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### c16s3

Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Word Count: 2764

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Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals 1. C appears to be a smooth curve, and since Theorem 2 says that the value of C f is continuous, we know f is differentiable. Then f dr is simply the difference of the values of f at the terminal and initial points of C . From the graph, this is 50 10=40. 2. C is represented by the vector function r(t)=(t +1)...

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Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals 1. C appears to be a smooth curve, and since Theorem 2 says that the value of C f is continuous, we know f is differentiable. Then f dr is simply the difference of the values of f at the terminal and initial points of C . From the graph, this is 50 10=40. 2. C is represented by the vector function r(t)=(t +1) i+(t +t) j , 0 t Since 3t +1 0 , we have r (t) 0 , thus C is a smooth curve. differentiable, so by Theorem 2 we have C 2 / 2 3 1 , so r (t)=2t i+(3t +1) j . / 2 f is continuous, and hence f is f dr= f (r(1)) f (r(0))= f (2, 2) f (1, 0)=9 3=6 . 2 3. (6x+5y)/ y=5= (5x+4y)/ x and the domain of F is R which is open and simply connected, so by Theorem 6 F is conservative. Thus, there exists a function f such that f =F , that is, f (x, y)=6x+5y x and f (x, y)=5x+4y . But f (x, y)=6x+5y implies f (x,y)=3x +5xy+g(y) and differentiating both sides y x 2 of this equation with respect to y gives f (x, y)=5x+g (y) . Thus 5x+4y=5x+g (y) so g (y)=4y and y / / / g(y)=2y +K where K is a constant. Hence f (x, y)=3x +5xy+2y +K is a potential function for F . 4. (x +4xy)/ y=4x , (4xy y )/ x=4y. Since these are not equal, F is not conservative. 5. (xe )/ y=xe , (ye )/ x=ye . Since these are not equal, F is not conservative. 6. (e )/ y=e = (xe )/ x and the domain of F is R . Hence F is conservative so there exists a function f such that y / y y y y 2 y y x x 3 3 2 2 2 f =F . Then f (x, y)=e implies f (x, y)=xe +g(y) and f (x, y)=xe +g (y) . But x y y y y / f (x, y)=xe so g (y)=0 g(y)=K . Then f (x, y)=xe +K is a potential function for F. 2 2 y 7. (2xcos y ycos x)/ y= 2xsin y cos x= ( x sin y sin x)/ x and the domain of F is R . Hence F is conservative so there exists a function f such that f =F . Then f (x, y)=2xcos y ycos x implies x f (x, y)=x cos y ysin x+g(y) and f (x, y)= x sin y sin x+g (y) . But f (x, y)= x sin y sin x so y y 2 2 / 2 g (y)=0 / g(y)=K . Then f (x, y)=x cos y ysin x+K is a potential function for F . 2 2 8. (1+2xy+ln x)/ y=2x= (x )/ x and the domain of F is { (x, y) | x>0 } which is open and simply connected. Hence F is conservative, so there exists a function f such that f =F . Then f (x, y)=1+2xy+ln x implies f (x, y)=x+x y+xln x x+g(y) and f (x, y)=x +g (y) . But f (x, y)=x so g (y)=0 x / y y 2 2 / 2 2 g(y)=K . Then f (x,y)=x y+xln x+K is a potential function for F . 1 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals 9. (ye +sin y)/ y=e +cos y= (e +xcos y)/ x and the domain of F is R . Hence F is conservative so there exists a function f such that x / y x x x 2 f =F . Then f (x, y)=ye +sin y implies f (x, y)=ye +xsin y+g(y) x x x y x x and f (x, y)=e +xcos y+g (y) . But f (x, y)=e +xcos y so g(y)=K and f (x, y)=ye +xsin y+K is a potential function for F . (xycosh xy+sinh xy) 2 (x cosh xy) 2 =x ysinh xy+xcosh xy+xcosh xy=x ysinh xy+2xcosh xy= 10. and y x the domain of F is R . Thus F is conservative, so there exists a function f such that f (x, y)=xycosh xy+sinh xy implies f (x, y)=xsinh xy+g(y) x 2 y 2 / y 2 2 f =F . Then f (x, y)=x cosh xy+g (y) . But f (x, y)=x cosh xy so g(y)=K and f (x, y)=xsinh xy+K is a potential function for F . 11. (a) F has continuous first order partial derivatives and y x open and simply connected. Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the value of C 2xy=2x= (x ) on R , which is 2 2 F dr depends only on the endpoints C of C . Since all three curves have the same initial and terminal points, value for each curve. (b) We first find a potential function f , so that x F dr will have the same 2 f =F . We know f (x, y)=2xy and f (x, y)=x . x y 2 Integrating f (x, y) with respect to x , we have f (x, y)=x y+g(y) . Differentiating both sides with respect to y gives f (x, y)=x +g (y) , so we must have x +g (y)=x y 2 2 / 2 / 2 g (y)=0 / g(y)=K , a constant. Thus f (x, y)=x y+K . All three curves start at (1, 2) and end at (3, 2) , so by Theorem 2, C F dr= f (3, 2) f (1, 2)=18 2=16 for each curve. / 12. (a) f (x, y)=y implies f (x, y)=xy+g(y) and f (x, y)=x+g (y) . But f (x, y)=x+2y so x y y g (y)=2y (b) C / g(y)=y +K . We can take K=0 , so f (x, y)=xy+y . 2 2 F dr= f (2, 1) f (0, 1)=3 1=2 . 1 4 4 4 3 / 4 3 x y +g(y) and f (x, y)=x y +g (y) . But f (x, y)=x y so y y 4 1 4 4 g(y)=K , a constant. We can take K=0 , so f (x, y)= x y . 4 3 4 2 13. (a) f (x, y)=x y implies f (x, y)= x g (y)=0 / Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals (b) The initial point of C is r(0)=(0, 1) and the terminal point is r(1)=(1, 2) , so C F dr= f (1, 2) f (0, 1)=4 0=4 . 2 2 2 14. (a) f (x, y)=y /(1+x ) implies f (x, y)=y arctanx+g(y) x f (x, y)=2yarctanx+g (y) . But y 2 / f (x, y)=2yarctanx so g (y)=0 y / g(y)=K . We can take K=0 , so f (x, y)=y arctanx. (b) The initial point of C is r(0)=(0, 0) and the terminal point is r(1)=(1, 2) , so C F d r= f (1, 2) f (0, 0)=4arctan1 0=4 4 = . 15. (a) f (x, y, z)=yz implies f (x, y, z)=xyz+g(y, z) and so f (x, y, z)=xz+g (y, z) . But f (x, y, z)=xz x y so g (y, z)=0 y g(y, z)=h(z) . Thus f (x, y, z)=xyz+h(z) and f (x, y, z)=xy+h (z) . But f (x, y, z)=xy+2z z z y / y , so h (z)=2z (b) C / h(z)=z +K . Hence f (x, y, z)=xyz+z (taking K=0 ). 2 2 F dr= f (4, 6, 3) f (1, 0, 2)=81 4=77 . 2 2 2 16. (a) f (x, y, z)=2xz+y implies f (x, y, z)=x z+xy +g(y, z) and so f (x, y, z)=2xy+g (y, z) . But x y y f (x, y, z)=2xy so g (y, z)=0 y y 2 2 / z g(y, z)=h(z) . Thus f (x, y, z)=x z+xy +h(z) and f (x, y, z)=x +h (z) . z 2 2 2 2 / But f (x, y, z)=x +3z , so h (z)=3z h(z)=z +K . Hence f (x, y, z)=x z+xy +z (taking K=0 ). 3 2 2 3 (b) t=0 corresponds to the point (0, 1, 1) and t=1 corresponds to (1, 2, 1) , so C F dr= f (1, 2, 1) f (0, 1, 1)=6 ( 1)=7 . 2 2 17. (a) f (x, y, z)=y cos z implies f (x, y, z)=xy cos z+g(y, z) and so f (x, y, z)=2xycos z+g (y, z) . But x y y f (x, y, z)=2xycos z so g (y, z)=0 y y 2 / z z g(y, z)=h(z) . Thus f (x, y, z)=xy cos z+h(z) and 2 / 2 f (x, y, z)= xy sin z+h (z) . But f (x, y, z)= xy sin z , so h (z)=0 (taking K=0 ). (b) r(0)= 0, 0, 0 , r( )= y 2 2 h(z)=K . Hence f (x, y, z)=xy cos z 2 , 0, so C y F dr= f ( , 0, ) f (0, 0, 0)=0 0=0 . y y 18. (a) f (x, y, z)=e implies f (x, y, z)=xe +g(y, z) and so f (x, y, z)=xe +g (y, z) . But f (x, y, z)=xe x y y y so g (y, z)=0 y g(y, z)=h(z) . Thus f (x, y, z)=xe +h(z) and f (x, y, z)=0+h (z) . But f (x, y, z)=(z+1)e z z y / z , so 3 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line (z)=(z+1)e / z h(z)=ze Integrals h +K (using integration by parts). Hence f (x, y, z)=xe +ze (taking K=0 ). C z y z (b) r(0)= 0, 0, 0 , r(1)= 1, 1, 1 so 2 F dr= f (1, 1, 1) f (0, 0, 0)=2e 0=2e . f =F 19. Here F(x, y)=tan yi+xsec y j . Then f (x, y)=xtan y is a potential function for F , that is, so F is conservative and thus its line integral is independent of path. Hence C tan ydx+xsec ydy= 2 C x F d r= f x 2, 4 f (1, 0)=2tan x 4 tan 0=2 . f =F so 20. Here F(x, y)=(1 ye ) i+e j . Then f (x, y)=x+ye is a potential function for F , that is, F is conservative and thus its line integral is independent of path. Hence C (1 ye )dx+e dy= 3/2 x x C F d r= f (1, 2) f (0, 1)=(1+2e ) 1=2/e . C 1 21. F(x, y)=2y i+3x y j , W = F d r . Since (2y )/ y=3 y = (3x y )/ x , there exists a 3/2 x 3/2 function f such that y f =F . In fact, f (x, y)=2y / f (x, y)=2xy +g(y) f (x, y)=2xy 3/2 But f (x, y)=3x y so g (y)=0 or g(y)=K . We can take K=0 W= C y 3/2 f (x, y)=3xy +g (y) . . Thus 1/2 / F d r= f (2, 4) f (1, 1)=2(2)(8) 2(1)=30 . y x 2 2 22. F(x, y)= 2y i j ,W= x C F dr . Since 2 2 y y x 2 2 = 2 2y x 2 = x 2y x y , there exists a / function f such that f =F . In fact, f =y /x x 2 2 f (x, y)= y /x+g(y) f = 2y/x+g (y) g (y)=0 , so / we can take f (x, y)= y /x as a potential function for F . Thus W= C F dr= f (4, 2) f (1, 1)= [( 2) /4]+(1/1)=0 . 23. We know that if the vector field (call it F ) is conservative, then around any closed path C , C F dr=0 . But take C to be some circle centered at the origin, oriented counterclockwise. All of the field vectors along C oppose motion along C , so the integral around C will be negative. Therefore the field is not conservative. 24. 4 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals From the graph, it appears that F is conservative, since around all closed paths, the number and size of the field vectors pointing in directions similar to that of the path seem to be roughly the same as the number and size of the vectors pointing in the opposite direction. To check, we calculate y x Thus F is conservative, by Theorem 6. (2xy+sin y)=2x+cos y , (x +xcos y)=2x+cos y 2 25. From the graph, it appears that F is not conservative. For example, any closed curve containing the point (2, 1) seems to have many field vectors pointing counterclockwise along it, and none pointing clockwise. So along this path the integral F dr 0 . To confirm our guess, we calculate x 2y y 1+x +y x 2 x 2 2 2 2 =(x 2y) y (1+x +y ) 2 2 3/2 2 1+x +y + 1 2 2 2 2 = 2 2x xy (1+x +y ) 2 2 3/2 2 2 , =(x 2) x 2 2 3/2 = 1+y +2x 2 2 3/2 . (1+x +y ) (1+x +y ) 1+x +y 1+x +y These are not equal, so the field is not conservative, by Theorem 5. 26. f (x, y)=cos (x 2y) i 2cos (x 2y) j C (a) We use Theorem 2: F dr= 1 C f dr= f (r(b)) f (r(a)) where C starts at t=a and ends at t=b . 1 1 So because f (0, 0)=sin 0=0 and f ( , )=sin ( (0, 0) to ( , ) ; that is, r(t)= t i+ t j , 0 t (b) From (a), C 2 2 )=0 , one possible curve C is the straight line from 1 1. 2 ,0 =1 , one possible F dr= f (r(b)) f (r(a)) . So because f (0, 0)=sin 0=0 and f 2 ti , 0 t 1 , the straight line from (0, 0) to 2 ,0 . curve C is r(t)= 2 5 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals 27. Since F is conservative, there exists a function f such that F= z f , that is, P= f , Q= f , and x y R= f . Since P , Q and R have continuous first order partial derivatives, Clairaut's Theorem says that P/ y= f = f = Q/ x , P/ z= f = f = R/ x , and Q/ z= f = f = R/ y . xy yx xz zx yz zy 28. Here F(x, y, z)=yi+x j+xyz k . Then using the notation of Exercise 27, P/ z=0 while R/ x=yz . Since these aren't equal, F is not conservative. Thus by Theorem 4, the line integral of F is not independent of path. 29. D={ (x, y) | x>0, y>0 }= the first quadrant (excluding the axes). (a) D is open because around every point in D we can put a disk that lies in D . (b) D is connected because the straight line segment joining any two points in D lies in D . (c) D is simply connected because it's connected and has no holes. 30. D={ (x, y) | x 0 } consists of all points in the xy plane except for those on the y axis. (a) D is open. (b) Points on opposite sides of the y axis cannot be joined by a path that lies in D , so D is not connected. (c) D is not simply connected because it is not connected. 31. D={ (x, y) | 1<x +y <4 }= the annular region between the circles with center (0, 0) and radii 1 and 2. (a) D is open. (b) D is connected. (c) D is not simply connected. For example, x +y =(1.5) is simple and closed and lies within D but encloses points that are not in D . (Or we can say, D has a hole, so is not simply connected.) 32. D={ (x, y) | x +y 2 2 2 2 2 2 2 1 or 4 x +y 2 2 2 9 }= the points on or inside the circle x +y =1 , together with 2 2 2 2 2 the points on or between the circles x +y =4 and x +y =9 . (a) D is not open because, for instance, no disk with center (0, 2) lies entirely within D . (b) D is not connected because, for example, (0, 0) and (0, 2.5) lie in D but cannot be joined by a path that lies entirely in D . (c) D is not simply connected because, for example, x +y =9 is a simple closed curve in D but encloses points that are not in D . P Q P Q x y x y x 33. (a) P= , and Q= , . Thus = . = = 2 2 2 2 2 22 2 22 y x y x x +y x +y (x +y ) (x +y ) (b) C : x=cos t , y=sin t , 0 t , C : x=cos t , y=sin t , t=2 to t= . Then y 1 2 6 2 2 2 2 2 2 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.3 The Fundamental Theorem for Line Integrals 1 2 cos t+sin t Since these aren't equal, the line integral of F isn't independent of path. (Or notice that C F dr= ( sin t)( sin t)+(cos t)(cos t) 0 2 2 dt= 0 dt= and C F dr= 2 dt= C F dr= 3 2 0 dt=2 where C is the circle x +y =1 , and apply the contrapositive of Theorem 3). This 3 2 2 2 doesn't contradict Theorem 6, since the domain of F , which is R except the origin, isn't simply connected. 34. (a) Here F(r)=cr/|r| and r=x i+y j+z k . Then f (r)= c/|r| is a potential function for F , that is, f =F . (See the discussion of gradient fields in Section 17.1). Hence F is conservative and its line integral is independent of path. Let P =(x , y , z ) and P =(x , y , z ) . 1 1 1 1 2 2 2 2 3 W= C F dr= f (P ) f (P )= 2 1 c (x +y +z ) 2 2 2 2 2 2 1/2 + c (x +y +z ) 1 1 1 2 2 2 1/2 =c 1 d 1 1 d . 2 (b) In this case, c= (mMG) 1 1 W = mMG 8 8 1.52 10 1.47 10 = (5.97 1024)(1.99 1030)(6.67 10 (c) In this case, c= qQ 1 1 W = qQ 12 10 5 10 11 )( 2.2377 10 10 ) 1.77 10 J 35 13 =(8.985 10 )(1)( 1.6 10 10 19 )( 10 ) 1.4 10 J. 12 4 7
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TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem1. (a)C : x=t1 2dx=dt , y=0 dx=0dt , y=tdy=0dt , 0 t dy=dt , 0 t2. 3. 2. 3.C : x=2 C : x=2 t3dx= dt , y=3 dx=0dt , y=3 tdy=0dt , 0 t dy= dt , 0 tC : x=04
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.2 Derivatives and Integrals of Vector Functions1. (a)r(4.5) r(4) =2[r(4.5) r(4)] , so we draw a vector in 0.5 the same direction but with twice the length of the (b) r(4.2) r(4) =5[r(4.2)
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.4 Motion in Space: Velocity and Acceleration1. (a) If r(t)=x(t) i+y ( t ) j+z(t) k is the position vector of the particle at time t , then the average velocity over the time interval 0,1 is
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.3 Arc Length and Curvature1. r (t)= 2cos t,5, 2sin t have10/L=10 /r (t) dt=10/1029 dt=./ tr (t) = (2cos t) +5 +( 2sin t) = 29 . Then using Formula 3, we10/29 t10
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.1 Functions of Several Variables1. (a) From Table 1, f ( 15,40)= 27 , which means that if the temperature is 15 C and the wind speed is 40 km / h, then the air would feel equivalent to a
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.2 Limits and Continuity1. In general, we can't say anything about f (3,1) !lim(x,y) (3,1)f (x,y)=6 means that the values of f (x,y)approach 6 as (x,y) approaches, but is not equal
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.4 Tangent Planes and Linear Approximations1. z= f (x,y)=4x y +2y z= 8x 2y .22f (x,y)=8x , f (x,y)= 2y+2 , so f ( 1,2)= 8 , f ( 1,2)= 2 . By Equation 2,x y x y x yan equation of
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations1. The auxiliary equation is r +3r+2=(r+2)(r+1)=0 , so the complementary solution is y (x)=c ec 1 2 2x2+c e . We try the particular
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.1 Second-Order Linear Equations1. The auxiliary equation is r 6r+8=0 is y=c e +c e1 2 4x 2x2(r 4)(r 2)=0r=4 , r=2 . Then by (8) the general solution.22. The a
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.7 Maximum and Minimum Values1. (a) First we compute D(1,1)= f (1,1) f (1,1) [ f (1,1)] =(4)(2) (1) =7 . Since D(1,1)&gt;0 andxx yy xy22f (1,1)&gt;0 , f has a local minimum at (1,1) by t
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.5 Applications of Double Integrals1.32Q=D 3(x,y)dA=10(2xy+y )dydx=1 2 3 123xy +21 3 y 3y=2 y=0dx=18 4x+ 38 dx= 2x + x 3=16+16 64 = C 3 32. Q=D 2 2 2
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.1 Double Integrals over Rectangles1. (a) The subrectangles are shown in the figure. The surface is the graph of f (x,y)=xy and A=4 , so we estimateV3 i =12 j=1f x ,yi( )jA A
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area1. Here z= f (x,y)=2+3x+4y and D is the rectangle 0,5 surface is A(S) =D1,4 , so by Formula 2 the area of the dAD[ f (x,y)] +[ f (x,y)] +1 dA=x y D223 +4 +1 dA=
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.3 Partial Derivatives1. (a) T / x represents the rate of change of T when we fix y and t and consider T as a function of the single variable x , which describes how quickly the temperatu
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.2 Iterated Integrals31.0 4 0 3( 2x+3x2y) dx=2x +x y223x=3 x=0 2= ( 9+27y ) ( 0+0 ) =9+27y ,y=4 y=0( 2x+3x y) dy=2.0 4 0y dx= yln x+2 x+2 y 23 1 2y 1 dy= x+2 x+
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates1. The region R is more easily described by polar coordinates: R= { (r, )|0 r2 22,02 } . Thusf (x,y)dA=R 0 0f (rcos ,rsin )r dr d .2. Th
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions1.1x21 0 1 0 3(x+2y)dydx =0 0xy+y2 y=x21 0y=0dx=x(x )+(x ) 0 0 dx1 0222= (x +x )dx=41 4 1 5 x+ x 4 5=9 202.22
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.4 Series Solutions1. Let y(x)=n=1n=0 nc x . Then y (x)=nn/n=1nc xnn 1and the given equation, yn=0 n/y=0 , becomesn+1nc xnn 1n=0 n n+1c x
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.9 Change of Variables in Multiple Integrals1. x=u+4v , y=3u 2v . ( x,y ) The Jacobian is = ( u,v ) 2. 3.x/ u y/ u x/ v = y/ vx/ v = y/ v 2u 2u1 34 =1( 2) 4(3)= 14 . 2( x,y ) = (
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.3 Applications of Second-Order Differential Equations100 is the spring constant and the differential equation is 3 10 10 / / 100 3x + x=0 . The general solution is x(t)=c
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.7 Triple Integrals1. xyz dV =0 10 1 2 1 23Bxyz dzdxdy=0 1 x=2 x= 121 2xy1 3 z 3z=3 z=01 2dxdy=1 09xydxdy0 1=09 2 x y 227 27 2 dy= ydy= y 1 4 0 2=27 42
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.6 Directional Derivatives and the Gradient Vector1. First we draw a line passing through Raleigh and the eye of the hurricane. We can approximate the directional derivative at Raleigh in
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.5 The Chain Rule1. z=x y+xy , x=2+t , y=1 t dz z dx z dy 2 = + = 2xy+y dt x dt y dt2243() ( 4t 3) + ( x2+2xy) ( 3t2) =4 ( 2xy+y2) 3 3( x2+2xy) t2 )1/22. z= x +y , x=e , y=
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.2 Line Integrals1. x=t and y=t , 0 tC22 , so by Formula 32y ds = =20 2t t2dx dt+dy dt2dt= t0 2 02(2t) +(1) dt 1 (17 17 1) 122204t +1 dt=1 2 3/2 (4t +1
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.5 Curl and Divergence1. (a) i curl F =2j / y 0F=/ x xyzk / z xy2=( x 0) i ( 2xy xy) j+(0 xz) k2= x i+3xy j xz k (b) div F= 2. (a) curlF = F=2 2F=x(xyz)+y(0)+z(
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.6 Parametric Surfaces and Their Areas1. r(u, v)=ucos v i+usin v j+u k , so the corresponding parametric equations for the surface are x=ucos v, y=usin v, z=u . For any point (x, y, z) on the
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.7 Surface Integrals1. Each face of the cube has surface area 2 =4 , and the points P are the points where the cubeij2*intersects the coordinate axes. Here, f (x, y, z)= x +2y +3z , so
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.8 Stokes' Theorem1. Both H and P are oriented piecewise smooth surfaces that are bounded by the simple, closed, smooth curve x +y =4 , z=0 (which we can take to be oriented positively for bo
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.9 The Divergence Theorem1. The vectors that end near P are longer than the vectors that start near P , so the net flow is inward1 1near P and divF(P ) is negative. The vectors that end ne
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.1 Three-Dimensional Coordinate Systems1. We start at the origin, which has coordinates ( 0,0,0 ) . First we move 4 units along the positive x axis, affecting only the x coo
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.2 Vectors1. (a) The cost of a theater ticket is a scalar, because it has only magnitude. (b) The current in a river is a vector, because it has both magnitude (the speed of
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.3 The Dot Product1. (a) a b is a scalar, and the dot product is defined only for vectors, so ( a b) c has no meaning. (b) ( a b) c is a scalar multiple of a vector, so it d
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.4 The Cross Producti j k 2 0 1 0 1 2 1. a b= 1 2 0 = i j+ k 3 1 0 1 0 3 0 3 1 =(2 0) i (1 0) j+(3 0) k=2 i j+3 k Now (a b) a= 2, 1,3 1,2,0 =2 2+0=0 and (a b) b= 2, 1,3 0,3,
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.5 Equations of Lines and Planes1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.6 Cylinders and Quadric Surfaces1. (a) In R , the equation y=x represents a parabola.22(b)In R , the equation y=x doesn't involve z , so any horizontal plane with e
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;12. Vectors and the Geometry of Space; 12.7 Cylindrical and Spherical Coordinates1. See Figure 1 and the accompanying discussion; see the paragraph accompanying Figure 3. 2. See Figure 5 and the accompanying discus
TN Tech - MATH - Cal I- Cal
Stewart Calculus ET 5e 0534393217;13. Vector Functions; 13.1 Vector Functions and Space Curves1. The component functions t , t 1 , and 5 t are all defined when t 1 0 t 5 , so the domain of r(t) is 1,5 . 2. The component functions2t1 and 5 t
FSU - HIST - 08758
Prcis 1-28-08The Chinese Exclusion Act of 1882 denied Chinese immigrants the right to obtain citizenship in America. It states that any captain of a ship facilitating Chinese immigrants into the country would be subject to criminal charges. Another
FSU - HIST - 08758
Prcis 2-11-08This passage of Harry Emerson Fosdick's &quot;Shall the Fundamentalists Win?&quot;(1922) shows Fosdick's distaste in the people who call themselves fundamentalists who feel there is no place for liberal Christians in the church. Evidence of this
FSU - HIST - 08758
Prcis 2-20-08The passage from Meridel Le Sueur's &quot;Women on the Breadlines&quot; (1932) displays how The Great Depression affected an endless number of people of any race, gender or religion. In this particular passage, Sueur describes a typical day of a
FSU - HIST - 08758
Prcis 2-22-08Franklin D Roosevelt's &quot;First Inaugural Address (1933)&quot;, gives a glimpse on how he planned to revive the nation through his presidency by using any tactics he felt that was necessary to restore American prosperity. He shows the desire
Kansas State - PHYS - 213
1. With speed v = 11200 m/s, we findK= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 22. (a) The change in kinetic energy for the meteorite would be1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2()()2= -5 1014 J
FSU - HIST - 08758
Prcis 3-26-08In George F. Kennan's &quot;The Sources of Soviet Conduct&quot; (1947), he states, &quot;it is easy to regard to the Soviet Union as a rival not a partner, in the political arena. It must continue to expect that Soviet policies will reflect no abstra
Kansas State - PHYS - 214
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
Kansas State - PHYS - 214
1. The amplitude of the induced emf in the loop is m = A 0 ni0 = (6.8 10-6 m 2 )(4 10 -7 T m A)(85400 / m)(1.28 A)(212 rad/s)= 1.98 10-4 V.2. (a) =d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dtchb g(b) Appealing to
Kansas State - PHYS - 214
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
FSU - HIST - 08758
Prcis 3-28-08The Supreme Court's decision on Brown v. Board of Education of Topeka (1954) served as a turning point and as a milestone in the start of the civil rights movement. It reviewed the &quot;separate but equal&quot; principle decided by the case of
FSU - HIST - 08758
Prcis 4-2-08George C. Wallace displayed his prominence among white conservatives as Governor of Alabama in a speech on&quot; The Civil Rights Movement: Fraud, Sham, and a Hoax&quot; (1964). In June of 1963, Wallace defiantly prevented the first black student
FSU - HIST - 08758
Prcis 4-4-08Martin Luther King's Letter from Birmingham Jail (1963)is a letter Dr. King wrote after being jailed in Birmingham Alabama after organizing a economic boycott of white businesses, using a smuggled pen and piece of paper. King's statemen
FSU - GLY - 01636
Origin of the Universe: The Big Bang: Supernova: Life elements: Leon Sinks: Karst topography: Caves, caverns: Ground water: Porosity: Permeability Water table: Springs: Sink holes: o Wet Sinks: o Dry sinks: o Sink vs. Swamp Natural Bridge: Evolution
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 1 ~ Eras of Roman History Reading: BHR 2-4.Spring 2008Roman history is traditionally divided into three phases: monarchy, republic and empire. The dates are as follows (keep in mind that the very earliest date