Hw 5 Solution

4. I : 5. 7,20 , 23,24

4. 2 : 1,3,

7, 16,1M

4. I Acceleration and Newton 's second law .

Tdd Ccitt ) t Cuts ] = fad -1 Cet, sint , t'

) t Ce't

,cost

,

-21-3 ]

= I [ Cette 't , sintt cost , - t 's ]= ( et - e-

t

,cost - Sint

,-3T

'

)

Ciel t cict ) = C et,

cost,

3T'

) t C - Et , - sint , - ft'

)

= Cet - it,

cost - sint,

-3T )

:.

We see that Tdd.

Ccitt ) tact ) ] = G'

Ct ) t cict ) .

→ → →

I k

Ci Ct ) x Cdt ) =I

It s.int t 'I

= C - at 's .int - Iast ) ?- C - stet - I e- t ) tcetcost-e-tsint.ITe-

t

cost -283

= ( - zt 's .int - Past,

ztettte - t,

et cost - e- taint )

add Ccitt ) x cuts ]

= ( - Gt 's .int - at'

cost - stcosttt 's .int,

ftettztettzte -t

.He - t

,

etcost-etsintte-ts.int - e- toast )I I I

Ci Ct ) X Catt ) =

yetcost ztz

)= ( - H' cost - start ) I - ( - 2T

' et - 3t2e ' t ) It( etcost - e- test ) I

e- t

cost - zf'

= ( - It'

cost - 3T'

cost,

ztettste - t,

et cost - e- tcost )

Alt ) x Ci CH =I t ,.si?nnt

, !t; ) = ( - lot 's .int - it's - int ) I - C - btettte - t ) It C- etsinti-e-ts.int ) -1

= C - lot 's .int t t 's .int,

ft'

et - te- t

,

- etsint-e-ts.int )

cicttxc.lt ) t C , Ct ) x cict )

= ( - H' cost - st'

cost - Gt 's .int t t 's .int , 2tettzte.tt ft'

et - te - t,

et cost - e- toast - etsint-e-ts.int )

.

.

'

.We have Idf Ccitt ) x Calf ) ) = Ciao ) x act ) t c. it ) x Ci Lt ) .

Suppose Krull achieves its local maximum or minimum at t = to .

Then the function truth'

also achieves its local maximum or minimum at t = to.

C Here we use the fact that Hrctsll is always nonnegative )

It follows that

o = at Clint'll ' ] It -- to

= Fdd [ ret ) . rct ) ] It - - to= [ I Lt ) o ret ) t rft ) . Nlt ) ] I t -- to= [ zit ) ont ) ] It -

- to

= 2 v'

Cto ) . Veto ).

We see that the vector r 't to ) is perpendicular to rcto ) .

c' it > = Ct,

et,

t'

) implies that

Cct ) = ( It 't Ci,

ett Ca,

It 't Cz )

for some constants Ci,

Cz,

and Cs.

Since

Cco ) = ( I o 't C.,

It Ca,

3h03 -1 Cz ) = ( C,

,

It Cz,

Cz ) = ( O , -5 , I ),

we have C , = o,

Cz = - 6,

Cz = I .

.

.

.

Cct ) = ( It'

,et - 6

,It 't I )

.

Let Ict) = Cacti , Yeti , 2- Ct ) ) .zero acceleration means I

"

Ct ) = To,

so we have

o = x' '

Ct ) = y' '

Ct ) = z"

Ct )

at all t where the path is defined . Clearly , x' '

Cti = o implies that

x' et ) is a constant function,

and therefore Xlt ) = ai t t b , for

some constants a , and bi.

In the same way ,we get

yet ) = Azt t be,

2- Ct ) = Ast + b ,

for some constants As .bz,

As,

and bs.

Then

Ict ) = C x Lt ), yet ) , 2- Ct ) )

= ( A , t t bi,

Azt t bz,

Azt t b 3 )= ( Ai , Az , As ) t + C b . , by b , ) .

. : Ict ) is a point if Cai

,as

,As ) = I

,

and a straight lineotherwise

.

4. 2 Arc length

S !"

It c' its Hdt = Jj"

It C - zsint,

zcost,

I > It dt

= Si"

T4 sin 't -14 cos 't t I dt

= Si"

is It - : sin 't -1 cos 't = I

= ( 21T ) rs

S ! H c' it ) Hdt = fo"

It C 3 cos 1st ),

-3 sin Cst ),

3t÷ ) Hdt= so

. T9 cos ' 4Ht9 sin'

Cst ) t 9T It= S

!Ttt dt

= S

!3 It dt

= [ 2 C Itt )÷ ]'

.

= 2 ( z÷ - I )

This path is not smooth b/c yet ) = Itt is not differentiable at o .It is

"

piecewise smooth"

.

In this case,

we compute the arc

length of c as the sum of arc lengths of smooth pieces .

Clt ) =

gC t

,

- t ) if - let to

Ct, t ) if o E t

E I

⇒ c' It ) =y

C I,

- I ) if - let so * c' it ) is not well - defined

( I,

I 7 if o s t a ,at endpoints .

.: LCC ) = f ! Hcl , -1711 dt t to Hcl , 17 Hdt

= f ! Edt t f ! rzdt = 2K.

(a) Note that HTC till'

= Ttt ) . Tft ) has constant value I . Hence,

o = ddt ( TCH . T Ct ) ) = TEE ) o Tct ) t Tct ) o Tht ) = 2 T ' it ) . . T C t ).

.

.

.T

'

Ct ) . TCH = O.

(b) Tct ) = II c' CEN⇒ T

'

et ) = ( c"

it ) It c' Hill - c' it > Edt C Hilts Il ) )mum

I+

ddt Clldctsll ) = ddt C Totti )= , ddt ( Clt ) . c' it ) ) . : chain rule= It ( 2C

"

Ct ) .c'

Ct ) )

= ¥I

H c' Hill

We have

Titi = Tint ( c' ' it ) He 't 'll - c' its 9i )= Y¥ - c

'

its tellc' It ) 113.

It's clear since

Lcc , = fab Has , Hds = fab Ids = b - a.

since Tcs ) = Tig = Cts ),

we have TEs ) = c"

cs ).

:.

k = It Tts , It = Hc "csoH.

Proof of cc ) is on the next page .

Note that

H c' it > It = Ht C - sint , cost , 1) It

= the #C- sin t ) 't C cost )-

+ 12

Tz= I

= I.

We see that the path c is parametrized by arc length . From Cbl ,K= It c' ' call = Htc - cost , - sint , o ) H= Ira -C - cost )

'

t C - sin t )'

=IBe

.

Let Clt ) be a curve defined on [ a ,b3 .Define let ) fat It c' cash doc

↳ let ) is the are length of the part of curve c-

between the points Cco )and act ) .Since Ctx > is never 0

,

It c' call is always positive .

⇒ lets ' is a strictly increasing function .⇒ The inverse function l

-

Is) is well - defined

.

C defined on Eo , Leo ] )

We define a path d : co , Lcc , ] → 1123 as the composition Col"

.

• Ccb )

" "

÷t÷:[ a , b ] l€ C o

, UI , ]t

'

It's easy to see thatc and d have the same image curve ,

and the interesting fact

is H d'

It = I for all s.

( Now you see why unit speed curves are said to be"

parametrized

by arc length"

).

The bottom line is that If c' it ) is never zero then the curve

at ) can be re parametrized to a unit speed curve des ) .For convenience

,we often write Ccs ) instead of des ) .

Here,

s is a new parameter satisfyings = let > = fat It c' call doc

.

WIS : k =l×¥k

First of all ;" " t ' "

. dat - = dats . ddt = Tess Holt 'llt

chain rule"

s = fat Hicxslldx implies ¥ = Hett 'll .⇒ HT 't > It = It Tis > 1111 c' it'll⇒ k = It TEs , It = 4i

.

Cx )-

From exercise lb,

we have

T'

Ct ) = Tic - c' its tellc' It ) 113= i - ai't iii. titer )

= Titi - Tct ) C Titi ° TCH ).

Put V Ct ) : = t ( then Ttt )= Vet ) - ( ult ) . Tch ) Ttt ) by above )

V Lt )

TV Lt )

Tn: ⇒ Ttt )> 7 . >Tv Ttt ) Tn TLE )

Recall that Tct ) is a unit vector .

I = projection of vet ) onto the line spanned by Tct )

= THT Ct )Il Tl till '

= ( ult ) . Tct ) ) TH )

Then T'

Ct ) = Vlt ) - C Vct ) . Tct ) ) Tct ) = Vct ) - Tv = →.

It follows that

H Vlt) x Tct ) H = the area of the parallelogram spanned by Vlt ) and TH )= It TCH 1111 Tht ) It

= It T'

Ct ) It.

'

i HIT till = I

Finally ,

KE,

't = "iY-=l¥÷¥I=

lt€It c' call

'.