Math 1320-9 Notes of 2/28/20 10.4 Velocity and Accelerationpa/1320/C30.pdf · Math 1320-9 Notes of 2/28/20 10.4 Velocity and Acceleration • This is another busy section! • We

Math 1320-9 Notes of 2/28/20

10.4 Velocity and Acceleration

• This is another busy section!

• We already discussed interpretation of a space curve asgiving our location at time t. Di!erentiating once or twicegives us velocity and acceleration. We (and the textbook)use the following notation.

r(t) = < f(t), g(t), h(t) > locationr!(t) = v(t) = < f !(t), g!(t), h!(t) > velocity

r!!(t) = v!(t) = a(t) = < f !!(t), g!!(t), h!!(t) > acceleration

• We also use the word speed for the norm of velocity anddenote it by v(t):

v(t) = !v(t)! =!"

f !(t)#2

+"

g!(t)#2

+"

h!(t)#2

Math 1320-9 Notes of 2/28/20 page 1

Uniform Circular Motion

• Assume a particle of mass m moves in a circle of radius a:

r(t) = a(cos!t)i+ a(sin!t)j

• ! is called the angular speed, i.e., the derivative of theangle with respect to time.

• Compute the force

F(t) = ma(t)

acting on the particle (by Newton’s Second Law of Mo-tion).

• So we get

F(t) = ma(t) = "am!2 < cos!t, sin!t >= "a!2r(t)


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Components of Acceleration

• When driving a car you can feel forces in the direction ofthe movement, and perpendicular to the movement. Theforce F = ma is the sum of these two forces. It is oftenuseful to be able to break the sum into its orthogonalconstituents.

• Let us define the unit tangent vector T, the unit normalvector N, and the binormal vector B

T =r!

!r!!, N =

T!

!T!!, and B = T#N.

• We know that these vectors are pairwise orthogonal unitvectors that form a local right handed coordinate system.

• Why is this?


T T ITo T To T TT T 2T T 1 0

• We will see that

a = v!T+ "v2N. (1)

where " is the curvature of the space curve, i.e., the recip-rocal of the radius of the osculating circle.

• First some notes on (1):

" The acceleration is contained in the plane spanned by Tand T! (the osculating plane).

" If " = 0, i.e., the curve is a line, we get the same equationas in the scalar case which we discussed in 1310.

" If we move at a constant speed, i.e., v! = 0, we get theterm we got for a circle, since in that case, with " = 1/awe have

r!!(t) = "a!2(cos!ti+ sin!tj)

= ""a2!2(cos!ti+ sin!tj)

= ""v2(cos!ti+ sin!tj)

" In other words, the formula (1) breaks the accelerationinto a part due to moving straight, and part due to movingin a circle. Very cool!

• So why is (1) true?

• On Wednesday we introduced the concept of curvature.It’s the derivative of the unit tangent vector with respectto arc length. It’s also equal to the reciprocal of the oscu-lating circle, and we obtained several formulas for it:

" =

$$$$

dT

ds

$$$$=

$$$$

T!(t)

s!(t)

$$$$=

!T!(t)!

!r!(t)!=

!T!(t)!

v=

!r!(t)# r!!(t)!

!r!(t)!3.

• The equation

" =!T!!

v


implies that!T!! = "v.

• Di!erentiating inv = vT

we get

a = v!T+ vT! = v!T+ v !T!!N% &' (

=!vN

= v!T+ "v2N

which is the formula (1) that we want.

• It would be nice to have expressions involving only r andits derivatives for the coe"cients

aT = v! and aN = "v2.

• To get aT note that

v · a = vT · (v!T+ "v2N)

= vv!T ·T+ "v3T ·N

= vv!

since T ·T = 1 and T ·N = 0.

• Hence

aT = v! =v · a

v=

r! · r!!

!r!!.

• To get aN we use the formula

" =!r! # r!!!

!r!!3

which gives

aN = "v2 =!r! # r!!!

!r!!


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• This gives us the formula

a =r! · r!!

!r!!T+

!r! # r!!!

!r!!N

. (2)

• Perhaps the following observations help to understand thisformula. Let # denote the angle between the velocity andthe acceleration vectors. Then, since

r! · r!! = !r!!!r!!! cos # and !r! # r!!! = !r!!!r!!! sin #

the equation (2) turns into

a = !r!!!(cos #)T+ !r!!!(sin #)N

or, perhaps more strikingly,

a = r!! = !r!!!

)

(cos #)T+ (sin #)N

*

(3)

• The acceleration r!! is proportional to the force that isapplied to our moving object. That acceleration, andthat force, is split into orthogonal components just likethe components of any vector on the unit circle.

• If velocity and acceleration point in the same direction,i.e., # = 0, then we keep going straight, and if velocityand acceleration are orthogonal then the acceleration inthe direction of our trajectory is zero and our speed alongthat trajectory is constant.

! We followed the textbook discussion and ended up withequation (3) (which is not in the textbook). In retrospect,we could argue that (3) applies to any vector r!!, regardlessof its physical interpretation. Thus we could have startedwith (3) and worked backwards, using the definitions ofthe dot and cross products in terms of the angle #.


• Example 7: A particle moves with position function

r(t) =< t2, t2, t3 >

Find the tangential and normal components of accelera-tion.


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Kepler’s Laws

1. Planets move in elliptical orbits, with the sun at one of itsfoci.

2. A line from the planet to the sun sweeps out equal areasin equal times.

3. The square of a planet’s orbital period is proportional tothe cube of its mean distance from the sun.

• Johannes Kepler (1571–1630) found these laws empirically,using data accumulated by his teacher Tycho Brahe (1546-1601). Isaac Newton (1642-1727) later derived these lawsmathematically using his law of motion (F = ma) andhis law of gravity (gravity is inversely proportional to thesquare of distance), using the Calculus that he had in-vented (as had, independently, Gottfried Leibniz, 1646-1716).

• The textbook has a lengthy derivation of Kepler’s first lawin section 10.4. It has hints on deriving the second andthird laws on page 726.

• In these notes, we’ll derive the second law, which is muchsimpler, using the approach in the Math 2210 textbook.


Kepler’s Second Law

2. A line from the planet to the sun sweeps out equal areasin equal times.

Figure 1. Kepler’s Second Law.

from: https://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Kepler’s law 2 en.svg/2000px-Kepler’s law 2 en.svg.png


• Let A(t) denote the area swept out by time t, startingwith A(0) = 0. We will show that A!(t) is constant. Thismeans precisely that the area swept out in a time intervalof given length is constant.

• Place the sun in the origin, let r(t) denote the positionvector of planet at time t, and let r(t+#t) be its position#t time units later. Let #A be the area swept out in thetime interval [t, t+#t].

• #A is approximately half the area of the parallelogramformed by r(t) and r(t+#t). This is the same as the areaof the triangle formed by r(t) and

#r = r(t+#t)" r(t)

which is half the magnitude of r(t)# r(t+#t):

#A $1

2!r(t)##r!.

• Dividing by #t:

#A

#t$

1

2

$$$$r(t)#

#r

#t

$$$$.

• Taking the limit as #t "% 0 gives

dA

dt=

1

2!r(t)# r!(t)!.

• The only force acting on the planet is the gravitationalattraction of the sun which is given by

F = "GMm

!r!2r

!r!

where M is the mass of the sun, m the mass of the planet,and G the gravity constant.

• By Newton’s Law of motion


"GMm

!r!2r

!r!= ma(t) = mr!!(t),

or, after dividing by m and simplifying:

r!!(t) = "GM

!r!3r.

• Now let’s compute

d

dtr(t)# r!(t) = r(t)# r!!(t) + r!(t)# r!(t)

% &' (

=0

="GM

2!r(t)!3r(t)# r(t)

= 0.

• This implies that r(t)#r!(t), and hence its norm and A!(t),are constant, as required.

! Note that this argument actually would have worked forany situation where r!! has the same direction as r.


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Math 1320-9 Notes of 2/28/20 10.4 Velocity and Accelerationpa/1320/C30.pdf · Math 1320-9 Notes of 2/28/20 10.4 Velocity and Acceleration • This is another busy section! • We