Math 375 – 12/13 Dec 13, 2017 Shawn Math 375 No comments yet Math 375 – 12/12 Dec 12, 2017 Shawn Math 375 No comments yet Math 375 – 12/11 Dec 11, 2017 Shawn Math 375 No comments yet Math 375 – 12/7 Dec 07, 2017 Shawn Math 375 No comments yet 1.1 – What is Physics? Dec 08, 2017 Shawn AP Physics C Mechanics No comments yet 1 … 41 42 43 44 45 … 84
![Question 1 • Find a basis in which the matrix (■8(3&0@3&−2)) becomes diagonalized • Let A=(■8(3&0@3&−2)) • det(A−λI)=|■8(3−λ&0@3&−2−λ)|=λ^2−λ−6=0 • ⇒λ_1=3, λ_2=−2 • ⇒Λ=(■8(3&0@0&−2)) • When λ_1=3 ○ A−λI=(■8(0&0@3&−5)) ○ ⇒v_1=k(5,3), k∈R • When λ_2=−2 ○ A−λI=(■8(5&0@3&0)) ○ ⇒v_2=k(0,1), k∈R • The basis is (5,3), (0,1) Exercise 8.17 Question 8 • Find a Cartesian equation for the tangent plane • to the surface xyz=a^3 at a general point (x_0,y_0,z_0 ). ○ ∇f=(█(yz@xz@xy)) ○ H={(x,y,z)∈R3│∇f(x_0,y_0,z_0 )⋅(█(x−x_0@y−y_0@z−z_0 ))=0} ○ ={(x,y,z)∈R3│y_0 z_0 (x−x_0 )+x_0 z_0 (y−y_0 )+x_0 y_0 (z−z_0 )=0} ○ ={(x,y,z)∈R3│xy_0 z_0+x_0 yz_0+x_0 y_0 z=3x_0 y_0 z_0=3a^3 } Question 2 • Let f:R2→R smooth • Let g(x,y)=f(u,v)=f(sin〖(x)y,x^y 〗 ) • Find g_x,g_y in terms of f_u,f_v ○ Let h(x,y)=[█(u(x,y)@v(x,y) )]=[█(sin(x)y@x^y )], Then T_g=T_f∘T_ℎ ○ T_f=[■8(∂f/∂u&∂f/∂v)]=[■8(f_u&f_v )] ○ T_ℎ=[■8(∂u/∂x&∂u/∂y@∂v/∂x&∂v/∂y)]=[■8(cos(x)y&sin(x)@y⋅x^(y−1)&x^y (y ln(x)+1) )] ○ T_g=T_f∘T_ℎ=[■8(f_u&f_v )][■8(cos(x)y&sin(x)@y⋅x^(y−1)&x^y (y ln(x)+1) )]=[█(f_u cos(x)y+f_v y⋅x^(y−1)@f_u sin(x)+f_v x^y (y ln(x)+1) )]^T ○ ⇒{█(g_x=f_u cos(x)y+f_v y⋅x^(y−1)@g_y=f_u sin(x)+f_v x^y (y ln(x)+1) )┤](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a31df1529981.png)
![Derivative of Vector Fields • Map ○ f:Rn→Rm ○ f(x_1,…,x_n )=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )] • Example 1 ○ n=1, m=2 ○ f:R→Rn ○ f:t↦[█(f_1 (t)@f_2 (t) )]=[█(x(t)@y(t) )] ○ Also called parametric curve • Example 2 ○ f:Rn→R, where n≥2 ○ f is a function of n variables f(x_1,…,x_n ) • Example 3 ○ Rn→Rm, where n≥2, m≥2 ○ f:R2→R2 ○ f:(u,v)⟼(x,y) ○ Polar Coordinates § f(r,θ)=[█(x(r,θ)@y(r,θ) )]=[█(r cosθ@r sinθ )] • Example 4 ○ f(u,v)=[■8(2&3@1&−1@1&1)][█(u@v)]=[█(2u+3v@u−v@u+v)] • Definition ○ A map f:Rn→Rm is differentiable at a∈Rn if ○ f(a+y)=f(a)+T_a (y)+E(a,y)‖y‖ ○ Where T_a:Rn→Rm is a linear transformation, and lim_(y→0)E(a,y)=0 • Theorem ○ f differentiable at a⇒f continuous at a • Theorem ○ If f(x)=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )], and f_1,…,f_m have continuous partial derivatives ○ Then f is differentiable and T_a (y)=[■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))][█(y_1@⋮@y_n )] • Example ○ f(r,θ)=[█(r cosθ@r sinθ )] ○ The linear transformation T_a has the matrix § mat(T_a )=[■8((∂f_1)/∂r&(∂f_1)/∂θ@(∂f_2)/∂r&(∂f_2)/∂θ)]=[■8(∂x/∂r&∂x/∂θ@∂y/∂r&∂y/∂θ)]=[■8(cosθ&−r sinθ@sinθ&r cosθ )] ○ At a=(√2,π/4) § mat(T_a )=[■8(√2∕2&−1@√2∕2&1)] § T_a [█(1@0)]=[█(√2∕2@√2∕2)] § T_a [█(0@1)]=[█(−1@1)] ○ In general § mat(T_((r,θ) ) )=[■8(cosθ&−r sinθ@sinθ&r cosθ )] § T_a [█(ε@0)]=ε[█(cosθ@sinθ )] § T_a [█(0@ε)]=εr[█(−sinθ@cosθ )] • Common notations ○ T_a (v)=df_a⋅v=Df(a)⋅v=f^′ (a)⋅v • Jacobian Matrix ○ [■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))] is called Jacobian Matrix • Chain Rule ○ Given f:Rn→Rm, g:Rm→Rk ○ Consider h(x)=g(f(x))=g∘f(x) ○ {█(f is differentiable at x@g is differentiable at f(x) )┤⇒h=g∘f is differentiable at x ○ And d(g∘f)_x=dg_f(x) ⋅df_x ○ Another notation: (g∘f)^′ (x)=g^′ (f(x))⋅f^′ (x) ○ In components § x=[█(x_1@⋮@x_n )]∈Rn § u=[█(u_1@⋮@u_m )]=[█(f_1 (x)@⋮@f_m (x) )]=[█(f_1 (x_1,…,x_n )@⋮@f_m (x_1,…,x_n ) )]∈Rm § v=[█(v_1@⋮@v_k )]=[█(g_1 (u)@⋮@g_k (u) )]=[█(g_1 (u_1,…,u_m )@⋮@g_k (u_1,…,u_m ) )]=[█(g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))@⋮@g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n )) )]∈Rn § mat[(g∘f)^′ ] § =[■8((∂v_1)/x_1 &⋯&(∂v_1)/x_n @⋮&⋱&⋮@(∂v_k)/x_1 &⋯&(∂v_k)/x_n )] § =[■8(∂/(∂x_1 ) g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))&⋯&∂/(∂x_n ) g_1 (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))@⋮&⋱&⋮@∂/(∂x_1 ) g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n ))&⋯&∂/(∂x_n ) g_k (f_1 (x_1,…,x_n ),…,f_m (x_1,…,x_n )) )] § =[■8((∂g_1)/(∂u_1 )⋅(∂f_1)/(∂x_1 )+…+(∂g_1)/(∂u_m )⋅(∂f_m)/(∂x_1 )&⋯&(∂g_1)/(∂u_1 )⋅(∂f_1)/(∂x_n )+…+(∂g_1)/(∂u_m )⋅(∂f_m)/(∂x_n )@⋮&⋱&⋮@(∂g_k)/(∂u_1 )⋅(∂f_1)/(∂x_1 )+…+(∂g_k)/(∂u_m )⋅(∂f_m)/(∂x_1 )&⋯&(∂g_k)/(∂u_1 )⋅(∂f_1)/(∂x_n )+…+(∂g_k)/(∂u_m )⋅(∂f_m)/(∂x_n ))] § =[■8((∂g_1)/(∂u_1 )&⋯&(∂g_1)/(∂u_m )@⋮&⋱&⋮@(∂g_k)/(∂u_1 )&⋯&(∂g_k)/(∂u_m ))][■8((∂f_1)/(∂x_1 )&⋯&(∂f_1)/(∂x_n )@⋮&⋱&⋮@(∂f_m)/(∂x_1 )&⋯&(∂f_m)/(∂x_n ))] § =mat(g^′ (f(x)))mat(f^′ (x))](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a308cf6ccaea.png)
![Quiz Question • Given ○ A:R3→R3 ○ det〖(A−2I)=0〗 ○ tr (A)=−2 ○ det(A)=6 • Question: Eigenvalue of A ○ det〖(A−2I)=0〗⇒2 is an eigenvalue ○ tr (A)=−2⇒∑_(i=1)^n▒λ_i =λ_1+λ_2+λ_3=−2 ○ det(A)=6⇒∏_(i=1)^n▒λ_i =λ_1 λ_2 λ_3=6 ○ {█(█(λ_1=2@λ_1+λ_2+λ_3=−2)@λ_1 λ_2 λ_3=6)┤⇒{█(λ_2=−3@λ_3=−1)┤ ○ Therefore eigenvalues are 2, −1, −3 • Question: Characteristic Polynomial ○ f(λ)=(λ−λ_1 )(λ−λ_2 )(λ−λ_3 )=(λ−2)(λ+1)(λ+3) • Note: f(λ)=det(λI−A) Question 1 • Given ○ f:R2\{0}→R ○ f(x,y)=xy/(x^2+y^2 ), ∀(x,y)∈R2 • Question: Find the direction of steepest decedent at (1,3) ○ ∇f(x,y)=[█(∂f/∂x@∂f/∂y)]=[█(y(y^2−x^2 )/(x^2+y^2 )^2 @x(x^2−y^2 )/(x^2+y^2 )^2 )] ○ ∇f(1,3)=[█(3(3^2−1^2 )/(1^2+3^2 )^2 @1(1^2−3^2 )/(1^2+3^2 )^2 )]=[█(6/25@−2/25)] • Question: Find the line best approximate the level set at (1,3) ○ ∇f(1,3)⋅n ⃗=0⇒n ⃗=[█(1@3)] ○ x+3y+c=0 ○ 1+3⋅3+c=0 ○ ⇒c=−10 ○ l: x+3y−10=0 ○ Alternative: ∇f(1,3)⋅[█(x−1@y−3)]=0 • Question: Estimate f(0.8,3.05) ○ f(0.8,3.05) ○ =f(1−0.2,3+0.05) ○ ≈f(1,3)+∇f(1,3)[█(−0.2@0.05)] ○ =3/(1^2+3^3 )+[█(6/25@−2/25)][█(−0.2@0.05)] ○ =0.248 Question 2 • Find a basis in which the matrix (■8(3&0@3&−2)) becomes diagonalized • Let A=(■8(3&0@3&−2)) • det〖(A−λI)=λ^2−λ−6=0〗 • ⇒λ_1=3, λ_2=−2 • When λ_1=3 ○ A−λI=(■8(0&0@3&−5)) ○ ⇒v_1=(5,3) • When λ_2=−2 ○ A−λI=(■8(5&0@3&0)) ○ ⇒v_2=(0,1) • The basis is (5,3), (0,1)](https://shawnzhong.com/wp-content/uploads/2017/12/img_5a2f00cd5f7f5.png)
