MA1506 Tutorial 3 Solutions (1a)

( ) x

xxx

exyBAByAy

eBxABeyeBxAy

yyy

3

333

2

t

)21(21310'11)0(

)(3')(3096

eySet 09'6"

−

−−−

+=→=⇒−=−⇒−=

=⇒=+−=→+=→

−=→=++

==++

λλλ

λ

(1b)

]2sin32cos2[322)13(2)13(2)0('

22)0(]2cos22sin2['

]2sin2cos[210)41(2 22

xxeyBBy

AyxBxAeyy

xBxAeyi

x

x

x

ππ

πππ

ππππ

ππ

πλπλλ

+−=⇒=⇒

+−=−⇒−=−=⇒−=

+−+=

+=→

±=→=++−

(2a) Try CBxAxy ++= 2

xx

ACBABAA

xCBxAxBAxA

yyy

−=→

===→=++=+=→

+=

+++++=

++

2

2

2

253102201042510

32510222

102

25y

0C -1,B ,/ , ,

)()('"

(2b) Try xeCBxAxy 32 )( ++=

x

xxxx

xx

exyCCCBCBBA

BBBABAAAAAA

eCBxAxeBAxeBAxAeyeCBxAxeBAxy

32

32333

323

)2(2081869332

0081812966118189

)(9)2(3)2(32")(3)2('

−−=

−=→=+−−+++=→=+−−++

−=→=+−+++++++=

++++=

(2c)

xxxyxxxzxxxixxxixz

iBBiA

xBAxBAxAizzBAx

exyy ix

sincossincosIm]sincos[sincose)(

0221

e2e)2("e)(iAeAiez"

B)ei(AxAez'B)e(AxzTry

above. thesatisfies Imz then 2xez"-zequation complex thesolvecan weifpart)imaginary (Im Im2"

ix

ixix

ixixix

ixix

ix

ix

−−=→−−=−−++−=−−=→

−=→=−−−=→

=−−−−=−

+−+=

++=

+=

=

==−

(2d)

xxzy

xxixixez

iBiB

AA

iBeBxeABxezziBee

ee

exyy

ix

ixixixixix

ixix

ix

2sin81

81Re

))2sin2cos(1(81

81

81

814

21

81

214

44444"4-4Bxz"

BxAzTry 21

214zz" Solve

)Re(21

21)2cos1(

214"

2

22222

22

2

−==

−+=+=

=→=−→

=→=→

+++−=+→+=

+=⇔−=+

−=−=+

(3a)

W

xx

W

xx

xa

RHSRHS

2cos])2cos1(21[

B' ,2sin])2cos1(

21[-

A'

satisfy chosen to are B(x) and A(x) if

)2cos1(214yy" ofsolution isB(x)sin2x )A(x)cos(2xThen

B(x). A(x), functions toB andA PromoteBsin2xAcos2xy04yy"solvefirst :parameters ofVariation

)()(

−+=

−=

−=++

+=→=+

sin2x x constant) (arbitrary cos2x x constant) (arbitrary solution general

the into absorbed be can cos- extra the and sin)cos(

sin)sinsinsincoscos(cos

since (2d) in as same the is which

]sin)sin(sincos)cos[(cosBsin2x2x Acos

is solution the so sinsin

coscos

)(coscos)(coscos'

sinsinsincossin41A'

so2sin2xs2x)'in2x)'-(co(cos2x)x(s Wwhere

+

−−=

−−+−

−−+−=+

−−=→

−=→

+−=−=

+−=+−=

==

xxxx

xxxxxxxx

xxxxxxx

xxxB

xxA

xxxxB

xxxxx

23212

82

411

81

28

2441224

412

81

2441224

412

81

84

3212

81

43212

81

14812

412

412

41

4812

4122

412

22

2

(3b)

xsinx|cosx|lncosx y

xB |,cos|lncosxsinx -A

1 (sinx)(cosx)' - x)'(cosx)(sin w

cosx [sec(x)]B' sinx [sec(x)]-A'

where)sin()(cos)(

+=→

===

+==

+==

+

∫ xdx

ww

xxBxxA

(4)

]2/)'[(' 22 y

dydy

dyd

dxdy

dxy

==2d so the given equation can be written as

]2/)'[( 2ydyd = F(y). Just integrate both sides with respect to y, and you get a

separable first-order ODE. Using this trick in the present case we have

]2/)[(2.

rdrd = - GM/ 2r [Sorry there should be a dot over the r on the left side, but

it’s not very clear.] From the problem statement it is clear that the Earth’s initial speed is zero when r = R, where we use R to denote the original radius of the Earth’s orbit

[150 billion metres]. Using that to fix the arbitrary constant, we get upon integrating both sides with respect to r:

RGM

rGMr −=2/)(

2..

Since the Earth will of course fall inward, the speed must always be negative after the first instant, so when we take the square root we must be careful to take the negative one. Hence

dt

RGM

rGM

dr=

−

−22

, where we have to integrate from r = R [Earth orbit] to r =2R/3

[Venus’ orbit radius, 100 billion metres]. It’s convenient to change the variable from r to x = r/R. You then get

1121

3/2

2/3

−= ∫

x

dxGM

Rt .

This is an improper integral, which is why the website recommended will sometimes refuse to do it [though usually it does work!] If it is in a bad mood, just integrate from 2/3 up to 0.999 or something like that. With the given data you should get about 44.74 days. So that’s how long it takes to reach the orbit of Venus. Remember that a day is 24 times 3600 seconds. 5. Set y = exp (λ t) as usual, and get

33 w−=λ So we need the cube roots of -1. Now - 1 = )exp( πi , )3exp( πi , )5exp( πi so the cube roots are )3/exp( πi , )3/3exp( πi , )3/5exp( πi , which are cos(π /3) + i sin(π /3) = (1/2) + i 2/3 cos(π ) + i sin(π ) = - 1 cos(5π /3) + i sin(5π /3) = (1/2) - i 2/3 . You should verify that the cubes of these three numbers are all equal to – 1. Following the usual rule that real parts yield exponentials, and imaginary parts yield cos and sin, we see that the general solution is

)2/3sin()2/3cos( 2/2/ wtCewtBeAe wtwtwt ++− , where A B C are the arbitrary constants. Notice that unless B and C are both exactly zero, this is not a bounded function. In the case of the solution you are asked to graph using graphmatica, you see that the solution behaves nicely for a while, but soon the oscillations get totally out of control. So apart from very exceptional cases, the solution will blow up --- even ordinary SHM would be unstable if Newton’s 2nd law

had a third derivative instead of a second derivative! Then we would not be here to talk about it.