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Unformatted text preview: Math 18.02 (Spring 2009): Lecture 6
Velocity & acceleration. Kepler’s second law February 13 Reading 1VIaterial: From Simimms: 17.4 and Lecture Notes K. Last time: Parametric equations for lines and curves
Today: Velocityr .54: acceleration. Derivatives of , x. Kepler’s second law. 2 Velocity and Acceleration We recall that a curve 7 in 3D is described by three parametric equations ‘1: = 1:0!)
9’ : .UU—l
.3 : :(i) where it belongs to an interval [n,b]. We can associate to the point PU] (.1:(t),y(t),:{t)) that
determines the curve the vector position Wen) = Te) = In)? + you“ + 3(t)k, that is the vector that starts at the origin 0 and ends on the point PU) of the curve: it is now clear that in order to describe a curve it is completely equivalent to either give parametric
equations or give the vector position FIG. Deﬁnition 1. Gwen e cm‘ne 7 described by (L vector FU.) = :L'(t)'i + y{l‘.)_;i + :(tlli' fort in [(1.1)], we
define Hie velocity oft—"(H to be the vector (TH) deﬁned us “‘3’ )i + @(nj + {lien} =i~1nyi+y'(t)j + 3(5):}. W) = a“ " I“. in We deﬁne the speed of F(t) to be sltl=17(t)l We usually also write “ ch
GU”) : or even simpler (l1T (ll . The vector (TU) describes the direction and speed of a particle moving following the vector position at). .1: Handy Fact: EU) is tangent, to the curve if placed at the point PU). To see this we jusL write , _ . F(l+Al)iF(l,)
“(Mimi—J— —.7 —_
v ( #61!)  PR) Deﬁnition 2. The acceleration of PU.) is Hui ’UEr:lt)1'ﬁ(l) deﬁned (15 NM? rlcv'rmlwr: of the ocular
ilclucily EU). llml. £5
(117 (12:1: 6 (lgu — (lg; a
fit. :—l :—liI—'l"+—llc.
() clt() (Ill) cll()'i (Lil) Exercise 1. Consider the curve 1r given by the vector position. F'(() = (all  1,t2,l”), for t real
number. F ian velocity, speed and acceleration.
Before starting solving the problem let's observe that the pammelric cquulluns of llrls cmve 7 an: E. .‘L' = ill+1
1 t
 l I
ll )
3 Q By deﬁnilion llmw
Fm = (lt+1)7i+t2j+l3.l‘
l" . A i, 
m) z in) = + 2:4 + 3m;
(ll
all} : lﬁ(l) = v16 + 4152 + 9:11 (Hf  n
7t =—t =2" no dto Mom [0 Consider now a different kind ol' question: Exercise 2. Comtrier the parabola { X Foal a tangent [me to the parabola. at the point Pu : (2, 2,11).
Solution: We ﬁrst notice that Pu : 13(2). Since we know that the velocity vector ﬁt) 75 always tangent to the curve at the point PU) we compute ‘17 : 17(2). We have an) = E H + 2a.
hence ﬂ _ A h
V = 17(2) = i +j + ills.
This vector“ gives as the direction for the tangent line. Since the tangent has to go through P“ we have the parametric equations for the tangent line: L=t+2
y=t+2
3=llt+zlr 3 Derivative rules for  and x products Recall for 1301 the product rule for l'uuCLious 'tt(t)"L'(t) of one variable: cl(uu) I (to _ (to.
dt at (It . Let .‘l[t),5(t) be vector functions of t. Product rules: (like 18.01 rule, but be careful about order when you deal wiLll x product!) {M E} _ (a? a; ﬁ
— = ‘4 . _ + . B
(tt (it {It
(rtT x E) a (18' ca? _
—~—— = 1 — ' _
alt f X m. f at XE Why true? If one expands the vector functions it) = Willi'1' Guilt)" + Bait) 3
so) = b1(t)i2 + ban} + as) and starts computing  or x, then one can see that the resulting object is a sum of products oi" two
scalar functions a,(t) and bj(t). On this one uses the rule from 18.01. 4 Kepler’s second law Nicholas Copernicus c. 15:10 solar system (sun at center)
Tycho Brahe c. 1570 careful observations of planets
Johannes Kepler c. 1605 numerologist (mystic) — found three “laws” linking the data —+ Kepler’s laws (see below)
Isaac Newton c. 1660 Derived Kepler’s laws from gravitation through mathematical reasoning
(calculus) —> revolution in our view of the universe. The universe is understands hie in terms
of mathematics. This is the beginning of physics and astronomy as sciences. I Kepler’s laws 1. The planetary motion is on an ellipsel with sun at focus. (Ellipse: c. + I) : const). P 2. Equal tirne sweeps out equal area. 3. (period)2 proportional to (semi majoraxis)d Goal: Prove 2. assuming only the the central force assumption: T he only force on a planet is in
direction of the sun. Theorem 1. The central force assumption implies Kepler’s 2””: law. Proof. To start we consider a 2D Cartesian coordinate system in which the sun is placed at the
origin 0. Let be the position vector of the planet at time 1?. Also We denote with 17(r‘) the lHe originally balanced circles, rejected ovals. velocity %’:(t) and with EU) the acceleration The Central force assumption in our case means that c? : cF for some c (negative here because it is in the direction of the sun!) 7 5,4 :A (Hoe) A(+) Second law =) Area is swept out at constant rate To write this mathematically we define AU) = area stvept out so far,
and we want to HlIOW that
(M t t (1 1)
—— : cons ,an '. :.
cit Let AA : area swept out in interval t to ti At: .. ,7 >
YﬁCf'Pbt'h‘fCtJ — A Y 76‘“) \ We)
AA {5 mppwmmﬁai ’96 i531 m 9f i343 Raga Recall that given two VECLDI'H 13" and C1 ‘13 x C' = Area pal"..1l]elogram spanned by B and C'.
So in our case I
AA 2 SiFX AF]. Divide both sides by At and let At a D. We get (1A _1 :x (F _ 1px,l
dz ‘2' (it _‘2’i
We are able to Silo“! (11.1) if we could prove that
ctHFx 5)
— : U. 41.2
Lit ( l (:51 Actually we will prove more: (MTx r7) _ (15 + dF _
—— = r x — — x t:
(I! (if. (H
= 1‘" x r? + 17 X If
\_v_1 \_V../
I) u
central force always
: (1 Remark. Above we proved that the macro!" F x ff does not change an time. As a consequence m
magnitude IF x dues not change m Mum. This as mmctiy Hui conicHI of (4.2) III Study Guide 1. Answer the foHomng questions: 0 Consider thefwncﬁzon of one variable v = "9:2. HS graph, 12; a curve. Curr), you “mitt: pu‘rumelrrc
equations‘ for UN}; curve? Arr: Hwy unique?  Cmtsiu'm' the when: :15") +313 + 32 = 5 and the helm: FU) : (cos Lain 1,, A). Fwd the intersecmm pom/.5 of the where with Um. helm. ‘~l ...
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