋. Find 𝑑 (𝐴. 𝐵) 𝑑𝑢 𝑑 𝑑𝑩 𝑑𝑨 𝑨. 𝑩 = 𝑨. +.𝑩 𝑑𝑢 𝑑𝑢 𝑑𝑢 = 5𝑢2 𝒊 + 𝑢 𝒋 + 𝑢3 𝒌. 𝑐𝑜𝑠𝑢 𝒊 + 𝑠𝑖𝑛𝑢 𝒋 + 10𝑢 𝒊 + 𝒋 − 3𝑢2 𝒌. ( 𝑠𝑖𝑛𝑢 𝒊 − 𝑐𝑜𝑠𝑢 𝒋) = [5𝑢2 cos 𝑢 + 𝑢 𝑠𝑖𝑛𝑢] + 10𝑢 𝑠𝑖𝑛𝑢 − 𝑐𝑜𝑠𝑢 = 5𝑢2 − 1 𝑐𝑜𝑠𝑢 + 11𝑢 𝑠𝑖𝑛𝑢