Vectors in motion Copyright 2008 by Evans M. Harrell II. MATH 2401 - Harrell- Uncle Si - John Saylor Coon, 1854-1938, Founder of GT School of Mechanical Engineering, “Engineering is common sense first, and mathematics next.”Any business to take care of?  date Wed, Aug 20, 2008 at 2:26 PM  subject Re: Math 2401 - homework question  Dr. Harrell,  Greetings! Thank you for all of the informative emails. I have a question in  regards to the homework: are we to submit it during lectures on MW or during the  recitation on T/Th? Thank you for your time. I look forward to hearing from you.  Best regards, ANSWER: Please submit your homework to the TA at recitation.Copyright 2008 by Evans M. Harrell II. But first - A bit more Vector Boot Camp!Why on earth would you differentiate a  dot product  cross product ? Examples  How fast is the angle between two vectors changing? cos θ(t) = v(t)•w(t) (You’ll need product and chain rule.)  How fast is the angular momentum changing? L = r × p.Why on earth would you integrate a vector function?Examples  Given velocity v(t) find position x(t).  Power = F•v . Work is the integral of this. If, say, v is fixed, you can integrate F and then dot it with v.The good news:  The rules of vector calculus look just like the rules of scalar calculusThe good news:  The rules of vector calculus look just like the rules of scalar calculus  Integrals and derivs of α f(t), f(t)+g(t), etc.The good news:  The rules of vector calculus look just like the rules of scalar calculus  Integrals and derivs of α f(t), f(t)+g(t), etc.  Also - because of this - you can always calculate component by component.Calculus is built on the idea of a limit. What does a limit mean for vector functions?  The limit means limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1Calculus is built on the idea of a limit. What does a limit mean for vector functions?  The limit means limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1Some kind of scalar that depends on tOne of the great tricks of vector calculus:  If you can rewrite a vector problem in some way as a scalar problem, it becomes “kindergarten math.”Calculus is built on the idea of a limit. What does a limit mean for vector functions?  The limit means limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1limt→t0f (t) = Llimt→t0!f (t) − L! = 0Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1 So if the left side → 0, each and every one of the contributions on the right → 0 as well. And conversely. limt→t0f (t) = L!f (t) − L!2= (f1− L1)2+ (f2− L2)2+ (f3− L3)2Since [H, G]uk= (H − λk)Guk,![H, G]uk!2=!Guk, (H − λk)2Guk"#dds, t$ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 2 #Guk, (H − λk)Guk$R(ζ) =%Ω(|∇ζ|2+ V (x)|ζ|2)dV%Ω|ζ|2dVλn∼ 4π&Γ(1 + d/2)n|Ω|'2/d+Vave2(z−λj) #uj, [G, [H, G]] uj$−2![H, G] uj!2=(k(z − λk)(λk− λj)| #uj, Guk$|2q(x) = +14)(jκj*2Write the test function asζ =1√ρ· (√ρζ)and use the product rule in the formψ(t) = e−iHtψ(0)−∇2#+ q(x) = −∆Ω+ q(x)1You can do calculus in terms of vectors or components. You choose.You can think in terms of vectors or components. A mystical picture:You can think in terms of vectors or components. You choose.  Some limit examples  spiral x(t) = t cos t, y(t) = t sin t  parabola x(t) = t2 , y(t) = - tThe good news:  The rules of vector calculus look just like the rules of scalar calculus  product rule(s)  (u(t) f(t))’ = u’(t) f(t) + u(t)