Lecture 8 MATH 2401 - Harrell Copyright 2008 by Evans M. Harrell II. Partial to partialsRecollections of last week  Cool shapes, like Möbius stripsRecollections of last week  Cool shapes, like Möbius stripsRecollections of last week  I can really get in to some of these shapes!Recollections of last week  The view from inside the hyperboloid.What is the derivative, really?  In KG calculus, it was the slope of the tangent line at a point.  For a vector function (same as curve), it still gave the direction of the tangent line at a point.  It tells us how to straighten out something that is curved, with the least error. !Ideal gas law. PV = nRT, so V(P,T) = nRT/P . If we fix P and change T to T+ ∆T, V(P,T+ ∆T) ≅ V(P,T) + (nR/P) ∆T ≅ V(P,T) (1 + ∆T/T) Extrapolating with the ideal gas lawIdeal gas law. PV = nRT, so V(P,T) = nRT/P . If we fix T and change P to P+ ∆P, V(P + ∆P,T) ≅ V(P,T) - (nRT/P2) ∆P ≅ V(P,T) (1 - ∆P/P) . Extrapolating with the ideal gas lawIdeal gas law. PV = nRT, so V(P,T) = nRT/P . What if both T and P change? Estimate the volume. We might guess V(P + ∆P,T + ∆T) ≅ V(P,T) (1 + ∆T/T - ∆P/P). Is this right? Extrapolating with the ideal gas lawThe partial derivative Just pretend y is a constant and differentiate with respect to x. Call this ∂F/∂x. If you pretend x is a constant and differentiate with respect to y, that’s ∂F/∂y.Anatomy of the partial derivative “Dee F by dee x” “F sub x”Why do we calculate partials? 1 Sometimes only interested in one variable. Example: If we only care how the concentration c(x,y) varies when we move in the x-direction, we want 󲰦c/󲰦x.This function still depends on y as well as x.Why do we calculate partials? 1 Sometimes only interested in one variable. 2 Nature loves partial derivatives: a Heat equation b Wave equation c Potential equationWhy do we calculate partials? 1 Sometimes only interested in one variable. 2 Nature loves partial derivatives: 3 All the Calculus I stuff (max-min, slopes, tangent planes rather than lines) will use partial derivatives when there is more than one variable.The partial derivative Important! ∂F/∂x and ∂F/∂y are still functions of 2 variables. Let’s do an example or two.Examples of partial derivatives (Let’s play Stump the """ !)!Examples of partial derivativesSets (something mathematicians are obsessed with) • Neighborhood of a point • Interior, boundary • Open • Closed • Neither open nor closedA set of wildebeest interior boundary isolated point exterior Copyright 2008 by Evans M. Harrell II.Limits for scalar fields Compare and contrast. limr→r0f(r) = Llimt→t0F(t) = Llimr→0x2− y2x2+ y2L(C) =!Cds =!ba"#dxdt$2+#dydt$2+#dzdt$2dtSince [H, G]uk= (H − λk)Guk,"[H, G]uk"2=%Guk, (H − λk)2Guk&'dds, t(ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 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 as1 Curves  scalar (time) in, vector (position) out  Scalar fields  vector (position) in, scalar outLimits for scalar fields What does it mean? limr→r0f(r) = Llimt→t0F(t) = Llimr→0x2− y2x2+ y2L(C) =!Cds =!ba"#dxdt$2+#dydt$2+#dzdt$2dtSince [H, G]uk= (H − λk)Guk,"[H, G]uk"2=%Guk, (H − λk)2Guk&'dds, t(ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 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 as1This kind of limit can depend on how you get where you are going.Example: What is limr→r0f(r) = Llimt→t0F(t) = Llimr→0x2− y2x2+ y2?L(C) =!Cds =!ba"#dxdt$2+#dydt$2+#dzdt$2dtSince [H, G]uk= (H − λk)Guk,"[H, G]uk"2=%Guk, (H − λk)2Guk&'dds, t(ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 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 as1Try this. First set y = 0 and let x → 0. You get the limit 1. But if you first set x = 0 and let y → 0. You get the limit -1. Weird. We don’t consider this function continuous.Second partial derivatives 1 Since fx(x,y) and fy(x,y) still depend on both variables, it makes sense to calculate fxx(x,y), fyy(x,y), fxy(x,y), fyx(x,y)Anatomy of the second partial derivative Isn’t there something funny about the x-y order? We write Fxy(x, y) or∂2F∂y∂xf(x + h) − f(x) = y · h + o(h)limr→0x2− y2x2+ y2?L(C) =!Cds =!ba"#dxdt$2+#dydt$2+#dzdt$2dtSince [H, G]uk= (H − λk)Guk,"[H, G]uk"2=%Guk, (H − λk)2Guk&'dds, t(ϕ = κnϕIn particular,#uk, [G, [H, G]] uk$ = 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·