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Course: MATH Cal I- Cal, Spring 2008
School: TN Tech
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Calculus Stewart ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations 1. The auxiliary equation is r +3r+2=(r+2)(r+1)=0 , so the complementary solution is y (x)=c e c 1 2 2x 2 +c e . We try the particular solution y (x)=Ax +Bx+C , so y =2Ax+B and y 2 p p 2 2 2 x 2 / / / p =2A . Substituting into the differential equation, we have (2A)+3(2Ax+B)+2(Ax +Bx+C)=x or 2Ax...

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Calculus Stewart ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations 1. The auxiliary equation is r +3r+2=(r+2)(r+1)=0 , so the complementary solution is y (x)=c e c 1 2 2x 2 +c e . We try the particular solution y (x)=Ax +Bx+C , so y =2Ax+B and y 2 p p 2 2 2 x 2 / / / p =2A . Substituting into the differential equation, we have (2A)+3(2Ax+B)+2(Ax +Bx+C)=x or 2Ax +(6A+2B)x+(2A+3B+2C)=x . Comparing coefficients gives 2A=1 , 6A+2B=0 , and 1 3 7 2A+3B+2C=0 , so A= , B= , and C= . Thus the general solution is 2 2 4 7 2x x 1 2 3 y(x)=y (x)+y (x)=c e +c e + x x+ . c p 1 2 2 2 4 2. The auxiliary equation is r +9=0 with roots r= 3i , so the complementary solution is y (x)=c cos (3x)+c sin (3x) . Try the particular solution y (x)=Ae c 1 2 p 3x 3x 3x 3x 2 , so y =3Ae p 3x / 3x 3x and y / / p =9Ae 3x . Substitution into the differential equation gives 9Ae +9(Ae )=e the general solution is y(x)=y (x)+y (x)=c cos (3x)+c sin (3x)+ c p 1 2 2 or 18Ae =e . Thus A= 1 and 18 1 3x e . 18 2x 3. The auxiliary equation is r 2r=r(r 2)=0 , so the complementary solution is y (x)=c +c e . Try c 1 2 the particular solution y (x)=Acos 4x+Bsin 4x , so y = 4Asin 4x+4Bcos 4x and p p / y / / p = 16Acos 4x 16Bsin 4x . Substitution into the differential equation gives ( 16Acos 4x 16Bsin 4x) 2( 4Asin 4x+4Bcos 4x)=sin 4x ( 16A 8B)cos 4x+(8A 16B)sin 4x=sin 4x . 1 1 and B= . Thus the general solution is Then 16A 8B=0 and 8A 16B=1 A= 40 20 1 2x 1 y(x)=y (x)+y (x)=c +c e + cos 4x sin 4x . c p 1 2 40 20 4. The auxiliary equation is r +6r+9=(r+3) =0 , so the complementary solution is y (x)=c e c 1 3x 2 2 +c xe 2 3x . Try the particular solution y (x)=Ax+B , so y =A and y p p / / / p =0 . Substitution into the differential equation gives 0+6A+9(Ax+B)=1+x or (9A)x+(6A+9B)=1+x . Comparing 1 1 coefficients, we have 9A=1 and 6A+9B=1 , so A= and B= . Thus the general solution is 9 27 1 3x 3x 1 y(x)=c e +c xe + x+ . 1 2 9 27 5. The auxiliary equation is r 4r+5=0 with roots r=2 i , so the complementary solution is 1 2 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations y (x)=e (c cos x+c sin x) . Try y ( x ) =Ae , so y = Ae and y =Ae . Substitution gives c 1 2 p p p Ae x 2x x / x / / x 4( Ae )+5(Ae )=e 2x 1 2 x x x 10Ae =e x x x A= 1 . Thus the general solution is 10 y(x)=e (c cos x+c sin x)+ x 1 e 10 . 2 x 6. y (x)=e (c x+c ) . Try y (x)=x ( Ax+B ) e so that no term in y is a solution of the complementary c 1 2 p p equation. Then y =[ Ax +(3A B)x +2Bx]e , y p 3 2 / 3 2 x / / p =[Ax +(B 6A)x +(6A 4B)x+2B]e and 3 2 3 2 3 2 x substitution gives [Ax +(B 6A)x +(6A 4B)x+2B]+2[ Ax +(3A B)x +2Bx]+(Ax +Bx )=x 1 1 3 x 2 x x 6Ax+2B=x . So y ( x ) =x x e and the general solution is y(x)=e c x+c + x e . p 1 2 6 6 ( ) 7. The auxiliary equation is r +1=0 with roots r= i , so the complementary solution is y (x)=c cos x+c sin x . For y c 1 2 / / 2 +y=e try y (x)=Ae . Then y =y p 1 x x / / / p 1 p =Ae and substitution gives 2 x 1 Ae +Ae =e / x x x A= 2 1 1 x , so y (x)= e . For y 2 2 p 1 / / p 2 / / +y=x try y (x)=Ax +Bx +Cx+D . p 2 3 3 Then y =3Ax +2Bx+C and y p 2 =6Ax+2B . Substituting, we have C= 6 , and 2B+D=0 D=0 . Thus 6Ax+2B+Ax +Bx +Cx+D=x , so A=1 , B=0 , 6A+C=0 y (x)=x 6x and the general solution is p 2 3 2 3 3 y(x)=y (x)+y (x)+y (x)=c cos x+c sin x+ c p 1 p 2 1 2 1 x 3 1 e +x 6x . But 2=y(0)=c + 1 2 2 c= 1 3 and 2 0=y (0)=c + 1 11 . Thus the solution to the initial value problem is 6 c= 2 2 2 2 3 11 1 x 3 y(x)= cos x+ sin x+ e +x 6x . 2 2 2 / 2 8. The auxiliary equation is r 4=0 with roots r= 2 , so the complementary solution is y (x)=c e +c e c 2x 2x / / 1 x 2 . Try y (x)=e (Acos x+Bsin x) , so y =e (Acos x+Bsin x+Bcos x Asin x) and p p x / x y e (2Bcos x 2Asin x) 4e (Acos x+Bsin x)=e cos x p x =e (2Bcos x 2Asin x) . Substitution gives x x (2B 4A)e cos x+( 2A 4B)e sin x=e cos x 2 x x x Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations 1 1 1 1 2x 2x x , B= . Thus the general solution is y(x)=c e +c e +e cos x+ sin x . But 1 2 5 10 5 10 1 1 9 3 / 1=y(0)=c +c and 2=y (0)=2c 2c . Then c = , c = , and the solution to the initial 1 2 5 1 2 10 1 8 2 40 9 2x 3 2x x 1 1 value problem is y(x)= e + e +e cos x+ sin x . 8 40 5 10 A= 9. The auxiliary equation is r r=0 with roots r=0 , r=1 so the complementary solution is y (x)=c +c e . Try y (x)=x(Ax+B)e so that no term in y is a solution of the complementary c 1 2 p p x x / 2 x / / p 2 x 2 equation. Then y =(Ax +(2A+B)x+B)e and y p 2 =(Ax +(4A+B)x+(2A+2B))e . Substitution into the x 2 x x differential equation gives (Ax +(4A+B)x+(2A+2B))e (Ax +(2A+B)x+B)e =xe 1 1 2 x x x A= , B= 1 . Thus y (x)= x x e and the general solution is (2Ax+(2A+B))e =xe p 2 2 1 2 x x / y(x)=c +c e + x x e . But 2=y(0)=c +c and 1=y (0)=c 1 , so c =2 and c =0 . The solution 2 1 2 1 2 2 1 2 1 2 1 2 x x x to the initial value problem is y(x)=2e + x x e =e x x+2 . 2 2 10. y (x)=c e +c e c 1 2 x 2x . For y / / +y / 2y=x try y (x)=Ax+B . Then y =A , y p 1 / / / p / 1 p =0 , and substitution 1 gives 0+A 2(Ax+B)=x 1 1 1 1 A= , B= , so y (x)= x . For y 2 4 2 4 p 1 / / / p 2 / / +y 2y=sin 2x try y (x)=Acos 2x+Bsin 2x . Then y = 2Asin 2x+2Bcos 2x, y p 2 p = 4Acos 2x 4Bsin 2x , and substitution A= 1 3 , B= . 20 20 2 gives ( 4Acos 2x 4Bsin 2x)+( 2Asin 2x+2Bcos 2x) 2(Acos 2x+Bsin 2x)=sin 2x Thus y ( x ) = 1 3 cos 2x+ sin 2x and the general solution is 20 20 p 2 1 1 3 1 1 x 2x 1 y(x)=c e +c e x cos 2x sin 2x . But 1=y(0)=c +c and 1 2 1 2 4 20 2 4 20 20 1 3 17 1 / 0=y (0)=c 2c c= and c = . Thus the solution to the initial value problem is 1 2 2 10 1 15 2 6 17 x 1 2x 1 1 1 3 y(x)= e+ e x cos 2x sin 2x . 15 6 2 4 20 20 11. y (x)=c e c 1 x/4 +c e 2 x . Try y (x)=Ae . Then 10Ae =e , so A= p x x x 1 and the general solution is 10 3 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations 1 x e . The solutions are all composed of exponential curves and with the 1 2 10 exception of the particular solution (which approaches 0 as x ), they all approach either 1 x as x . As x , all solutions are asymptotic to y = e . p 10 y(x)=c e x/4 +c e + x or 12. The auxiliary equation is (2r+1)(r+1)=0 , so r= 1 , 2y / / / p p 1 x and y (x)=c e +c e c 1 2 2 / / / x/2 . For +3y +y=1 , try y (x)=A ; substituting gives y (x)=1 . For 2y 1 1 +3y +y=cos 2x try y =Acos 2x+Bsin 2x p 2 y = 2Asin 2x+2Bcos 2x , y p 2 / / / p 2 = 4Acos 2x 4Bsin 2x . Substituting into the differential equation gives cos 2x= ( 6B 7A) cos 2x+ ( 7B 6A) sin 2x . 7 6 , B= . Thus, Then solving the equations 6B 7A=1 and 7B 6A=0 gives A= 85 85 7 6 7 6 x x/2 y ( x) = cos 2x+ sin and 2x the general solution is y ( x ) =c e +c e +1 cos 2x+ sin 2x 1 2 85 85 85 85 p 2 . The graph shows y =y +y and several other solutions. Notice that all solutions are asymptotic to y p p 1 p 2 p as x . 13. Here y (x)=c cos 3x+c sin 3x . For y c 1 2 2 2 / / +9y=e 2x try y (x)=Ae p 1 2x and for y / / +9y=x sin x try 2 y (x)=(Bx +Cx+D)cos x+(Ex +Fx+G)sin x . Thus a trial solution is p p 2 y (x)=y (x)+y (x)=Ae +(Bx +Cx+D)cos x+(Ex +Fx+G)sin x . p 1 2x 2 2 p 2 4 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations 14. Since y (x)=c +c e c 1 2 9x , try y (x)=(Ax+B)e cos x+(Cx+D)e sin x . p x x 15. Here y (x)=c +c e c 1 2 9x . For y / / +9y =1 try y (x)=Ax (since y=A is a solution to the p 1 / complementary equation) and for y x 4x / / +9y =xe 3 / 9x try y (x)=(Bx+C)e p 2 9x . 16. Since y (x)=c e +c e c 1 2 try y (x)=x(Ax +Bx +Cx+D)e so that no term of y (x) satisfies the p p 2 x complementary equation. 17. Since y (x)=e (c cos 3x+c sin 3x) we try y (x)=x(Ax +Bx+C)e cos 3x+x(Dx +Ex+F)e sin 3x c 1 2 p x 2 x 2 x (so that no term of y is a solution of the complementary equation). p 18. Here y (x)=c cos 2x+c sin 2x . For y c 1 2 p / / +4y=e 3x try y (x)=Ae p 1 3x and for y / / +4y=xsin 2x try y (x)=x(Bx+C)cos 2x+x(Dx+E)sin 2x (so that no term of y is a solution of the complementary 2 p 2 equation). 19. (a) The complementary solution is y (x)=c cos 2x+c sin 2x . A particular solution is of the form c 1 2 y (x)=Ax+B . Thus, 4Ax+4B=x p A= 1 and B=0 4 y (x)= p 1 x . Thus, the general solution is 4 1 x. 4 (b) In (a), y (x)=c cos 2x+c sin 2x , so set y =cos 2x , y =sin 2x . Then y=y +y =c cos 2x+c sin 2x+ c p 1 2 c 1 2 2 2 1 2 y y / 1 2 y y =2cos 2x+2sin 2x=2 so u = 2 1 1 / / 1 xsin 2x 2 1 1 1 / 1 xsin 2xdx= xcos 2x+ sin 2x and u = xcos 2x 1 2 2 2 4 2 1 1 1 u (x)= xcos 2xdx= xsin 2x+ cos 2x . Hence 2 2 4 2 1 1 1 1 1 y (x)= xcos 2x+ sin 2x cos 2x+ xsin 2x+ cos 2x sin 2x= x . Thus p 4 2 4 2 4 1 y(x)=y (x)+y (x)=c cos 2x+c sin 2x+ x . c p 1 2 4 u (x)= 20. (a) Here r 3r+2=0 2 r=1 or 2 and 5 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations y (x)=c e +c e . We try a particular solution of the form y (x)=Acos x+Bsin x c 2x x and y 1 / / 2 p y = Asin x+Bcos x p / p = Acos x Bsin x . Then the equation y / / 3y +2y=sin x becomes A= 3 1 and B= . Thus, 10 10 / (A 3B)cos x+(B+3A)sin x=sin x y (x)= A 3B=0 and B+3A=1 3 1 cos x+ sin x . Therefore, the general solution is p 10 10 1 2x x 3 y(x)=y (x)+y (x)=c e +c e + cos x+ sin x . c p 1 2 10 10 (b) From (a) we know that y (x)=c e +c e . Setting y =e c 1 2 1 2x x 2x , y =e , we have 2 / x y y / 1 2 y y =e 2 1 2x / 3x 2e = e 1 e 5 3x 3x . Thus u = 1 / sin xe e 3x x =sin xe 2x and u = 2 sin xe 2x u (x)= e 1 sin xdx= 2x ( 2sin x cos x ) and u2(x)= e 1 x x e sin xdx= e ( sin x cos x) . Thus 2 3x = sin xe x . Then 1 1 1 3 ( 2sin x cos x)+ (sin x+cos x)= sin x+ cos x and the general solution is p 5 2 10 10 3 2x x 1 y(x)=y (x)+y (x)=c e +c e + sin x+ cos x . c p 1 2 10 10 y (x)= 21. (a) r r=r(r 1)=0 2 r=0 , 1 , so the complementary solution is y (x)=c e +c xe . A particular c 2x p x x solution is of the form y (x)=Ae c . Thus 4Ae x 1 2x 4Ae +Ae =e x 2x x 2x 2x 2x Ae =e 2x 1 2x 2 A=1 y (x)=e p 2x . So a general solution is y(x)=y (x)+y (x)=c e +c xe +e p 2 x x x / c 1 2 1 2 x . / x 1 2 (b) From (a), y (x)=c e +c xe , so set y =e , y =xe . Then, y y u = xe 1 / x y y =e (1+x) xe =e 2 1 / 2x 2x 2x and so 2x 2x u ( x) = 1 xe dx= (x 1)e and u =e 2 c p x x u (x)= e dx=e . Hence y ( x ) =(1 x)e +xe =e 2 p x 2x 2 x 2x and the general solution is y(x)=y (x)+y (x)=c e +c xe +e 1 x . 22. (a) Here r 2r+1=(r+1) =0 x p 2 2 r= 1 and y (x)=c +c e and so we try a particular solution of the c 1 2 / / x form y (x)=Axe . Thus, after calculating the necessary derivatives, we get y Ae (2+x) Ae (1+x)=e x x x x p x y =e 1 / x x x A=1 . Thus y (x)=xe and the general solution is y(x)=c +c e +xe . x / c x x 1 2 1 2 1 2 (b) From (a) we know that y (x)=c +c e , so setting y =1 , y =e , then y y u = e /e = e and u =e /e =1 . Then u (x)= 1 2 1 / 2x x x / x x 2 y y =e 0=e . Thus 2 1 x x p / 2 x x e dx= e and u (x)=x . Thus y (x)= e +xe and the 6 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations general solution is y(x)=c +c e e +xe =c +c e +xe . 1 2 1 3 x x x x x 23. As in Example 6, y (x)=c sin x+c cos x , so set y =sin x , y =cos x . Then c 1 y y 2 / 1 2 sec xcos x =1 y y = sin x cos x= 1 , so u = 2 1 1 1 / 2 2 2 / 1 2 u (x)=x and u = 1 2 / sec xsin x = tan x 1 u (x)= . Hence y (x)=xsin x+cos xln (cos x) and the p 2 general solution is y(x)=(c +x)sin x+ c +ln (cos x) cos x . 1 2 tan xdx=ln cos x =ln (cos x) on 0<x< 24. Setting y =sin x , y =cos x , then y y 1 2 / / 1 2 cot xcos x cos x y y = sin x cos x= 1 . Thus u = = 2 1 1 1 sin x / 2 2 / 2 2 cot xsin x cos x = cos x . Then u (x)= dx= (csc x sin x)dx=ln (csc x cot x)+cos x and and u = 1 2 1 sin x u (x)= sin x . Thus y (x)=[cos x+ln (csc x cot x)]sin x+( sin x)(cos x) and the general solution is 2 p y(x)=c sin x+c cos x+sin xln (csc x cot x) . 1 2 25. y =e , y =e 1 2 x 2x and y y / 1 2 y y =e 2 1 x / / 3x . So u = 1 / e 2x x 3x = e x x and (1+e )e = x 1+e so u (x)= 1 e x x dx=ln (1+e ) . u = 2 e x x 3x e 3x x 2x 1+e e 3x x (1+e )e x e +e x u (x)= 2 e +e 2x dx=ln e +1 e x x e =ln (1+e ) e x x . Hence y (x)=e ln (1+e )+e [ln (1+e ) e ] p x x 2x x x 2x x x and the general solution is y(x)=[c +ln (1+e )]e +[c e +ln (1+e )]e 1 2 x 2x / / 3x / . x 26. y =e 1 / , y =e 2 x x and y y x 1 2 y y = e 2 1 . So u = 1 x x (sin e )e e 3x x 2x =e sin e and x x u = 2 (sin e )e e 3x 2x = e sin e . Hence u ( x ) = e sin e dx= cos e and 1 x x x x x x 2x p 2x u (x)= 2 e sin e dx=e cos e sin e . Then y (x)= e cos e e x x x x x 2x 1 2 / [sin e e cos e ] and the general x x x solution is y(x)=(c cos e )e +[c sin e +e cos e ]e 27. y =e 1 x . , y =e and y y 2 x / 1 2 y y =2 . So 2 1 7 Stewart Calculus ET 5e 0534393217;17. Second-Order Differential Equations; 17.2 Nonhomogeneous Linear Equations e / e u = ,u = and 1 2 2x 2x / x x y (x)= e p x e x e dx+e dx . Hence the general solution is 2x 2x e dx 2x , y =xe 2 2x x x x y(x)= c 1 e + x e c+ dx 2 2x / x e . e 2x 3 x 28. y =e 1 / 2x and y y 1 2 y y =e 2 1 / 4x . Then u = 1 / xe 4x 2x = 2x 1 x 2 so u (x)=x 1 1 and xe xe 2x 2 u = 2 e 2x 3 e 2x 4x = 1 x 3 so u (x)= 2 1 2x 2 . Thus y (x)= p e 2x xe 2x x = 2x e and the general solution is 2x y(x)=e [c +c x+1/(2x)] . 1 2 8
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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
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
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books I-IVSpring 2008Virgil's Aeneid, strongly modeled on Homer's Iliad and Odyssey, was considered even in antiquity the great epic of Rome. It tells the story of the Trojan Aeneas who, afte
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books V-VIIISpring 2008Book V provides a transition between the high emotion of Book IV and the sombre majesty of the descent to the underworld in Book VI. Most of the book is taken up with g
FSU - CLA - 2123
Classics 2123: The Roman Way Notes to Virgil's Aeneid, Books IX-XIISpring 2008Book IX. War finally breaks out, the full-scale battles spoken of in Book VII. The book divides into three sections: (1) Turnus and the Rutulians attack the Trojan ship
FSU - CLA - 2123
Classics 2123: The Roman Way Lecture 5. Roman Imperialism and Expansion Readings: BHR 44-67; Shelton 291-293 I. Roman Imperialism - definition of `imperialism' - older notion of `defensive imperialism': how realistic? - was Rome more warlike/aggressi
FSU - CLA - 2123
Classics 2123: The Roman Way Roman Names A Roman had three names: praenomen (first name) nomen (name of the gens or clan) cognomen (family branch) Thus for: Publius praenomen Cornelius nomen Scipio cognomenSpring 2008his given name was &quot;Publius,&quot;
Stevens - PEP - 111
AC CIRCUITS35.1. Model: A phasor is a vector that rotates counterclockwise around the origin at angular frequency w. Solve: (a) Refemng to the phasor in Figure Ex35.1, the phase angle isU? = 180'n rad - 30&quot; = 150 x -= 2.618 rad180&quot;w=2*618ra
Stevens - PEP - 111
15.1. Solve: The density of the liquid is=m 0.120 kg 0.120 kg = = = 1200 kg m 3 V 100 mL 100 10 -3 10 -3 m 3Assess: The liquid's density is more than that of water (1000 kg/m3) and is a reasonable number.15.2. Solve: The volume of the helium