Given three vectors A, B, andc. We list three products with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B);

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1 Prouct of three vectors. Given three vectors A, B, anc. We list three proucts with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B); a 1 a 2 a 3 (A B) C = b 1 b 2 b 3 c 1 c 2 c 3 where the entries are the coorinates of the three vectors. The first two proucts are calle vector triple proucts, the thir is calle scalar triple prouct. The proof for the formulas for the vector triple proucts are complicate. But the proof for the formula for the scalar triple prouct is straightforwar. The reaer shoul be able to o it alone. To remember the formulas for the two vector triple proucts, there is a quick way. You see that the final prouct of the first vector triple prouct will be perpenicular to A B, so it will lie in the plane spanne by A an B. It is perpenicular to C, so there will be no component in the C irection. So the first vector triple prouct is a linear combination of A an B, notc. The coefficients are the inner proucts of the remaining two vectors, with a minus sign for the secon term; while the mile vector B is the first term. Recall that the magnitue (length) of A B is the area of the parallelogram spanne by A an B, an the inner prouct with C is this magnitue times C cos φ, which is exactly the height of the parallelepipe with a slante height C an a bottom parallelogram spanne by A an B. Thus the magnitue of the scalar triple prouct is the volume of the parallelepipe forme by the three vectors. See Figure : Volume of the parallelepipe forme by three vectors. One can form other triple proucts, but they all can be reuce quickly to one of the three mentione here. One may notice that the secon vector triple prouct can be reuce to the first vector prouct easily. So essentially there is only one vector triple prouct an one scalar triple prouct.

2 AxB height = projection of C. C B Area A Figure The volume of the parallelepipe is the magnitue of (AxB) \centerot C Variable vectors Vector functions of a scalar argument. Example 1.2.1a. A(t) =(cost, sin t, ), ( <t< ). The graph (the collection of all the tips of the vector A(t)) is a circle. Example 1.2.1b. A(t) =(cost, sin t, t), ( <t< ). The graph is a helix. Example 1.2.1c. A(t) =(3t, 2t, t) =(3, 2, 1)t. It represents a straight line. The general formula for a straight line is A(t) =tα + β where α an β are numerical vectors inepenent of t The erivatives of a vector function. Let A(t) =(A 1 (t),a 2 (t),a 3 (t)) = A 1 (t)i 1 + A 2 (t)i 2 + A 3 (t)i 3. Then A(t) =( A 1(t), A 2(t), A 3(t) )= A 1(t) i 1 + A 2(t) i 2 + A 3(t) i 3. Formal efinition: A (t) = lim t A(t + t) A(t). t 2

3 (Figure Derivative: pay attention to the irection of the ifference quotient.) A(t) A(t+ ) Figure Derivative: tangent irection. Example 1.2.2a. Let r(t) be the position vector of a moving particle. Then r(t) = v(t) is the velocity; 2 r(t) 2 = v(t) = a(t) is the acceleration. Example 1.2.2b. Consier r(t) =(cost, sin t). Then r (t) =( sin t, cos t). Note that the erivative is tangent to the graph of r(t). Consier R(t) = 2(cost, sin t). Then R (t) =2( sin t, cos t). Notice that the magnitue of the erivative is twice as large. See Figure (Figure Derivative: irection an magnitue.) R r r R 1 2 Figure Derivative: irection an magnitues. 3

4 Simple rules: (A ± B) = A ± B, (ca) = c A + c A, (A B) = A B + A B (A B) =. A B + A B. (1) The integral of a vector function. The integral of a vector function is efine also component-wise. Let A(t) =(A 1 (t),a 2 (t),a 3 (t)) = A 1 (t)i 1 + A 2 (t)i 2 + A 3 (t)i 3. Then B(t) = A(t) =( A 1 (t), A 2 (t), A 3 (t) ). So far we have covere in the lectures the following sections of our text book (Vector an Tensor Analysis with Applications, by Borisenko etc.) 1.2.3, 1.4, 1.5, an Vector fiels We have mentione the magnetic fiel, which is efine as a omain in which a vector of magnetism is efine at every point. Another example is the velocity fiel in a stream: each water roplet has a velocity. See Figure 1.3.1: Velocity fiel. Figure The velocity fiel of a stream. Generally, a vector fiel is a omain Ω an a vector function A(r) efineinit. Furthermore, a vector fiel may be time-epenent: A(r,t). 4

5 Line integrals an circulation. We introuce an integral that gives work one by a force fiel or circulation of velocity aroun a loop. Let A(r) be a vector fiel with omain Ω. Let M 1 M 2 be a curve in the omain irecte from M 1 to M 2. Chop the curve into many small pieces, say n pieces. One typical piece is enote by the en points r i an r i+1. See Figure The work one in this piece is approximately A(r i ) r i,where r i = r i+1 r i, if we imagine that the vector fiel is a force fiel. This can also be interprete as the flow of the vector fiel in the irection of r i. We sum over all such pieces an take the limit as all r i toefinetheline integral: lim n ni=1 A(r i ) r i = M 1 M 2 A(r) r = M 1 M 2 A 1 x 1 + A 2 x 2 + A 3 x 3. (2) Here the notation is A = A 1 i 1 + A 2 i 2 + A 3 i 3. Line integrals give either total work one by the vector fiel, or total flow of the vector fiel along the curve M 1 M 2 in the irection specifie. Total circulation aroun a contour L is efine as Γ= A r. (Figure Definition of the line integral.) L A M 2 r i M 1 r i+1 r i O Figure Definition of the line integral. 5

6 Example 1.3.1a. Let A =( x 2,x 1, ). Let L be the unit circle: x x2 2 = 1,x 3 = an counter-clockwise. Then Γ = L A r = L x 2x 1 + x 1 x 2 (L : x 1 =cosθ, x 2 =sinθ, θ 2π) sin θ cos θ +cosθ sin θ sin 2 θθ +cos 2 θθ θ =2π. See Figure (Figure Examples of circulation an line integrals.) (3) Γ = Γ = 2π (a) (b) Figure Examples of circulation an line integrals. Example 1.3.1b. Let A =(x 1,x 2, ) an L as before. Then Γ = L A r = L x 1x 1 + x 2 x 2 (L : x 1 =cosθ, x 2 =sinθ, θ 2π) cos θ cos θ +sinθ sin θ cos θ sin θθ +sinθ cos θθ θ =. Example 1.3.1c. Let A =(x 1,x 2, ) an L be the line segment: x 1 1,x 2 = irecte towar the x 1 -axis. Then 1 Γ= A r = x 1 x 1 = 1 2 x2 1 1 = 1 2. (5) En of Lecture 3 L 6 (4)