( k) Integrating by using (8.2) and taking the constant of integration to be 0: ( ) C = m. v dv = v = 0.1 t = +

Transcription

1 Solutions 13 1 Complete solutions to Miscellaneous Exercise 13 Throughout these solutions I hae assumed the argument, x, in ln () x to be positie. 1. Rearranging gies d( pv) dm C = pv m Integrating by using (8.2) and taking the constant of integration to be 0: C Cln ( pv) = ln ( m) gies ln ( pv) = ln ( m ) Taking exponentials of both sides pv ( ) C = m 2. Substituting for acceleration, a, we hae 2 = Separating ariables 2 = which gies = 2 = t+ C, thus = t+c Putting t = 0, = 10 gies 0.1 = C. Hence 1 = t 0.1 multiplying by 1 gies 1 = 0.1 t 1 = 0.1 t 3. We hae =k2. Separating ariables gies =k () 2 We can rewrite this and integrate 2 2 = ( k), = ( k) Thus 1 = kt + C Substituting the gien initial condition t = 0, = 0.1 gies C =10. Therefore 1 1 kt 10, multiplying through by 1 gies kt 10 = = + 1 We hae =. The graphs for different alues of k are kt k = k = 1 k = 10 0 t The larger the alue of k the larger the drop in elocity,, with time,t. du (8.2) ln u u =

2 Solutions Separating ariables ( k) = Integrating, = () k, gies ln ( ) = kt+ C (*) Substituting t = 0, = V 0 yields Putting this, ln ( V 0 )= C, into (*) ln ln ( V0 ) = C ( ) ln ( V ) =kt 0 ln = kt V0 Taking exponentials gies = e kt. Multiplying both sides by V 0 gies = V0e kt V 0 5. We hae dp P =k dv. Integrating by (8.2) yields V ln P = kln V + C rearranging ln P + kln V = C ( ) ( ) ( ) ( ) By using the properties of logs we hae k ln P + ln V = C ( ) ( ) k ln ( PV ) k C Taking exponentials gies the required result PV = e = constant 6. We can rearrange the gien differential equation to the same differential equation of question 5 and sole as aboe. 7. Substituting = 240sin( 100t) and L = gies 3 di di 240 ( 1 10 ) = 240sin ( 100 t) rearranging = sin 3 ( 100t) di = ( ) sin ( 100t ) Integrating both sides 3 i = ( ) sin ( 100t) 3 cos( 100t ) = ( ) + C = 2400cos( 100t) + C = i by (8.39) 100 Using the initial condition t = 0, i = 0 gies C = 0, C = 2400 Substituting C = 2400 into i = 2400cos( 100t)+ C yields the required result = C ( t) ( ) i = 2400cos = cos 100t du (8.2) = ln u u (8.39) sin kt =cos kt ( ) ( ) k

3 Solutions Subtracting V from both sides we hae RC dv = E V. Separating ariables RC dv = E V dv Integrating RC = E V gies Substituting t = 0, V = 0 Putting this into (*) RC ln E V = t RC ln E RCln ( E V) = t+ C (*) RCln ( E) = C ( ) ( ) ln ( ) ln ( ) ln ( ) ln ( t = RC EV RC E = RC EV E) t EV = ln RC E Taking exponentials t t EV RC RC e = gies Ee = E V E Hence V = E Ee trc = E( 1 e trc ). 9. (i) Diiding both sides by 100 gies = Separating ariables = Integrating, = 2 0.1, by using (8.42) on the left hand side we hae 10 ln ( 2 0.1) = t+ C Substituting t = 0, = 0 gies 10ln 2 10 ln ( 2 0.1) = t10 ln ( 2) ( )= C. We hae t = 10 ln ( 2 0.1) 10 ln ( 2) = 10 ln ( 2 0.1) ln ( 2) Diiding both sides by 10 and using the properties of logs gies t = ln we hae 0.1t = ln ( 10.05) 10 2 Taking exponentials of both sides 0.1t 0.1t ln( 10.05) 1 e 0.1t e = e = rearranging = = 20( 1 e ) t (ii) As t, e 0 so 20. (iii) Plotting = 20 1 e 0.1t gies ( ) (8.42) ( ) ( x) f x dx = ln f x f ( )

4 Solutions (1 e -0.1t ) t 10. Diiding through by m gies = g k. Separating ariables m = k g m Integrating g k = gies m m k ln g k m = t + C by (8.42 ) Using the initial condition t = 0, = 0 gies m k ln ()= g C. So we hae m k m ln g = t ln ( g) k m k m k m m k t = ln g ln g = ln g ln k m k k m k g kt ln m = m g ( ) ( g) g k k = ln = ln 1 g mg mg Taking exponentials of both sides k k mg e = 1, = 1 e rearranging = 1e mg mg k As t, e kt m 0, hence mg k. f ( x) (8.42) dx = ln f() x f x ( ) kt m kt m kt m ()

5 Solutions Diiding the gien differential equation by m and separating ariables k 2 k 2 g, = = g m m We can factorize the bracket term as g k m 2 = k mg m k 2. Thus k mg 2 k = rearranging = m k mg (*) 2 m k Need to integrate both sides. How do we integrate the left hand side? du By using the hint in question = 1 u 2 a 2 2a ln u a. Denominator in (*) is u + a 2 mg k mg 2 = 2, let α = mg we hae mg k k k 2 = α 2 2 =( 2 α 2 ) Integrating (*) becomes k = 2 α. Using hint 2 m 1 α k α k ln = t+ C, ln = 2 α t 2 αc 2α + α m + α m Taking exponentials of both sides gies k k k α 2α t2αc 2α t 2α t m m 2αC m 2αC = e = e e = Ae where A= e (constant) + α The index can be simplified as 2 k 2 mg k 2 gk 2 where gk α = = = β β = m k m m m using rules of indices Putting this into the aboe we hae α 2βt = Ae + α Rearranging to make the subject gies 2βt 2βt 2βt α = Ae ( + α) = Ae + αae 2βt 2βt 2βt 2 t Ae = α + αae factorizing ( 1 Ae ) = α( 1+ Ae β ) 2βt 2βt 1+ Ae mg 1+ Ae Hence we hae the required result = α 2βt = 2βt 1Ae k 1Ae. As t, e 2βt 0, so mg k. 12. Diiding by L gies di + R L i = t. Using the integrating factor method, L i ( I.F. ) = ( I.F. ) f ( t), yields R R t L L t I.F. = e = e we hae ie = e (*) L t How do we find e d L t?

6 Solutions 13 6 Use integration by parts formula, (8.45), with u = t L = e e Le u = 1 L = e = = R L R by (8.41) Applying (8.45) we hae t t Le Le 1 e = L L R R L te e t L by (8.41) = + C = e C 2 + R R( R L) R R Substituting into (*) gies t L ie = e C 2 + R R Diiding through by e yields e t L C t L i= C 2 + = + e 2 e R R e R R Using the initial condition t = 0, i = 0 gies C = LR 2. Hence we hae t L L t L i = + e = ( e 1) R R R R R 13. Let be the oltage across the capacitor, then ( )i + = 5e t cos( 100πt ) (*) where i is the current flowing through the circuit. We hae i = C 6 = ( ) Substituting this into (*) gies 3 6 t ( )( ) + = 5e cos( 100πt) 3 t ( ) = 5e cos( 100πt) t = e cos ( 100πt) Diiding through by L we hae di + R L i = E L cos ( ω t ). Using the integrating factor method, with I.F.= e we hae E ie = e cos ( ω t) ( ) L How do we find [ e cos( ω t) ]? (8.41) kt kt e e = k u = u u (8.45) ( )

7 Solutions 13 7 Use (8.33) e R e cos( ωt) = cos 2 ( ωt) + ωsin ( t) 2 ω + C + ω L ( RL) 2 Le R R ω L L = cos( ωt) + ωsin ( ωt) + C + Substituting this into ( ) gies 2 E Le R CE ie = cos ( ωt) + ωsin ( ωt) + LR + ω L L L Ee CE = Rcos( ωt) + ωlsin ( ωt) R ω L L multiplying by L Diiding through by e yields E CE i Rcos ( t) Lsin ( t) e = ω + ω ω R + ω L + L Substituting t = 0, i = 0 0 = ER CE LR, C R + ω L + L R + ω L = The last term on the right hand side of (*) becomes CE LR E E e = e = Re L R + ω L L R + ω L Thus putting this into (*) and factorizing E i = Rcos ( ωt) + ωlsin ( ωt) Re R ω L + For Maple Solutions see next page. ( ) (*) au e a + b (8.33) au e cos( bu) du = acos( bu) + bsin ( bu) 2 2

8 Solutions We hae de_15:=diff(y(x),x)=(x^2)+(y(x)^2); d de_15 := = dx y( x ) x2 + y( x) 2 soln:=dsole({de_15,y(.5)=0},y(x)): e_soln:=dsole({de_15,y(0.5)=0},y(x),type=numeric,method= classical,output=array([1,1.2,1.4,1.6,1.8]),stepsize=0.1) ; [ x, y( x) ] e_soln := rk_soln:=dsole({de_15,y(0.5)=0},y(x),type=numeric,method =classical[rk4],output=array([1,1.2,1.4,1.6,1.8]),stepsiz e=0.1); [ x, y( x) ] rk_soln := with(plots): p_1:=plots[odeplot](e_soln,[x,y(x)],linestyle=4,color=blu e): p_2:=plots[odeplot](rk_soln,[x,y(x)],linestyle=6,color=bl ack): p_3:=plot(rhs(soln),x=1..1.8): plots[display]({p_1,p_2,p_3});

9 Solutions 13 9 Clearly the graphs show that the Runge-Kutta solution is closer to the actual solution then the Euler solution. 16. The Maple solution is de_16:=diff(y(x),x)=(x^2)+(y(x)*exp(x)); d de_16 := = dx y( x ) x2 + y( x) e x soln:=dsole({de_16,y(0)=-1},y(x)): e_soln:=dsole({de_16,y(0)=- 1},y(x),type=numeric,method=classical[heunform],alue=arr ay([0.2,0.4,0.6,1.2]),stepsize=0.2); Warning, the 'alue' option is deprecated, use 'output' instead [ x, y( x) ] e_soln := rk_soln:=dsole({de_16,y(0)=- 1},y(x),type=numeric,method=classical[rk4],alue=array([0.2,0.4,0.6,1.2]),stepsize=0.2); Warning, the 'alue' option is deprecated, use 'output' instead

10 Solutions with(plots): Warning, the name changecoords has been redefined [ x, y( x) ] rk_soln := p_1:=plots[odeplot](e_soln,[x,y(x)],linestyle=4,color=bla ck): p_2:=plots[odeplot](rk_soln,[x,y(x)],linestyle=6,color=bl ue): p_3:=plot(rhs(soln),x= ): plots[display]({p_1,p_2,p_3});

11 Solutions We hae de_17:=(2*10^(-3))*diff(i(t),t)+(1*10^3)*i(t)=10; 1 de_17 := d + = 500 i( t ) 1000 i( t ) 10 soln:=dsole({de_17,i(0)=0},i(t)); 1 1 soln := i( t ) = e ( t) plot(rhs(soln),t=0..10*10^(-6),labels=['t','i']);

12 Solutions We hae de_18:=(10*10^3)*c*diff((t),t)+(t)=9; de_18 := C d + = ( t ) ( t ) 9 sol:=dsole({de_18,(0)=0},(t)); sol := ( t ) = 9 9e C_ls:=seq(10*i*10^(-6),i=1..5); 1/ t C C_ls :=,,,, solns:={seq(subs({c=c_ls},rhs(sol)),c=c_ls)}; solns := { 9 9e ( t ), 9 9e ( 5 t), 9 9e ( 52t / ), 9 9e ( 2 t ), 9 9e ( 10/ 3 t ) f_1:=op(1,solns); f_2:=op(2,solns); 10 } f_1 := 9 9e ( 10 t ) f_2 := 9 9e ( 5 t ) f_3:=op(3,solns); f_4:=op(4,solns); f_5:=op(5,solns); f_3 := 9 9e ( 52t / ) f_4 := 9 9e ( 2 t ) f_5 := 9 9e ( 10/ 3 t)

13 Solutions plot({f_1,f_2,f_3,f_4,f_5},t=0..1.5,color=[blue,green,red,coral,black],labels=['t','']); (i) As t gets large the oltage tends to 9 olt, (the 'final alue'). ()The graphs show that the smaller the alue of C the quicker the oltage () reaches its final alue of 9 olt. 19. The Maple solution is de:=(10*10^3)*i(t)+l*diff(i(t),t)=9; de := i( t) + L d = i( t ) 9 sol:=dsole({de,i(0)=0},i(t)); sol := i( t ) = e L_ls:=seq(1/2^k,k=1..5); L_ls :=,,,, t L solns:={seq(subs({l=l_ls},rhs(sol)),l=l_ls)}; solns e ( t) := {, e ( t), e ( t), e ( t) } e ( t), p_1:=op(1,solns); p_2:=op(2,solns); p_1 := e ( t)

14 Solutions p_3:=op(3,solns); p_4:=op(4,solns); p_5:=op(5,solns); p_2 := e ( t) p_3 := e ( t) p_4 := e ( t) p_5 := e ( t) plot({p_1,p_2,p_3,p_4,p_5},t= ,color=[blue,green, red,coral,black],labels=['t','i']);

15 Solutions (i) AS t gets larger i tends to 0.9 ma () The graph shows that the larger the inductance alue (L), the longer it takes the current (i) to reach its final alue.