Dr. Nestler - Math 11 - Some Definitions and Theorems from Calculus 1 and 2. lim

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1 Dr. Nestler - Math 11 - Some Definitions and Theorems from Calculus 1 and 2 Definition. A function 0 is continuous at a number + if lim 0ÐBÑ œ 0Ð+Ñ. Extreme Value Theorem. A continuous function on a closed interval has an absolute maximum value and an absolute minimum value on that interval. Definition. The derivative of a function 0 at a number + is the number BÄ+ 0Ð+Ñœ lim 2Ä! 0Ð+ 2Ñ 0Ð+Ñ 0ÐBÑ 0Ð+Ñ 2 œ lim BÄ+ B + if this limit exists. Theorem & Definition. Suose 0 is continuous on Ò+ß,Ó. Divide Ò+ß,Ó into 8 subintervals ÒB ß B Ó of equal length? B, and choose a number A in ÒB ß B Ó for each 3 œ "ß á ß 8. 3 " " 3 Then lim 8 8Ä 3œ" 0ÐA Ñ? Bexists, and this number is called the definite integral of 0 from + 3 to,, and is denoted by, + 0ÐBÑ.B. B Fundamental Theorem of Calculus, Part I. If 0 is continuous on Ò+ß,Ó, then JÐBÑ œ + 0Ð>Ñ.> is an antiderivative of 0, meaning J ÐBÑ œ 0ÐBÑ for all Bin Ò+ß,Ó. Fundamental Theorem of Calculus, Part II. If J is any antiderivative of 0 on Ò+ß,Ó, then +, 0ÐBÑ.B œ JÐ,Ñ JÐ+Ñ. Mean Value Theorem for Differentiable Functions. If 0 is continuous on Ò+ß,Óand differentiable 0Ð,Ñ 0Ð+Ñ, + on Ð+ß,Ñ, then there exists at least one - in Ð+ß,Ñ such that œ 0 Ð-Ñ. That is, 0Ð,Ñ 0Ð+Ñ œ 0 Ð-ÑÐ, +ÑÞ " " Inverse Function Theorem. If C œ 0ÐBÑ is differentiable ith inverse 0, and 0 Ð0 Ð+ÑÑ Á!, " then 0 is differentiable at +, and " œ.0.0.b Bœ+ ".B Bœ0 " Ð+Ñ Shorthand:.B ".C œ.c.b

2 Dr. Nestler - Math 11 - Essential Trigonometric and Hyerbolic Identities Fundamental trigonometric identities: " " sin B cos B " sin B cos B cos B sin B tan B csc Bœ sec Bœ tan Bœ cot Bœ œ sin B cos Bœ" tan B "œsec B " cot Bœcsc B sinð BÑ œ sin B cosð BÑ œ cosðbñ Double-angle identities: sin B œ sin B cos B cos B œ cos B sin B œ " sin B œ cos B " Half-angle identities: " cos B " cos 2B 2 Bœ 2 sin Bœ cos Trigonometric derivatives:....b.b.b Ðsin BÑ œ cos B Ðcos BÑ œ sin B Ðtan BÑ œ sec B....B.B.B Ðsec BÑœsec Btan B Ðcsc BÑœ csc Btan B Ðcot BÑœ csc B Hyerbolic functions definitions and fundamental identities: B B B B sinh / / / / B cosh B cosh Bœ sinh Bœ tanh Bœ " " cosh B " cosh B sinh B sinh B tanh B sech Bœ csch Bœ coth Bœ œ cosh B sinh B œ " sinhð BÑ œ sinh B coshð BÑ œ coshðbñ Hyerbolic derivatives:....b.b.b Ðsinh BÑœcosh B cosh B œsinh B Ðtanh BÑœsech B....B.B.B Ðsech BÑ œ sech Btanh B csch B œ csch Bcoth B coth B œ csch B

3 Dr. Nestler - Math Surfaces A surface is a to-dimensional grah of an equation in BCD,,. 1. shere: 2. lane: 3. cylinder: a surface consisting of all lines ( rulings) that are arallel to a fixed line and assing through a fixed lane curve ( directrix) Examle: C D œ "' Bis arbitrary: cross-sections erendicular to the B-axis are circles of radius 4. Thm. The grah of an eqn. in 2 of the 3 variables BCD,, is a cylinder hose rulings are arallel to the axis of the missing variable. Examle: DœB Cross-sections erendicular to the C-axis are arabolas. 4. Excet for degenerate cases, a quadric surface is a grah of a quadratic equation in 3 variables EB FC GD HBC ICD JBD KB LC MD N œ! hich by translation and rotation can be ut into standard form EB FC GD N œ! or EB FC MD œ!. Traces in lanes arallel to the coordinate lanes are conic sections. Some degenerate examles: B œ!, B œ +, B C D œ!

4 Examle: B Examle: B C D +, - œ " Bœ! : trace in CD-lane is ellise Cœ! : trace in BD-lane is ellise Dœ! : trace in BC-lane is ellise The surface is called an ellisoid. C D +, - œ " Dœ! : trace in BC-lane is ellise Bœ! : trace in CD-lane is hyerbola Cœ! : trace in BD-lane is hyerbola C +, - Let Dœ5: B œ " 5 B C + 5 ", 5 " - - œ " Cross-sections arallel to the BC-lane are ellises. As l5l Ä, ellises get larger. The surface is called a hyerboloid of one sheet or an ellitic hyerboloid. The D-axis, corresonding to the minus sign in the equation, is the axis of the hyerboloid. Examle: B C D +, - œ " This is a hyerboloid of one sheet ith the C-axis as its axis. B C D Similarly, the grah of +, - œ " is a hyerboloid of one sheet ith the B-axis as its axis.

5 B C D +, - Examle: œ " Examle: B Dœ! : no trace in BC-lane Bœ! : trace in CD-lane is hyerbola Cœ! : trace in BD-lane is hyerbola The surface is called a hyerboloid of to sheets. The D-axis, corresonding to the lus sign in the equation, is the axis of the hyerboloid. C D +, - œ " The axis of this hyerboloid of to sheets is the B-axis, and there is no trace in the CD-lane. (Try letting B œ!.) Examle: B C D +, - œ! Dœ5: Cross-sections arallel to the BC-lane are ellises. Dœ - B +, C The surface is a ( double-naed) ellitic cone ith the D-axis as its axis. Examles: The grah of B C D +, - œ! is a cone ith the C-axis as its axis. The grah of B C D +, - œ! is a cone ith the B-axis as its axis.

6 Examle: B +, D œ! C Bœ! : trace is arabola Dœ C, oening u Cœ! : trace is arabola Dœ B + oening u Dœ5! : Cross-sections arallel to BC-lane are ellises. The surface is called an ellitic araboloid. The D-axis, corresonding to the minus sign, is the axis of the araboloid. B C +, Examle: D œ! Bœ!: trace is arabola oening u Cœ!: trace is arabola oening don Dœ5Á! : Cross-sections arallel to BC-lane are hyerbolas. The surface is called a hyerbolic araboloid. [Can do , 11-28, (classify only); Ch 12 T/F Quiz all 1-22; Ch 12 Revie 1-7, 11-13, 15-22, 24, 25, 27-36]

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9 Dr. Nestler - Math 11 - Chater 13 - Vector-valued Functions and Sace Curves Introduction Defns. Let H be a set of real numbers. A vector-valued function < ith domain H assigns to each number > H a vector <Ð>Ñ. Write < ÀH ÄZ, $ < Ð>Ñ œ 0Ð>Ñ3 s 1Ð>Ñ4 s 2Ð>Ñ5 s œ Ø0Ð>Ñß 1Ð>Ñß 2Ð>ÑÙ. For > H, e dra <Ð>Ñas a osition vector ST, and call T the endoint of <Ð>Ñ. Examle: Find the domain of <Ð>Ñ œ ln >ß " & >ß. > $ Defns. A (sace) curve determined by < Ð>Ñ œ Ø0Ð>Ñß 1Ð>Ñß 2Ð>ÑÙ ith 0, 1 and 2 continuous on an interval Mis a set GœÖÐBßCßDÑ $ ÀBœ0Ð>Ñ, Cœ1Ð>Ñ, Dœ2Ð>Ñ, > M. The equations B œ 0Ð>Ñ, C œ 1Ð>Ñ and D œ 2Ð>Ñ are arametric equations for G. The orientation of the curve is the direction determined by increasing values of the arameter >. The curve is smooth if 0, 1 and 2 are continuous and not all zero excet ossibly at an endoint.

10 Examle: Describe and sketch the curve G determined by < Ð>Ñ œ Ø sin >ß & cos >ß $>Ù, >!. Thm. (Math 8, section 10.2,. 693) The arc length of a smooth lane curve G given by arametric equations B œ 0Ð>Ñ, C œ 1Ð>Ñ, > Ò+ß,Ó is P œ Ð0 Ð>ÑÑ Ð1 Ð>ÑÑ.>. +, Thm. If a smooth sace curve G given by < Ð>Ñ œ Ø0Ð>Ñß 1Ð>Ñß 2Ð>ÑÙ is traced exactly once as > increases from + to,, then the length of G is P œ Ð0 Ð>ÑÑ Ð1 Ð>ÑÑ Ð2 Ð>ÑÑ.>, +, and this value is indeendent of the choice of arametric equations for G. This result can be extended to ieceise-smooth curves.

11 Examle: Find the length of the curve defined by arametric equations B œ " >, C œ %>, D œ $ >, > Ò!ß Ó. Examle: Describe and sketch the curve given by $ <Ð>ÑœØ>ß>ß>Ù> Ò!ß%Ó,. [Can do , 2, 7-13, 21-26, 29; ]

12 Dr. Nestler - Math Curvature (lecture 1 of 2) For the rest of Chater 13, e assume that the curve G is smooth and simle (it does not intersect itself excet ossibly at its endoints). Definition. Suose the curve G is given by the vector-valued function < Ð>Ñ œ Ø0Ð>Ñß 1Ð>Ñß 2Ð>ÑÙ, > Ò+ß,Ó. The arc length function for G is > > =Ð>Ñ œ Ð0 Ð?ÑÑ Ð1 Ð?ÑÑ Ð2 Ð?ÑÑ.? œ ll< Ð?Ñll.?. + + Think of =Ð>Ñ as "the length so far.".= By the Fundamental Theorem of Calculus, =Ð>Ñ is differentiable and œ ll< Ð>Ñll. Definition. The unit tangent vector is given by " X Ð>Ñ œ < Ð>Ñ at oints here < Ð>Ñ Á!. ll< Ð>Ñll.> X indicates the orientation of the curve, and llx Ð>Ñll œ " for all >. Definition. The curvature of G at a oint is a nonnegative scalar given by the function OÐ=Ñ œ.x.=. That is, curvature is the magnitude of the rate of change of the unit tangent vector ith resect to arc length. This definition uses arametrization by arc length, hich is a natural, intrinsic feature of the curve. Therefore curvature is defined to be indeendent of choice of arametrization. Suose that G has a smooth arametrization <Ð>Ñ here the arameter > need not reresent arc.= length. Since.> œ ll< Ð>Ñll! for all >, =Ð>Ñ is an increasing function, and therefore one-toone. Since =Ð>Ñ is one-to-one ith a nonzero derivative, e can conclude by the Inverse.> " Function Theorem (. 404) that there exists a differentiable inverse function >Ð=Ñ, and œ. Thus OÐ>Ñ œ.x.= œ.=.=.>

13 Examle: Calculate the curvature of a circle of radius +! at an arbitrary oint. Put its center at Ð!ß!Ñ and use the arametrization < Ð>Ñ œ Ø+ cos >ß + sin >Ùß > Ò!ß 1Ó. < Ð>Ñ œ Ø + sin >ß + cos >Ù ll< Ð>Ñll œ XÐ>Ñ œ XÐ>Ñœ <Ð>Ñ ll< Ð>Ñll œ llx Ð>Ñll œ So OÐ>Ñ œ Examle: Calculate the curvature of a line at an arbitrary oint. Let G be a line through a oint ÐB! ßC! ßD! Ñand arallel to a vector Ø+ß,ß-Ù. Then G BœB! +> is given by arametric equations CœC!,> DœD ->! So G is traced by the vector-valued function < Ð>Ñ œ ØB +>ß C,>ß D ->Ù. Its derivative is the constant vector-valued function magnitude ll< Ð>Ñll œ +, -.!!! <Ð>ÑœØ+ß,ß-Ù, hich has constant The unit tangent vector at any oint on the line is therefore constant: <Ð>Ñ XÐ>Ñ œ œ ß ß ll< Ð>Ñll +, - +, - +, - +, -. Thus X llx Ð>Ñll Ð>Ñ œ!, and so the curvature at an arbitrary oint on the line is OÐ>Ñ œ œ ll< Ð>Ñll

14 Theorem. OÐ>Ñ œ ll< Ð>Ñ < Ð>Ñll ll< Ð>Ñll$ Proof: By definition, XÐ>Ñ œ <Ð>Ñ ll< Ð>Ñll, so < œ ll<.= llx œ X. Differentiate this equation using the roduct rule:.> < œ.=.= X X. Take the cross-roduct:.>.> < <.=. =.=.=. =.= œ X X X œ ÐX XÑ ÐX XÑ œ.>.>.>.>.>.>.=.> ÐX X Ñ Since llx Ð>Ñll is constant for all >, X ¼ X so ll<.= < ll œ llx X ll œ œ So llx ll œ.> llx ll llx ll sin.= 1.>.=.> llx ll ll< < ll ll< < ll œ.= ll< ll.> llx ll ll< ll ll< < ll ll< ll$ Finally, Oœ œ. Examle: Find the curvature of the tisted cubic $ < Ð>Ñ œ Ø>ß > ß > Ù. < Ð>Ñ œ Ø"ß >ß $> Ù, so < Ð>Ñ œ Ø!ß ß '>Ù. < < œ ll < < ll œ OÐ>Ñ œ ll< < ll ll< ll$ œ [Can do , 33a]

15 Dr. Nestler - Math Curvature (lecture 2 of 2) Curvature measures ho quickly a smooth, simle curve is bending. We defined curvature by O œ ll.x.= ll, the magnitude of the rate of change of the unit tangent vector ith resect to arc length. At each oint on a smooth, simle curve, there is a ell-defined nonnegative curvature value. We roved that if a curve is given by a vector-valued function <Ð>Ñ, then OÐ>Ñ œ llx Ð>Ñll ll< Ð>Ñll œ ll< Ð>Ñ < Ð>Ñll ll< Ð>Ñll$ Examle: Find the curvature of the circular helix defined by Bœ+ cos >, Cœ+ sin >, Dœ,> Ð+!Ñ at an arbitrary oint. < Ð>Ñ œ Ø+ cos >ß + sin >ß,>Ù < Ð>Ñ œ Ø + sin >ß + cos >ß,Ù < Ð>Ñ œ Ø + cos >ß + sin >ß!Ù < < œ ll< < ll œ ll< ll œ OÐ>Ñ œ

16 Secial case (42): Suose that G is a lane curve ith arametric equations B œ 0Ð>Ñ, C œ 1Ð>Ñ. Then G is defined in $ by vector function < Ð>Ñ œ Ø0Ð>Ñß 1Ð>Ñß!Ù <Ð>ÑœØ0ß1ß!Ù < Ð>Ñ œ Ø0 ß1 ß!Ù < < œ Oœ To hel remember this formula, notice that the numerator is the absolute value of this 0 1 determinant: 0 1 Examle: Find the curvature of the curve BÐ>Ñ œ > sin >, CÐ>Ñ œ " cos > at the oint 1 TÐ "ß"Ñ. T corresonds to the value > œ BÐ>Ñœ BÐ>Ñœ CÐ>Ñœ CÐ>Ñœ 1 OÐ Ñ œ An even more secial case: Suose that G is a lane curve ith equation C œ 1ÐBÑ. Then < ÐBÑ œ ØBß 1ÐBÑß!Ù ith arameter B, so OÐBÑ œ Examle: Find the curvature of the arabola C œ B at Ð!ß!Ñ, Ð"ß "Ñ and Ðß %Ñ.. C œb, C œêoðbñœ

17 Examle: Find the oints on the curve Cœ/ B at hich the curvature is a maximum. B B C œ / C œ / lc l " ÐC Ñ OÐBÑ œ œ To maximize the function OÐBÑ, e find its critical OÐBÑœ $ oints using single-variable calculus: B $ B B $ B " B B $ œ! Ð" / Ñ Ð / Ñ / Ð" / Ñ Ð / Ñ Ð" / Ñ B B $B / Ð" / Ñ $/ " / B & œ! B / / " / $B B & œ! " " B œ ln œ ln œ ln So the only critical oint is B œ B B / Ð" / Ñ œ! B / œ " B " / œ Using the 1st or 2nd derivative test, e can verify that O has a maximum at this critical oint. [Can do 27-32, 38, 39, 43-46]

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19 Dr. Nestler - Math Differentiable Functions [art 1] Facts from Calculus 1: If a real-valued function (1) 0 is continuous at +, 0ÐBÑ is differentiable at a number +, then (2) the curve C œ 0ÐBÑ has a nonvertical tangent line at Ð+ß 0Ð+ÑÑ, and (3) a small change (increment) in 0 near + can be aroximated using a differential. Math 11 Goal: Define "differentiable" so that if a function 0ÐBßCÑ is differentiable at Ð+ß,Ñ, then (1) 0 is continuous at Ð+ß,Ñ, (2) the surface D œ0ðbßcñhas a nonvertical tangent lane at Ð+ß,ß0Ð+ß,ÑÑ, and (3) a small change (increment) in 0 near Ð+ß,Ñcan be aroximated using a differential. Consider the case Cœ0ÐBÑ. If? Bis an increment of B, let? Cœ0Ð+? BÑ 0Ð+Ñ. If 0 is differentiable at +, then 0 Ð+Ñ œ lim exists.? C? B C BÄ! B Given an increment? B, let % œ 0 Ð+Ñ, so % is a function of? Band lim % œ!. Rerite as:? C? B œ0ð+ñ %? C œ Ð0 Ð+Ñ % Ñ? B BÄ! Thus? C œ Ð0 Ð+Ñ % Ñ? B here % is a function of? B and % Ä! as? B Ä!. % It can be shon that, conversely, the existence of such a number 0Ð+Ñand function is equivalent to the definition of differentiability for 0ÐBÑ. Definitions. Let D œ 0ÐBßCÑ. An increment of 0 is? D œ 0ÐB? BßC? CÑ 0ÐBßCÑ. The function 0 is differentiable at a oint Ð+ß,Ñ in its domain if (1) the artial derivatives 0 Ð+ß,Ñ and 0 Ð+ß,Ñ exist, and B (2) an increment?d at Ð+ß,Ñcan be exressed in the form C? Dœ 0 Ð+ß,Ñ %? B 0Ð+ß,Ñ %? Chere % and % are functions of Band B " C "? C such that %, % Ä! as? Band? C aroach 0. "

20 Examle: Sho that, according to the definition, the function each oint Ð+ß,Ñ in the lane. 0ÐBßCÑ œ B C is differentiable at Theorem. If 0ÐBßCÑ is differentiable at +ß,, then 0 is continuous at Ð+ß,Ñ. Proof: Since 0 is differentiable, given increments? Band? C, an increment of 0 at Ð+ß,Ñ can be ritten? D œ 0Ð+? Bß,? CÑ 0Ð+ß,Ñ Let Bœ+? B, Cœ,? C. Then œ 0 Ð+ß,Ñ %? B 0 Ð+ß,Ñ %? C. B " C? D œ 0ÐBß CÑ 0Ð+ß,Ñ œ 0 Ð+ß,Ñ % ÐB +Ñ 0 Ð+ß,Ñ % ÐC,Ñ B " C No let ÐBß CÑ Ä Ð+ß,Ñ, so? Bß? C Ä!: lim 0ÐBß CÑ 0Ð+ß,Ñ œ! ÐBß CÑÄÐ+ß,Ñ So lim ÐBß CÑÄÐ+ß,Ñ 0ÐBßCÑ œ 0Ð+ß,Ñ, as desired.

21 Theorem. If 0 and 0 exist in an oen disk containing Ð+ß,Ñand are continuous at Ð+ß,Ñ, then 0 B is differentiable at +ß,. C Proof: Next lecture. See also Aendix F. Examle: Prove that 0ÐBßCÑ œ BC C B is differentiable on its domain. if, Examle: We can sho that the function 0ÐBßCÑ œ " B C! has first artial! otherise derivatives at the origin, but the function is not continuous there, and so it is not differentiable there. The next examle shos that the converse of the first theorem above is false. Examle: We can sho that the function Dœ0ÐBßCÑœ B C is continuous at the origin, but the function does not have first artial derivatives there, and so it is not differentiable there. [Can do "Exlain hy the function is differentiable at the given oint," 43, 44, 46]

22 Dr. Nestler - Math Differentiable Functions (Part 2) Thm. Let 0ÐBßCÑ be a function of to variables. If 0 and 0 exist in an oen disk containing a oint Ð+ß,Ñ and they are continuous at +ß,, then 0 is differentiable at +ß,. Proof: An increment of D œ 0ÐBß CÑ at Ð+ß,Ñ is? D œ 0Ð+? Bß,? CÑ 0Ð+ß,Ñ. We must sho that e can rite this in the form? Dœ 0 Ð+ß,Ñ %? B 0Ð+ß,Ñ %? C B C B " C here the functions %", % Ä! as? B,? CÄ!. We can rite the increment as a sum (*)? D œ 0Ð+? Bß,? CÑ 0Ð+ß,? CÑ 0Ð+ß,? CÑ 0Ð+ß,Ñ For simlicity, assume? B,? C!. Let 1B œ0ðbß,? CÑ, hich is the function 0ith the C-comonent of its domain restricted to the value,? C. For small enough? Band? C, the single-variable function 1ÐBÑ is differentiable on the interval Ò+ß +? BÓ, and 1 ÐBÑ œ 0 ÐBß,? CÑ. (The size of the increments? Band? C is determined by the size of the disk in hich the first artials of 0 exist.) By the Mean Value Theorem,? Ð+ß+ 1Ð+? BÑ 1Ð+Ñ œ 1 Ð?Ñ? B? BÑ. Thus the first art of the sum (*) is 0Ð+? Bß,? CÑ 0Ð+ß,? CÑ œ 0BÐ?ß,? CÑ? B B for some number Similarly, the function 2ÐCÑ œ 0Ð+ß CÑ is differentiable on Ò,ß,? CÓ for small enough increments, and 2 ÐCÑ œ 0C Ð+ß CÑ. By the Mean Value Theorem, 2Ð,? CÑ 2Ð,Ñ œ 2 Ð@Ñ? C for some Ð,ß,? CÑ. Thus the second art of the sum (*) above is 0Ð+ß, CÑ 0Ð+ß,Ñ œ 0 C?? C

23 Rerite (*) using (1) and (2):? Dœ0 Ð?ß,? CÑ? B 0Ð+ß@Ñ? C B C œ 0 Ð+ß,Ñ 0 Ð?ß,? CÑ 0 Ð+ß,Ñ? B 0 Ð+ß,Ñ 0 0 Ð+ß,Ñ? C. B B B C C C As? BÄ! and? CÄ!,?Ä+ and so % Ä! and % Ä! since 0, 0 are continuous at Ð+ß,Ñ. Thus 0 is differentiable at Ð+ß,Ñ. " B C Notes: 1. There is a stronger result that says if both first artial derivatives are defined in an oen disk containing the oint and if at least one of them is continuous at the oint, then the function must be differentiable at the oint. 2. The converse of the theorem is false. See the class homeage for an examle of a differentiable function ith discontinuous first artial derivatives.

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25 Math 11 - Dr. Nestler - Revie of Definite Integrals from Math 7 (Calculus 1) Definitions. Suose a function 0 is continuous on a closed interval Ò+ß,Ó. Divide Ò+ß,Óinto 8 subintervals ÒB ß B Ó of equal length? B, and choose a number ( samle oint) A in 3 " 3 3 ÒB ßBÓfor each 3œ"ßáß8. The exression 0ÐAÑ? B is called a Riemann sum 3 " 3 3 3œ" for 0 on Ò+ß,Ó. If it exists, the number lim 0ÐA Ñ? Bis called the definite integral +, 8 8Ä 3œ" of 0 on Ò+ß,Ó, denoted by 0ÐBÑ.B, and e say 0 is integrable on Ò+ß,Ó. 8 3 Theorem. If 0 is continuous on Ò+ß,Ó, then 0 is integrable on Ò+ß,Ó., If 0 is continuous and nonnegative on Ò+ß,Ó, then + 0ÐBÑ.B equals the area under the grah of Cœ0ÐBÑ from + to,. Fundamental Theorem of Calculus: Each continuous function 0 on a closed interval Ò+ß,Ó has an, antiderivative J on Ò+ß,Ó, and + 0ÐBÑ.B œ J Ð,Ñ J Ð+Ñ.

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31 Dr. Nestler - Math Indeendence Of Path $ Definitions. A region H in ( ) is oen if for each oint : H, there exists a disk (ball) centered at : contained in H. An oen region is connected if each air of oints in H can be joined by a ieceise-smooth curve lying in H. From no on, the domain H of a vector field is an oen, connected region. Theorem. If J is a continuous vector field on a region H, then line integrals of J in H are indeendent of ath if and only if J is conservative. Proof: ( É ) This imlication follos from the Fundamental Theorem of Line Integrals: if J œ f0 for a scalar function 0, then f0.< œ 0Ð< Ð,ÑÑ 0Ð< Ð+ÑÑ. G ( Ê ) Suose J ÐBß CÑ œ ØT ÐBß CÑß UÐBß CÑÙ, and line integrals of J in H are indeendent of ÐBß CÑ ath. Let ÐB ß C Ñ be a fixed oint in H. Define a scalar function by 0ÐBß CÑ œ J.<!! for ÐBß CÑ H. We ill sho that f0 œ J. ÐB ß C Ñ Given a oint ÐBß CÑ H, H contains a disk centered at ÐBß CÑ, since H is oen. Choose a oint ÐB" ß CÑ in the disk ith B" Á B, and let G consist of a ath G" from ÐB! ß C! Ñ to ÐB" ß CÑ folloed by the horizontal line segment from to. Then G ÐB ßCÑ ÐBßCÑ "!!

32 0ÐBßCÑœ J.< J.< ÐB œ J.< J " ßCÑ.< Differentiate both sides ith resect to B: G G ÐB ß C Ñ G "!! œ J.< J.< `0 ` ÐB" ßCÑ ` `B `B ÐB ß C Ñ `B `0 `B!! œ J.< ` `B On G, C is constant so.c is!. Rerite the revious equation using the comonent form of the line integral: `0 `B œ T.B U.C ` `B G ` œ TÐ>ßCÑ.> œ TÐBßCÑby the Fundamental Theorem of Calculus. `B B B " G G Similarly, e could sho `0 `C œuðbßcñ, using a vertical line segment instead of a horizontal one. Thus J œ ØT ß UÙ œ Ø0Bß 0CÙ œ f0, as desired.

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36 Dr. Nestler - Math Proof of a secial case of the Stokes Theorem Suose the surface W is the grah of D œ 1ÐBßCÑ ith ÐBßCÑ in a lane region H, and 1 has continuous 2nd artials. If W is oriented uard, then the ositive orientation of G corresonds to the ositive orientation of its rojection G", the boundary of H. Let ØBÐ>Ñß CÐ>ÑÙ, > Ò+ß,Ó be a arametrization of G". Then BÐ>Ñß CÐ>Ñß 1 BÐ>Ñß CÐ>Ñ, > Ò+ß,Óis a arametrization of G. Let J œøtßußvùbe a vector field hose comonent functions have continuous 1st artials on a region containing W. Then, (1) J.< œ J Ð< Ð>ÑÑ < Ð>Ñ.> G +,.B `D.B `D +.>.> `B.> `C.> (2) œ.c T U V.C.>, `D.B `D + `B.> `C.> (3) œ T V U V.C.> (4) œ `D T V.B `D U V.C G " `B ` (5) œ `D U V ` `D T V `B `C `C `B.E H (6) œ U U V V V H `D ` D `D `D `D B D `B `B`C B`C D `B `C `D ` D `D `D `D T T V V V.E C D `C `C`B C `B D `C `B `D `D (7) œ ÐVC UDÑ `B ÐTD VBÑ `C ÐUB TCÑ.E H (8) œ ØV U ßT V ßU T Ù Ø `Bß `Cß"Ù.E H `C C D D B B C (9) œ curl J 8.W. W `D `D

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38 Dr. Nestler - Math Conservative Vector Fields Theorem. Given a vector field J ith continuous first artial derivatives on a simly connected domain H, the folloing are equivalent: (1) J is conservative. (2) Line integrals of J are indeendent of ath. (3) G J.<œ! for all closed curves G. (4) The cross artial derivatives of J are equal. (5) J is irrotational. Proof: We roved (1) Í (2) Í (3) and (1) Í (4). (1) Ê (5): roved on in (5) Ê (3): This is a corollary of the Stokes Theorem. If G is a simle closed smooth curve in H, then an advanced theorem states that G is the boundary of a smooth oriented surface W in H. By the Stokes Theorem, J.< œ Ðcurl JÑ.W œ G!.W œ!. W W Note: Imlications (4) Ê (1) and (5) Ê (1) require the simle connectivity of the domain.

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