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11 Pages

### c16s4

Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Word Count: 1952

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Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem 1. (a) C : x=t 1 2 dx=dt , y=0 dx=0dt , y=t dy=0dt , 0 t dy=dt , 0 t 2. 3. 2. 3. C : x=2 C : x=2 t 3 dx= dt , y=3 dx=0dt , y=3 t dy=0dt , 0 t dy= dt , 0 t C : x=0 4 Thus xy dx+x dy C C +C +C +C 1 2 3 4 2 3 xy dx+x dy 2 0 3 0 2 0 3 2 3 = 0dt+ 8dt+ =0+24 18+0=6 (b) xy dx+x dy = C D 2 0 2 2 3 9(2 t)dt+ 0dt 0 x (x ) 3 y...

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Calculus Stewart ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem 1. (a) C : x=t 1 2 dx=dt , y=0 dx=0dt , y=t dy=0dt , 0 t dy=dt , 0 t 2. 3. 2. 3. C : x=2 C : x=2 t 3 dx= dt , y=3 dx=0dt , y=3 t dy=0dt , 0 t dy= dt , 0 t C : x=0 4 Thus xy dx+x dy C C +C +C +C 1 2 3 4 2 3 xy dx+x dy 2 0 3 0 2 0 3 2 3 = 0dt+ 8dt+ =0+24 18+0=6 (b) xy dx+x dy = C D 2 0 2 2 3 9(2 t)dt+ 0dt 0 x (x ) 3 y (xy ) dA= 0 0 2 2 3 (3x 2xy)dydx 2 = (9x 9x)dx=24 18=6 2 2 2. (a) x=cos t , y=sin t , 0 t (b) C 2 . Then C ydx xdy= [sin t( sin t) cos t(cos t)]dt= 0 0 2 dt= 2 . ydx xdy= D x ( x) y (y) dA= 2 D dA= 2A(D)= 2 (1) = 2 3. (a) 1 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem C : x=t 1 2 dx=dt , y=0 dx=0dt , y=t dy=0dt , 0 t dy=dt , 0 t 1 2. 1. C : x=1 C : x=1 t 3 dx= dt , y=2 2t dy= 2dt , 0 t Thus xydx+x y dy = C C +C +C 1 2 3 2 3 xydx+x y dy 1 0 2 0 3 1 0 2 0 2 3 2 3 = 0dt+ t dt+ =0+ (b) xydx+x y dy = C D 1 2 3 (1 t)(2 2t) 2(1 t) (2 2t) dt + 2 3 8 6 (1 t) + (1 t) 3 3 1 0 1 4 t 4 =4 10 2 = 3 3 x (x y ) 2 3 1 2x y y=2x y=0 (xy) dA= 0 0 1 0 5 2 (2xy x)dydx 4 3 2 2 = 3 3 3 C :x=0 1 dx=0dt= 0 , y=1 t dy= dt , 4. (a) 1 0 t C :x=t dx=dt 2 1 4 xy xy 2 dx= (8x 2x )dx= , y=0 dy=0dt , 0 t 1 C :x=1 t 3 dx= dt 2 2 , y=1 (1 t) =2t t dy=(2 2t)dt ,0 t 1 Thus 2 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem xdx+ydy = C C +C +C 1 2 3 xdx+ydy 1 0 1 0 1 1 1 0 1 0 2 = (0dt+(1 t)( dt))+ (t dt+0dt)+ ((1 t)( dt)+(2t t )(2 2t)dt) 1 2 t t 2 1 1 = + + 2 2 = 1 2 1 4 3 5 2 t + t 2t + t t 0 2 0 2 2 1 5 2+ 1 =0 2 2 + (b) C xdx+ydy= D x (y) y (x) dA= D 0dA=0 ,0 7 5. We can parametrize C as x=cos 2 , y=sin 2 2 . Then the line integral is sin 6 Pdx+Q dy= cos C 0 4 sin 5 ( sin )d + ( cos 0 2 )cos d = 29 , according to a CAS. 1024 The double integral is Q x P y 1 1 x dA= 1 1 x 2 ( 7x y 5x y )dydx= 6 6 4 4 D 29 , 1024 verifying Green's Theorem in this case. 6. Since y=x along the first part of C and y=x along the second part, the line integral is 1 2 0 1 Pdx+Q dy = C 0 x sin x+x sin (x )(2x) dx+ (x sin x+x sin x)dx 4 2 2 2 2 = 16cos 1 23sin 1+28 according to a CAS. The double integral is 3 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem R Q x P y 1 x dA= 0 x (2xsin y 2ysin x)dydx= 16cos 1 23sin 1+28 2 7. The region D enclosed by C is [0, 1] [0, 1] , so e dx+2xe dy = C D 1 0 1 0 y y x y (2xe ) 1 y y 0 (e ) dA= 0 0 y 1 1 (2e e )dydx y y = dx e dy=(1)(e e )=e 1 8. The region D enclosed by C is given by { (x, y) | 0 x 1, 3x y 3 } , so x y dx+4xy dy = C D 1 2 2 3 x 4 (4xy ) 3 y 1 0 (x y ) dA= 2 4 2 2 1 3 0 3x (4y 2x y)dydx 72 318 = 5 5 3 2 = 0 y x y 2 2 y=3 y=3x dx= (81 9x 72x )dx=81 3 9. (y+e C x )dx+(2x+cos y )dy = D 1 y 2 2 x (2x+cos y ) 1 0 1/2 2 y 2 (y+e 1 3 x ) dA = 0 y (2 1)dxdy= (y y )dy= 10. xe C 2x dx+(x +2x y )dy = D 4 2 2 x 2 (x +2x y ) 2 2 2 4 2 2 y (xe 2x ) dA= D 2 (4x +4xy 0)dA 3 2 =4 D 2 x(x +y )dA=4 0 1 2 1 4 (rcos )(r )rdr d 2 0 =4 cos d 0 r dr=4 sin 1 5 r 5 2 1 =0 11. 4 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem y dx x dy = C D 2 3 3 x 2 0 ( x) 3 3 y (y ) dA= D 3 ( 3x 3y )dA= 0 0 2 2 2 2 ( 3r )rdr d 2 = 3 d 0 r dr= 3(2 )(4)= 24 12. sin ydx+xcos ydy= C 3 2 D x (xcos y) y (sin y) dA= D (cos y cos y)dA= D 0dA=0 13. F(x, y)= x +y , x + y and the region D enclosed by C is given by { (x, y) | 0 x , 0 y sin x }.C is traversed clockwise, so C gives the positive orientation. F dr = C ( 0 0 x +y dx+ x + y dy= D 2 3 ) ( 2 ) C sin x ( x2+ y ) x y=0 ( y x +y 3 ) dA = = 0 (2x 3y )dydx= 0 3 2xy y 3 y=sin x dx 2 (2xsin x sin x)dx= 0 (2xsin x (1 cos x)sin x)dx 1 3 cos x 3 = = 2sin x 2xcos x+cos x 2 2 0 2+ 2 3 = 2 4 2 3 14. F(x, y)= y cos x, x +2ysin x and the region D enclosed by C is given by { (x, y) | 0 x 2, 0 y 3x } . C is traversed clockwise, so C gives the positive orientation. F dr = C C (y cos x)dx+(x +2ysin x)dy (x +2ysin x) 2 2 2 = D x y (y cos x) dA 2 3x 2 = D (2x+2ycos x 2ycos x)dA= 0 0 2xdydx 5 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem 2 = 0 2x y y=3x y=0 2 0 y 2 dx= 6x dx= 2x 2 3 2 0 = 16 15. F(x, y)= e +x y, e xy and the region D enclosed by C is the disk x +y C is traversed clockwise, so C gives the positive orientation. F dr = C C x 2 2 2 25 . (e +x y)dx+(e xy )dy y 2 x 2 2 2 x 2 y 2 = D 2 x 2 (e xy ) 2 5 2 y (e +x y) dA= D 2 0 5 0 3 ( y x )dA 1 4 r 4 5 0 = D (x +y )dA= 0 0 (r )rdr d = d r dr=2 = 625 2 y and the region D enclosed by C is the disk with radius x 1 centered at (2, 3).C is oriented positively, so 16. F ( x, y ) = y ln (x +y ), 2tan 2 2 1 C F d r = (y ln (x +y ))dx+ 2tan C 2 2 1 y x y dy (y ln (x +y )) dA dA= 2y D 2 2 = D x 2 2tan yx 1 y x 1 2 2 = 2y x +y 2 2 D 1+(y/x) x +y 2 2 1+ 2y x +y 2 2 dA = D dA= ( area of D)= 17. By Green's Theorem, W = F dr= x(x+y)dx+xy dy= C 11 x 1 0 C y=1 x y=0 1 0 D 2 (y x)dydx where C is the path described 2 in the question and D is the triangle bounded by C . So W= 0 0 (y x)dydx= 2 1 3 y xy 3 dx= 1 3 (1 x) x(1 x) dx 3 6 Stewart Calculus ET 5e 0534393217;16. Calculus; Vector 16.4 Green's Theorem = 1 4 1 2 1 3 (1 x) x+ x 12 2 3 1 0 = 1 1 + 2 3 3 2 1 12 = 1 12 2 2 18. By Green's Theorem, W = F dr= xdx+(x +3xy )dy= C 2 C 2 0 D (3x +3y 0)dA , where D is the semicircular region bounded by C . Converting to polar coordinates, we have W =3 0 0 r rd dr=3 2 1 4 r 4 =12 . 2 , and let 19. Let C be the arch of the cycloid from (0, 0) to (2 , 0) , which corresponds to 0 t 1 C be the segment from (2 , 0) to (0, 0) , so C is given by x=2 2 2 t , y=0 , 0 t 2 . Then C=C 1 C is traversed 2 clockwise, so C is oriented positively. Thus C encloses the area under one arch of the cycloid and from (5) we have 2 2 0 2 A= C ydx = ydx+ ydx= (1 cos t)(1 cos t)dt+ 0( dt) C 2 0 1 C 2 0 = (1 2cos t+cos t)dt+0= t 2sin t+ 1 1 t+ sin 2t 2 4 2 0 =3 20. 2 A= C xdy= (5cos t cos 5t)(5cos t 5cos 5t)dt 0 2 0 = (25cos t 30cos tcos 5t+5cos 5t)dt = 25 =30 21. (a) Using Equation 17.2.8 , we write parametric equations of the line segment as x=(1 t)x +tx , 1 2 2 2 1 1 t+ sin 2t 2 4 30 1 1 sin 4t+ sin 6t +5 8 12 1 1 t+ sin 10t 2 20 2 0 y=(1 t)y +ty , 0 t 1 2 1 1 . Then dx=(x x )dt and dy=(y y )dt , so 2 1 2 1 xdy ydx = C 0 [(1 t)x +tx ](y y )dt+[(1 t)y +ty ](x x )dt 1 2 2 1 1 2 2 1 7 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem 1 = 0 1 (x (y y ) y (x x )+t [(y y )(x x ) (x x )(y y )])dt 1 2 1 1 2 1 2 1 2 1 2 1 2 1 = 0 (x y x y )dt=x y x y 1 2 2 1 1 2 2 1 (b) We apply Green's Theorem to the path C=C i i i+1 i+1 1 C 2 n C , where C is the line segment that n i n n 1 1 joins (x , y ) to (x , y ) for i=1 , 2 , ... , n 1 , and C is the line segment that joins (x , y ) to (x , y ) . From (5), 1 xdy ydx= 2C dA , where D is the polygon bounded by C . Therefore D area of polygon =A(D)= D dA= 1 xdy ydx 2C xdy ydx+ xdy ydx n 1 = 1 2 xdy ydx+ xdy ydx+...+ C 1 C 2 C C n To evaluate these integrals we use the formula from (a) to get A(D)=[(x y x y )+(x y x y )+...+(x y x y )+(x y x y )]. 1 2 2 1 2 3 3 2 n 1 n n n 1 n 1 1 n (c) A= 1 [(0 1 2 0)+(2 3 1 1)+(1 2 0 3)+(0 1 ( 1) 2)+( 1 0 0 1)] 2 1 9 = (0+5+2+2)= 2 2 1 2A 2 22. By Green's Theorem, 1 2A y dx= C 2 x dy= C 1 2A 2xdA= D 1 A xdA=x and D 1 2A ( 2y)dA= D 1 A ydA=y . D 1 1 (1)(1)= and C=C +C +C , where C : x=x , y=0 , 0 x 1 ; 1 2 3 1 2 2 C : x=x , y=1 x , x=1 to x=0 ; and C : x=0 , y=1 to y=0 . Then 23. Here A= 2 3 1 1 2 2 2 2 2 x= x dy= x dy+ x dy+ x dy=0+ (x )( dx)+0= . Similarly, 2A C 3 1 C 1 0 C 2 C 3 8 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem 1 1 2 2 2 2 2 y= y dx= y dx+ y dx+ y dx=0+ (1 x) ( dx)+0= . 2A C 3 1 C 1 0 C 2 C 3 Therefore (x, y)= 2 1 1 , 3 3 . a 2 2 1 1 24. A= so x= x dy and y= y dx. 2 2 2 a C a C Orienting the semicircular region as in the figure, x = 1 a = 1 a and 2 2 C +C 1 2 x dy 2 0+ (a cos t)(acos t)dt 0 2 2 =0 y= 1 a 2 a a 0 0dx+ (a sin t)( asin t)dt 4a 3 . 1 3 2 2 = a 0 sin tdt= 3 a cos t+ 1 3 (cos t) 3 0 = 4a . 3 Thus (x, y)= 0, 25. By Green's Theorem, 1 3 x dy= C 3 y dx= C 2 3 1 3 y ( 3y )dA= D D 2 y dA=I and x 2 1 3 (3x )dA= D D 2 x dA=I . 26. By symmetry the moments of inertia about any two diameters are equal. Centering the disk at the origin, the moment of inertia about a diameter equals 9 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem I = 1 y 3 = 1 4 a 3 C 1 x dy= 3 3 2 1 4 (a cos t)dt= a 3 0 4 4 2 0 3 1 1 + cos 2t+ cos 4t dt 8 2 8 3(2 ) 1 4 = a 8 4 27. Since C is a simple closed path which doesn't pass through or enclose the origin, there exists an open region that doesn't contain the origin but does contain D . Thus P= y/(x +y ) and Q=x/(x +y ) have continuous partial derivatives on this open region containing D and we can apply Green's Theorem. But by Exercise 17.3.33(a) , P/ y= Q/ x , so C 2 2 2 2 F dr= D 0dA=0 . y d } where f and f are 1 2 28. We express D as a type II region: D={ (x, y) | f (y) x 1 f (y), c 2 continuous functions. Then D Q dA= x c d f 2(y) f (y) 1 Q dxdy= Q( f (y), y) Q( f (y), y) dy by 2 1 x c Qdy= C C +C +C +C 1 2 3 d the Fundamental Theorem of Calculus. But referring to the figure, c d 1 Qdy. Then 4 Qdy= Q( f (y), y)dy , C 1 Qdy= Qdy=0 , and C 2 Qdy= Q( f (y), y)dy . Hence C 3 d d C 4 c 2 Qdy= C c Q( f (y), y) Q( f (y), y) dy= 2 1 D ( Q/ x)dA. 29. Using the first part of (5), we have that R dxdy=A(R)= xdy . But x=g(u, v) , and R h h dy= du+ dv , and we orient S by taking the positive direction to be that which corresponds, u v under the mapping, to the positive direction along R , so 10 Stewart Calculus ET 5e 0534393217;16. Vector Calculus; 16.4 Green's Theorem xdy = R S g(u, v) h h du+ dv u v g(u, v) h v 2 = g(u, v) S h h du+g(u, v) dv u v h u dA [using Green's Theorem in the uv = S u v g(u, v) plane] = S g u x u h h +g(u, v) v u v y v x v y u g v h h g(u, v) u v u 2 dA (x, y) dudv (u, v) = S dA [by the equality of mixed partials] = S The sign is chosen to be positive if the orientation that we gave to S corresponds to the usual positive orientation, and it is negative otherwise. In either case, since A(R) is positive, the sign chosen must be the same as the sign of (x, y) . Therefore A(R)= (u, v) dxdy= R S (x, y) (u, v) dudv. 11
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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
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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
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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
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 2: Roman Society and Roman Values Readings: BHR 129-131; Shelton 128, 163-4, 171, 194-5, 197Spring 2008Roman Values and Virtues Roman virtues public, not private: virtuous man acts on behalf of the state Virtu