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,
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];
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> subs(t=490/32,r);evalf(%);
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> r2:=diff(r,t);
So at max height, t = 490/32
> subs(t=490/32,r2);evalf(%);
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> with(linalg):
Warning, the protected names norm and trace have been
redefined and unprotected
> solve(-16*t^2+490*t=0,t);evalf(%);
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> subs(t=245/8,490*sqrt(3)*t);evalf(%);
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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];
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> rs:=solve(r[2]=0,t);
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> rt:=subs(t=rs[2],r[1]);
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> angle:=solve(rt=20000,theta):evalf(%);
>
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> angle*180/Pi:evalf(%);
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>
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
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>
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