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# Calculus III Exam 1 study guide MATH- 2326-004

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This 9 page Study Guide was uploaded by Chandon Lim on Thursday February 18, 2016. The Study Guide belongs to MATH- 2326-004 at University of Texas at Arlington taught by Christopher Kribs in Spring 2016. Since its upload, it has received 84 views. For similar materials see Calculus III in Applied Mathematics at University of Texas at Arlington.

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Date Created: 02/18/16

Calculus Study Guide Formula and Theorems 11.1 Vector Value Functions R(t) = <f(t),g(t),h(t)> = f(t)i + g(t)j +h(t)j The Limit of a vector function Let r be a defined as lim ????(????) = − [im???? ???? ???? ] lim[ ???? ???? + [l]mℎ ???? ]???? ( ) ????→???? ????→???? ????→???? ????→???? Continuity Vector function r is continuous if [????→???????? ???? )] = ???? ???? ) A vector function r is continuous on an interval if it is continuous at every number in I is continuous 11.2 Differentiation and integration of Vector Value Functions Derivative of a vector function ( ) r’(t) =???? = lim ???? ????+ℎ −????(????) ???????? ℎ→0 ℎ r’(t)=f’(t)i + g’(t)j+h’(t)k The process is the same for integration in that the integral of r found by integrating each function ???? ???? ???? ???? ∫ ???? ???? ???????? = [∫ ???? ???? ????????]???? + [∫ ???? ???? ????????]???? + [∫ ℎ ???? ????????]???? ( ) ???? ???? ???? ???? 11.3 Arc Length and Curvature Length can be found with this equation if the function is continuous throughout the length ???? ???? ???? = ∫ √ [????′(????)] + [????′(????)] + [ℎ′(????)] ???????? = ∫ |???? ???? |????????) ???? ???? If the curve is a smooth curve then the function can be rewritten as s(t)=∫???? |????′ ????|???????? ???? Curvature If Curvature is represented as κ and T represents the unit tangent vector then κ(s)=|???????? | = |T’(s)| ???????? The curvature of any point on a curve at a certain time is given by |???? ???? ???? ????(???? | κ(t) = 3 ????′(????) If the curve is twice differentiable then the curvature at a point (x,y) where y = f(x) is given by |????′(????)| |???? | κ(x)= 3= [1+(???? ) ]3/2 [1+ ???? ????)]22 Surfaces in 3-D spaces Traces: Example x =4???? + 9???? 2 2 Z=0 → x=4???? a parabola that opens along the axis Y=0 → x=9???? a parabola that opens along the x axis 2 2 X=0 →4???? + 9???? =0 a point at the origin X=1→4???? +9???? = 1 2 → ????2 ????2 1 1 1 + 1 = 1 Ellipse with major axis at ≤ ???? ≤ the 3D shape becomes an elliptic ( )2 ( )2 2 2 3 3 paraboloid ????2 Ex ???? −2 = 1 4 Traces ???? 2 Z=0 → ???? −2 = 1 hyperbola 4 Y=0 → ???? = 1 → ???? ± 1 Hyperbolic Cylinder X=0 has no solution Cases Traces ????2 ????2 ????2 2 + 2 + 2 = 1 if they are all positive → all ellipses →ellipsoid ???? ???? ???? 2 2 2 ???? + ???? − ???? = 1 if only one is negative → 1 ellipse and 2 hyperbola→ Hyperbola of 1 sheet ????2 ????2 ????2 ????2 ????2 ????2 2 − 2 − 2 = 1 if there are two negatives → 2 hyperbolas → Hyperbola of 2 sheets ???? ???? ???? 2 2 2 ???? ???? ???? ????2 + ????2 − ????2 = 0 there would be a point→ an hour glass shape ????2 ????2 + = ???? there would be an ellipse and 2 parabolas → elliptic paraboloid ????2 ????2 ????2 ????2 2 2 − 2 = ???? there would be a hyperbola and 2 paraboloid ???? ???? 12.1 Functions of Two Variables Let D={(x,y)|x,y ɛ R} be a subset of the xy-plane. A function of two variables is a rule that assigns to each ordered pair of real numbers (x,y) in D a unique number z. the set D is called the domain of f and the set of corresponding values of z is called the range of f. This rule is similar to a function of three variables. Level Curves The level curves of a function of two variables are the curves in the xy-plane with equations f(x,y) = k, where k is a constant in the range of f. 12.2 Limits and Continuity Ex 1 show that the limit does not exist 3???? − 2???? 2 lim ( ) (????,????)→(0,0) ???? + ???? 2 3???? −2????2 Let f(x,y)=(3 2 2 ) ???? +???? First, we approach toward (0,0) along the x-axis. 2 Then y=0 gives f(x,0) = 3???? =3for all x≠ so f(x,y)→3 as (x,y)→(0,0) along the x axis ????2 New lets try approaching along the y axis by setting x=0. −2???? 2 Then f(0,y) = =-2 for all y≠0, so f(x,y)→ -2 as (x,y)→(0,0) along the y-axis ????2 Since the two limits are different and they were along two different lines, the given limit can not exist. Continuity a point Let f be a function that is defined at all points (x,y) close to the point (a,b), Then f is continuous at that point if the lim ????(????,????) = ????(????,????) (????,????)→(????,????) Continuity on a region if R is a region in a plane then f is continuous on R if f is continuous at every point in R Properties of Continuous Function of Two Variables If f and g are continuous at a point then the following functions are also continuous 1. F ± g 2. Fg 3. Cf, c being a constant 4. f/g, if g(a,b) does not equal 0 Partials f(x,y)=cos(???? ), calculate???? ???????????? Ə???? 1+???? Ə???? Ə???? Ə???? = -sin(???? )* Ə ( ???? )=-( ???? )* 1 Ə???? 1+???? Ə???? 1+???? 1+???? 1+???? Ə???? ???? Ə ???? ???? ???? Ə???? = -sin1+????)*Ə???? 1+???? ) = −sin( 1+????) ∗ (1+???? )2 Suppose z is a differentiable function of x and y that is defined implicitly by Ə Ə???? ???? + ???? − ???? + 2???????? = 5 ???????????????? ???????????? ???????? Ə???? ???????????????????????????????????????????????????????? ????????????ℎ ???????????????????????????? ???????? ????: Ə???? 2 3 2 Ə???? Ə???? (???? + ???? − ???? + 2???????? ) = Ə???? (5) Ə???? Ə???? 2x-Ə????+ 2????(2???? Ə????)=0 Ə???? ( ) Ə???? 4???????? − 1 + 2???? = 0 Ə???? 2???? Ə????= 1−4???????? Higher Order Derivatives 2 2 2 2 Ə ???? Ə Ə???? Ə ???? Ə Ə???? Ə ???? Ə Ə???? Ə ???? Ə Ə????2 = Ə???? Ə???? ),Ə????Ə???? = Ə???? Ə???? ),Ə????2 = Ə???? Ə???? ) ,???????????? Ə???? 2(Ə????) Differentials Increments 2 2 Ex z=f(x,y(=???? + 3???????? − ???? Find dz and compare it to Δz when x changes to 2.08 and y changes to 2.95 Ə???? Ə???? Dz= + = 2???? + 3???? ???????? + 3???? − 2???? ???????? Ə???? Ə???? Setting x=2, dx =Δx = .08, y= 3, and dy=Δy=-.05, we get dz =[13]*.08+[3(2)-2(3)](-.05)= 1.04 Then the increment of Z is Δz =f(2.08, 295) –f(2.3) = [2.08 + 3 2.08 2.95 − 2.95 ]-[2 +3(2)3 ] = 1.0319 Chain Rule 3 3 3 ???? 2 −???? 2 U=???? ???? + ???? ???? ???? = ???????????? ,???? = ???????? ???? ,???? = ???? ???????????????? ???? ( ) When r=4, s=1,t=0 Ə???? Ə????Ə???? Ə????Ə???? Ə????Ə???? = + + Ə???? Ə???? Ə???? Ə???? Ə???? Ə???? Ə???? = 3???? ????(???????? )+(???? + 2???????? ) 2????????????−????)+ (3???? ???? )(???? sin(????) Ə???? When r=4, s=1 and t=0, x =4, y=4,zƏ????,(192)(4)(64)(8))=1280 Implicit differentiation Ə???? Ə???? Ə???? Ə???? = + =0 Ə???? Ə???? Ə???? Ə???? Ə???? ???????? Ə???? ???????? ???????? = − Ə????= ???????? ???????? ???? ≠ 0 Ə???? Gradient f(x,y,z)=xsin(zy) find the gradient of f and the directional derivative of f at (2,5,0) in the direction of v=i+4j-k The gradient is ∇f(x,y,z) = <???? ????,????,???? ,????????????,????,???? ,???? ????,????,???? > ) =<sin(yz),xzcos(yz),xycos(yz)> At (2,5,0) we have ∇f(2,5,0)=<0,0,10> The unit vector in the direction of v=i+4-k is U= 1 ???? + 4 − 1 ???? √18 √18 √18 Therefore the equations outputs ???????????? 2,5,0 = ????f(2,5,0) ∙ U 1 4 1 5√2 10k∙ ( ???? + − ????) = − √18 √18 √18 3 Tangent Planes and Normal Lines Find the equations for the normal line and tangent plane at point (-2,1,-5) 2 2 ???? 2 ???? 4 + ???? + 25 = 3 The function is the level surface (with k=3) of the function 2 2 F(x,y,z)=???? + ???? + ???? 4 25 This becomes ???? ????,????,???? = ???? ???? 2 ????????????,????,???? = 2???? 2???? ????????????,????,???? = 25 ???? −2,1,−5 = −1 ???? ????????−2,1,−5 = 2 ????????−2,1,−5 = −2/5 The Tangent plane at (-2,1-5) -1(x+2)+2(y-1)-(2/5)(z+5)=0 This simplifies to 5x-10y+2z+30=0 To find the normal line ????+2 ????−1 5 −1 = 2 = − 2???? + 5) Extrema of multivariable functions Find the distance that is shortest from a point (1,0,-3) to the plane x+2y+z=9 Using the distance formula we get ???? = √ (???? − 1) + ???? + (???? + 3) 2 But if (x,y,z) lies on the plane x+2y+z=9 then z = 9-x-2y and so we √av???? − 1)2 + ???? + ???? + 3 )2 We can minimize d by minimizing the simpler expression 2 2 2 2 ???? = ???? ????,???? = (???? − 1) + ???? +(12 − x − 2y) By solving the equation we get ???? =2(x-1)-2(12-x-2y)=4x+4y-26=0 ???? ????????= 2???? − 4 12 − ???? − 2???? = 4???? + 10???? − 48 = 0 We find that the only critical point ,1,(1) 6 3 Since ???????? = 4,???????????? = 4 ???????????? ???????????? = 10 we have D(x,y)=???????????? ????????(???? ) ????????24 > 0 ???????????? ???? ????????> 17 11 0,???????? ???????? ????ℎ???? ???????????????????????? ???????????????????????????????????????? ???????????????? ???? ℎ???????? ???? ???????????????????? ???????????????????????????? ???????? ( , 6 3 Intuitively, we can observe that the local minimum is actually an absolute minimum because there has 17 11 to be a point in the plane that is the closest to (1,0,-3)and y = 3 then 2 2 2 ???? = √ (???? − 1) + ???? + 12 − ???? − 2???? ) ???? 11 2 112 11 2 = ( ) + + ( ) 6 3 6 11 = ???? = √ ∙ 6 11 The smallest distance from(1,0,-3) to the plane x+2y+z=√ is 6 ∙ 6 Relative Extrema of a function of Two Variables Let f be a function in a region R that have points (a,b). Then f has a relative maximum at (a,b) if f(x,y) ≤ f(a,b) for all points (x,y) in an open disk containing (a,b). The number f(a,b) is called a relative maximum value. By the same token f has a relative minimum at (a,b) with relative minimum value f(a,b) if f(x,y) ≥ f(a,b) for all points (x,y) in an open disk containing (a,b) A saddle point is a critical point that does not give an extremum The second derivative test Suppose that f has continuous second order partial derivatives on an open region containing a critical point (a,b) of f let D(x,y)-???? (????,???? ???? (????,???? − ???? 2(????,???? ) ???????? ???????? ???????? a) if D(a,b)>0 and ???? ????????(????,???? < 0 ????ℎ???????? ???? ????,???? ???????? ???? ???????????????????????????????? ???????????????????????????? b) if D(a,b)>0 and ???? ????????(????,???? > 0 ????ℎ???????? ???? ????,???? ???????? ???? ???????????????????????????????? ???????????????????????????? ???????????????????? c) if D(a,b)<0 then (a,b,f(a,b)) is a saddle point d) if D(a,v)=0 then the test is inconclusive Lagrange Multipliers Find the extremum of the function f(x,y)=3???? + 4 on the circle ???? + ???? = 1 2 Using Lagrange multipliers because we are asked to find extremum and are given a constraint 2 2 g(x,y)= ???? + ???? = 1 We can try and solve the equations ∇f = λ∇g and g(x, y) = 1 Which can be described as fx= λgx fy= λgy g(x, y) = 1 2 2 or as 6x=2xλ, 8y=2yλ, ???? + ???? = 1 from the first equation we have 0 or λ =3. If x= 0, then the third equations gives you y=± 1. Therefore f has possible extreme values at the points (0,1),(0,-1),(1,0), and (-1,0). Evaluating f at these four points, we find that F(0,1)=4 F(0,-1)=3 F(-1,0)=3 2 2 There fore the maximum value of f on the circle ???? + ???? = 1 is f(0,±1)=4 And the minimum value is f(±1,0)=3

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