Merced College; Don Power

 

MULTIVARIATE CALCULUS - CH 12, LECTURE

 

VECTOR VALUED FUNCTIONS

 

12.1  Introduction to Vector Valued Functions

 


For Maple work:  use the commands with(plots);  followed by ?tubeplot   Copy the first example from the tubeplot help file into your worksheet and modify it;  Recall, for p, use Pi.

 

Intro:  Consider the vector form for the equation of a line:

      Example:  <x,y,z> = <3+t, -1+2t, 2t>

      Note that x, y, and z are linear functions of the parameter t

      We write x(t), y(t), z(t) instead of x,y,z to highlight this fact

      Denote the vector <x,y,z> as r, then r is also a function of t and we write r(t)

      In function notation, the eqn of the line becomes:

            r(t) = <x(t),y(t),z(t)> = <3+t, -1+2t, 2t>

                  or r(t) = x(t)i + y(t)j + z(t)k = <(3+t)i + (-1+2t)j + (2t)k

 

Extension to curves in space:

      We still have r(t) = <x(t),y(t),z(t)> or x(t)i + y(t)j + z(t)k

      But now one or more of the component functions are not linear.

 

Helix

      Ex:  r(t) = <3sin(t),3cos(t),t>

            Projection onto xy-plane is a circle (verify)

            As t ranges from 0 to 2p, z-coord increases linearly from 0 to 2p

            Resulting shape is like a spring

 

Curves in space as intersections of two surfaces

      Ex:  Twisted cubic as intersection of cylinders z=x^3 and y=x^2:  Pick x as parameter

            Illustration in text pg 861

      Simult eqns:  intersec of z=x^2 + y^2 and z=8-x^2-y^2

 

Domain / Natural domain

 

Maple example:

 

Using set notation, you can plot two or more space curves on the same screen:

 

 

12.2  Calculus of Vector-Valued Functions

 

 

Geometric interpretation of limits:  r must approach L in both length and direction

epsilon-delta def of limit:  interval about the target point is a sphere of radius epsilon

Calculate limits component-wise

Def of continuity at a point - as expected

Def of derivative r'(t);  domain of the deriv

Thm 6 Deriv exists iff deriv of each component function exists; calculate deriv component-wise

Thm 7:  Deriv of const, scalar mult, sums, differences as expected

            Deriv of scalar func * vector func as expected by product rule

Tangent vectors and tangent lines at a point r0 - Formula 4

      Notation v0 for r'(t0)

Deriv of dot products and cross products - expected, but watch order of func for cross product

Thm 9:  if the norm of r is const, then r dot r' is 0, that is r and r' are orthogonal

      Curves on the surface of circles, spheres

Integrals

      Indef integr still needs +C, now a vector C

      Def integr still by Fund Thm of Calc

      Thm 10 - scalar mult, sums, diffs, as expected

Solving differential equation with initial condition - as expected

 

12.3  Change of Parameter, Arc Length

 

 

r(t) is smooth if r'(t) is (1) continuous and (2) non-zero

i.e. we must not have any t for which all two/three components are 0 simultaneously

Ex:  r(t) = <cos t , cos2t> for 0£t£2p in spite of fact that it is y=x2 ("corners"at ends of graph)

Arc Length

      Review formula for parameterized functions

      Extend to R3

      Note that it's the same as L = integral of norm of dr/dt on a£t£b

            Ex (#7) <cos3t,sin3t> on 0£t£p/2

      Arc length parameterization

            Set upper limit to t

            integrate:  s is a function of t; solve for t and replace t in original function

            Same example

Chain rule

      If t=g(τ) is a change of parameter in which g is differentiable with respect to τ,

      then dr/dτ = dr/dt * dt/dτ

 

12.4  Unit Tangent, Normal, and Binormal Vectors

 

 

 

Def:  unit tangent vector:  T(t) = r'(t) / ||r'(t)||

      Ex 1:  Find T for r'(t) = <t2,t3>

            a.  Find T(2).  Note:  it's easiest to find r'(2) and normalize the result without finding T(t) first

            b.  Find T(t)

Geometry:  where would accel vector r''(t) go? 

      if t marks off increasing increments of arc length (that is, r' is increasing)

      if t marks off decreasing increments of arc length

      if t marks off equal increments of arc length

      note:  Points toward inside of curve

 

Def:  [principal] unit normal vector:  N(t) = T'(t) / ||T'(t)||

      By using T, which has constant length, we get that N is orthogonal to T

      N points toward inside of the curve

            More specifically, toward the center of the circle of best fit (osculating circle)

      Calculation by the definition can be a challenge (our example:  requires quo rule, with radical)

 

Computing formulas for N:

      These are especially efficient for finding N at a particular point,
            but are also more practical for calculating N(t)

 

      1.   For R2, note that a normal to <v1,v2> is ±<-v2,v2>

            Why?  m1 = -1/m2

            Select the sign that directs vector to inside of the curve

            normalize for N

2.   N = (v´a)´v, normalized          This way avoids all normalization until the end

            Why (geometrically)?

3.   For R2, N = k´v, normalized.  This is the previous formula, with the observation that k is in the same direction as (v´a).

      4.   N = a - projva, normalized       Recall projva has length of a dot v/||v||, direction of v/||v||

Why (geometrically)

 

Binormal vector B = T´N

      App:  this is normal to the "osculating plane," the plane of best fit for a curve

            (next lesson)

      Computing formula:  B = v´a, normalized

 

12.5  Curvature

 

 

 

Def:  magnitude of the rate of change of the unit tangent vector, when parameterized by arc length:

 

      κ = || dT / ds ||  =  || r''(s) ||

 

      What can change when the unit tangent vector changes?  Only the direction.

 

      Why parameterize by arc length?  Example:  Say as t increases from 1 to 2, we travel 1 inch along a curve, and the unit tangent vector changes direction 15 degrees to the left; and as t increases from 2 to 3, we travel 2 inches along a curve, and the unit tangent vector changes direction another 15 degrees to the left.  dT/dt is still 15 degrees per unit change of t, but clearly, the second segment is straighter -- less curved.

 

Fact:  Ex 1 in text shows that a circle of radius a has curvature 1/a.  Therefore:

      1.  The radius and the curvature are reciprocals.

      2.  For any curve, the "radius of curvature ρ" is  ρ = 1/κ.

            This is the radius of the circle of best fit, also known as the "osculating circle"

 

Ex 2 in the text demonstrates that the curvature of a straight line is 0.

 

Thus, higher curvatures are associated with tighter curves.

 

Ex 3 in text calculates the curvature of a helix.  See lab 2.

 

When parameterized by t, the formula becomes κ = || (dT / dt) / (dr / dt) ||  =  || T'(t) / r'(t) ||

 

      Proof:  κ = || dT / ds ||  =  || (dT / dt) / (ds / dt) ||  =  || (dT / dt) / (dr / dt) || 

      The last step comes from differentiating both sides of  s = int ( r(u), with respect to u, from  t0 to t)

            (by the fundamental theorem of calculus)

 

Computing formula for vector functions in general:   or    (Derived in the textbook)

 

Computing formula for "ordinary" functions of a single variable, in the form y = f(x):

 

     

 

Notice the formula summary in the text at the end of this lesson.

 

12.6  Motion Along a Curve

 

 

 

Let r(t) be the position function.  Then:

 

      v(t) = r'(t) is the corresponding velocity function;

      || v(t) || is the speed

      a(t) = v'(t) = r''(t) is the corresponding acceleration function

 

Displacement vs arc length  [integral of velocity vs integral of speed]

 

      Displacement = distance between initial point and teminal point:  || r(t2) - r(t1) ||

      Displacement vector = vector between initial point and teminal point:   r(t2) - r(t1)

            Note that this is the sum of the changes of position, integral of dr/dt, i.e. integral of v(t)

      Arc length is the sum of the changes in the distances traveled along the curve, i.e. of   ||dr/dt||

            So arc length is the integral of ||dr/dt||, i.e. integral of || v(t) ||

 

Normal and tangential components of acceleration:

 

It is often useful to write the acceleration vector as a linear combination of a vector parallel to the tangent vector and a vector parallel to the normal vector:

a = aTT + aNN

aT is the norm of the projection projva:        aT = (v dot a) / ||v||

 

aN can be calculated by:                              aN = ||v cross a|| / ||v||

      or, by the Pythagorean theorem:            aN = sqrt ( ||a||2 - aT2)

 

Notice the relationship between aN and curvature κ:

 

      From the previous lesson                            κ = ||v cross a|| / ||v||3

 

      Therefore, κ = aN / ||v||2;  or  aN = κ ||v||2

 

      and if we write arc length "s" as a function of the parameter t,

            aT = d2s/dt2

            aN = κ (ds/dt)2        since ||v|| = ds/dt

 

Projectile Motion:

 

      Parametric equations:

      x  = (v0 cosθ)t

      y  = –1/2 gt2 + (v0 sinθ)t + s0

 

Ex:  A shell is fired from ground level with a muzzle speed of 980 ft/sec and an elevation angle of 30o.  Find:

Equation(s) for the path of the shell

Max height

Range (horizontal distance)

Speed of the shell at the max height

 

Solution, per Maple:

 

> r:=[490*sqrt(3)*t,-16*t^2+490*t];

> subs(t=490/32,r);evalf(%);

> r2:=diff(r,t);

  So at max height, t = 490/32

> subs(t=490/32,r2);evalf(%);

> with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

 

> solve(-16*t^2+490*t=0,t);evalf(%);

> subs(t=245/8,490*sqrt(3)*t);evalf(%);

Let us modify the problem:  with the same muzzle velocity, what angle of elevation would be needed to hit a target at a range of 20,000 ft.?

 

Maple solution:

 

> r:=[980*cos(theta)*t,-16*t^2+980*sin(theta)*t];

> rs:=solve(r[2]=0,t);

> rt:=subs(t=rs[2],r[1]);

> angle:=solve(rt=20000,theta):evalf(%);

>

> angle*180/Pi:evalf(%);

>

 Note:  either of the two positive angles will solve the gunnery problem.

 

 

 

12.7  Kepler's Laws of Planetary Motion

 

Caution:  work through the complete proof before attempting the homework

 

The central point of this lesson is to show that Kepler's laws of planetary motion can be derived from Newton's law of universal gravitation.  Historically, Kepler discovered his laws first, from Tycho Brahe's observations of the motions of the planets.  Newton subsequently showed that all three laws would result from a single inverse-square law of gravitation (along with his second law of motion).

 

Kepler's Laws:

      1.   Each planet has an elliptical orbit, with the sun at one focus.

      2.   A ray from the sun to a planet sweeps out equal areas in equal times

3.   The square of a planet's period (its "year") is proportional to the cube of the semimajor axis of its ellipse

 

Newton's Laws, in vector form

      Motion:  F = ma

      Gravitation:  ||F|| = GMm / ||r||2

 

Kepler's First Law

 

The proof starts by combining the Newtonian laws to get a as a scalar multiple of r:

Since the only force operating on the planet is gravitation, r and F have opposite directions.

Hence, F = ||F|| (−r / ||r|| ), and

ma = F = (GMm / ||r||2) (−r / ||r|| )

Solving for a, and defining r = ||r||, we get a = (−GM/r3) r

 

Preliminary:  the orbit of the planet must lie in a plane containing the sun:

We can prove this by demonstrating that  rXv = constant vector.

The derivative of (rXv), using the product rule, is rXa + vXv

Again, the only force operating on the planet is gravitation, so r and a have opposite directions.

And the cross product of parallel vectors is 0

So, (d/dt) (rXv) = rXa + vXv = 0

Integrating both sides, we get rXv = constant vector, call it b.

Hence, the orbit is in a plane containing the sun.

 

A polar coordinate system is introduced, with the sun (the mass M) at the pole, with q=0 when t=0, with the planet's closest point to the sun along the polar axis (the positive x-axis), and with the motion of the planet given by r = <rcosq,rsinq> = ru, where u  = <cosq,sinq>

 

Initial conditions:

      Position:  we define r0 = ||r(0)||, so that r(0) = <r0cos(0), r0sin(0) = <r0,0> (q = 0 when t = 0)

      Unit vector:  u(0) = <cos(0),sin(0)> = <1,0>

      Velocity:  we define v0 = ||v(0)||, so that v(0) = <0,v0>

      Normal to plane:  b = b(0) (since b is constant) = r0Xv0 = r0v0k, or <0,0, r0v0>
     

The triple scalar product  r . (vXb)  is calculated two ways, and the results equated:

 

      1.  First way:

 

            b = rXv = r0Xv0 = r0v0k or <0,0, r0v0>.  (b is constant, so we use r and v at t=0 to calculate it)

            r . (vXb)  = rXv .b = b .b = (r0v0)2

 

      2.   Second way:  We will show      r . (vXb)  = GMr + r(r0v02 - GM)cosq   .  Derivation:

 

            a.   Combine a = (-GM/r3)r, r = ru, and u  = <cosq,sinq>, to get a = (-GM/r2) <cosq,sinq>.

 

            b.   Combine    v = dr/dt = (d/dt)(ru) = (by product rule) r du/dt + dr/dt u   and...

                                    du/dt = (by chain rule) du/dq dq/dt = <-sinq,cosq> dq/dt   and...

                                    u X du/dt = <cosq,sinq,0> X <-sinq,cosq,0> dq/dt = dq/dt k

                  to get b = r X v = ru X (r du/dt + dr/dt u) (now distribute)

                              = r2u X du/dt + r dr/dt u X u (and since  u X u = 0 ...)

                              = r2u X du/dt

                              = r2 (u X du/dt)

                              = r2 dq/dt k

           

            c.  From a. and b., we get

                  a X b = (-GM/r2) <cosq,sinq,0> X <0, 0, r2 dq/dt> = GM du/dt

                 

            d.  Note that the result in c. is (d/dt) v X b = v X db/dt + dv/dt X b = a X b = GM du/dt

 

            e.   Integrating both sides in d., we get v X b = GMu + C

 

            f.    Apply initial condition (t = 0):

                  v(0) X b(0) = GMu(0) + C

                  <0,v0,0> X <0,0, r0v0> = GM <1,0,0> + C

                  < r0v02,0,0> - <GM,0,0> = C

 

                  We get C = <r0v02 - GM, 0, 0>

 

                  So rewrite the result of e. as v X b = GMu + <r0v02 - GM, 0, 0>

 

            g.   Thus r . (vXb)  = ru . GMu + <r0v02 - GM, 0, 0> = r (GM + r0v02 cosq- GM cosq )

 

      3.  We equate the two expressions for r . (vXb)  from 1. and 2., and solve for r:

            r . (vXb)  = (r0v0)2 = r (GM + r0v02 cosq - GM cosq )

            r = r02v02 / (GM + (r0v02 - GM) cosq)

           

            Force the first term in the denom to equal 1 by dividing both the num and denom by GM:

            r = (r02v02 / GM) / (1 + (r0v02 / GM - 1) cosq)

 

            This is an equation in the polar form of a conic section,

            r = k / (1 + e cosq), where k = r02v02 / GM  and  e = r0v02 / GM - 1

            Note that e is the eccentricity: 

e = 1 for parabola;

e = c/a for ellipse, typically x2/a2 + y2/b2 = 1, with a2 = b2 + c2

e = c/a for hyperbola, typically x2/a2 − y2/b2 = 1, with c2 = a2 + b2

 

Conclusion:  Objects subject only to the sun's gravitation could have trajectories that are:

      elliptical (if 0 < e < 1), parabolic (if e = 1), or hyperbolic (if e >1).

 

Note:  If  e = 1, we get "escape velocity"  (solve for v0)

 

See text for Kepler's 2nd and 3rd laws.

 

Return to:  Merced College; Don Power               Updated 03/01/10 by Don Power