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Contents. 9.1 Vector Functions 9.2 Motion in a Curve 9.3 Curvature and Components of Acceleration 9.4 Partial Derivatives. 9.1 Vector Functions.

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Contents
Contents

  • 9.1 Vector Functions

  • 9.2 Motion in a Curve

  • 9.3 Curvature and Components of Acceleration

  • 9.4 Partial Derivatives


9 1 vector functions
9.1 Vector Functions

  • IntroductionA parametric curve in space or space curve is a set of ordered coordinates (x, y, z),wherex = f(t), y = g(t), z = h(t)(1)

  • Vector-Valued FunctionsVectors whose components are functions of t, r(t) = <f(t), g(t)> = f(t)i+ g(t)jor r(t) = <f(t), g(t), h(t)> = f(t)i+ g(t)j + h(t)kare vector functions. See Fig 9.1



Example 1 circular helix
Example 1: Circular Helix

Graph the curve byr(t) = 2cos ti+ 2sin tj + tk, t 0

Solutionx2 + y2 = (2cos t)2 + (2sin t)2 = 22See Fig 9.2. The curve winds upward in spiral or circular helix.



Example 2
Example 2

Graph the curve byr(t) = 2cos ti+ 2sin tj + 3k

Solutionx2 + y2 = (2cos t)2 + (2sin t)2 = 22, z = 3See Fig 9.3.



Example 3
Example 3

Find the vector functions that describes the curve Cof the intersection of y = 2x and z = 9 – x2 – y2.

SolutionLet x = t, then y = 2t, z = 9 – t2 – 4t2 = 9 – 5t2Thus, r(t) = ti+ 2tj +(9 – 5t2)k. See Fig 9.4.



DEFINITION 9.1

Limit of a Vector Function

If exist, then


THEOREM 9.1

If , then (i) , c a scalar(ii) (iii)

Properties of Limits


DEFINITION 9.2

Continuity

A vector function r is said to be continuous at t = a if

(i) r(a) is defined, (ii) limta r(t)exists, and

(iii) limta r(t) = r(a).

DEFINITION 9.3

Derivative of Vector Function

The derivative of a vector function r is (2)

for all t where the limits exists.


THEOREM 9.2

Proof

If , where f, g, and h are

Differentiable, then

Differentiation of Components


Smooth curve
Smooth Curve

  • When the component functions of r have continuous first derivatives and r’(t) 0 for t in the interval (a, b), then ris said to be a smooth function in (a, b),and the corresponding curve is called a smooth curve.




Example 4
Example 4

Graph the curve by r(t) = cos 2t i + sin t j, 0  t  2. Graph r’(0) and r’(/6).

Solutionx = cos 2t, y = sin t, then x = 1 – 2y2, −1  x  1andr’(t) = −2sin 2ti + cos tj,r’(0) = j, r’(/6) =



Example 5
Example 5

Find the tangent line to x = t2, y = t2 – t, z = −7t at t = 3

Solutionx’ = 2t, y’ = 2t – 1, z’ = −7 when t = 3,and r(3) = 9i + 6j – 21kthat is P(9, 6, –21), then we havex = 9 + 6t, y = 6 + 5t, z = –21 – 7t


THEOREM 9.3

If r is a differentiable vector function and s = u(t) is a

differentiable scalar function, then the derivatives ofr(s) with respect to t is

Chain Rule


Example 7
Example 7

If r(s) = cos2si + sin2sj + e–3sk, s = t4, then


THEOREM 9.4

If r1 and r2 are differentiable vector functions and u(t)

is a differentiable scalar function.

(i)

(ii)

(iii)

(iv)

Chain Rule



Example 8
Example 8

If r(t) = 6t2i + 4e–2t j + 8cos 4t k, thenwhere c = c1i + c2j + c3k.


Length of a space curve
Length of a Space Curve

  • If r(t) = f(t)i + g(t)j + h(t)k is a smooth function, then the length of this smooth curve over (a,b) is (3)


Arc length as a parameter
Arc Length as a Parameter

  • A curve in the plane or in space can be parameterized in terms of the arc length, r(s).

  • ||r’(s)||=1, i.e. r’(s) is a unit tangent vector. (hint: the length of the curve from r(0) to r(s) is s).


Example 9
Example 9

Consider the curve in Example 1. Since , from (3) the length from r(0) to r(t) is

Using then (4)Thus


9 2 motion on a curve
9.2 Motion on a Curve

  • Velocity and AccelerationConsider the position vector r(t) = f(t)i + g(t)j + h(t)k, then


Example 1
Example 1

Position vector: r(t) = t2i + tj + (5t/2)k. Graph the curve defined by r(t) and v(2), a(2).

Solutionso that See Fig 9.7.



Particle moves with constant speed
Particle moves with constant speed

‖v(t)‖2 = c2or v‧v = c2a(t)‧v(t) = 0


Example 21
Example 2

Consider the position vector in Example 2 of Sec 9.1. Graph the velocity and acceleration at t = /4.

SolutionRecall r(t) = 2cos ti + 2sin tj + 3k.then v(t) = −2sin ti + 2cos tja(t) = −2cos ti −2sin tjand



Centripetal acceleration
Centripetal acceleration

  • See Fig 9.9. For circular motion, a(t) is called the centripetal acceleration.

  • Fig 9.9


Curvilinear motion in the plane
Curvilinear Motion in the Plane

  • See Fig 9.10. Acceleration of gravity : −gjAn initial velocity: v0 = v0 cos i + v0 sin j from an initial height s0 = s0j, then where v(0) = v0, then c1 = v0. Thereforev(t) = (v0cos )i + (– gt + v0sin )j


Integrating again and using r(0) = s0, Hence we have (1) See Fig 9.11




Example 31
Example 3

A shell is fired from ground level with v0 = 768 ft/s at an angle of elevation 30 degree. Find (a) the vector function and the parametric equations of the trajectory, (b) the maximum attitude attained, (c) the range of the shell (d) the speed of impact.

Solution(a) Initially we have s0= 0,and (2)


Example 3 2
Example 3 (2)

Since a(t) = −32j and using (2) gives (3)Integrating again,Hence the trajectory is (4)

(b) From (4), we see that dy/dt = 0 when −32t + 384 = 0 or t = 12. Thus the maximum height H is H = y(12) = – 16(12)2 + 384(12) = 2304 ft


Example 3 3
Example 3 (3)

(c) From (4) we see that y(t) = 0 when −16t(t – 24) = 0, or t = 0, 24.Then the range R is

(d) from (3), we obtain the impact speed of the shell


9 3 curvature and components of acceleration
9.3 Curvature and Components of Acceleration

  • Unit TangentWe know r’(t) is a tangent vector to the smooth curve C, then (1)is a unit tangent. Since the curve is smooth, we also have ds/dt = ||r’(t)|| > 0. Hence (2)See Fig 9.19.



DEFINITION 9.4

Curvature

From (2) we have T = dr/ds, then the curvature of C

at a point is (3)


Example 11
Example 1

Find the curvature of a circle of radius a.

SolutionWe already know the equation of a circle isr(t) = a cos ti + a sin tj, then We getThus, (5)



Tangential and normal components
Tangential and Normal Components

  • Since Tis a unit tangent, then v(t) = ||v(t)||T = vT, then (6)Since T  T = 1 so that T  dT/dt = 0 (Theorem 9.4),we have T and dT/dt are orthogonal. If ||dT/dt||  0, then (7)is a unit normal vector to C at a point P with the direction given by dT/dt. See Fig 9.18.



  • The vector N is also called the principal normal. However  =║dT / dt║/ v, from (7) we havedT/dt = vN. Thus (6) becomes (8)By writing (8) as a(t) = aNN + aTT (9)Thus the scalar functions aN and aT are called the normaland tangential components.


The binormal
The Binormal

  • A third vector defined byB = T  Nis called the binormal. These three vectors T, N, Bform a right-hand set of mutually orthogonal vectors called the moving trihedral. The plane of T and N is called the osculating plane, the plane of N and B is called the rectifying plane. See Fig 9.19.



Example 22
Example 2

The position vector r(t) = 2cos ti + 2sin tj + 3tk, find the vectors T, Nand B, and the curvature.

SolutionSincefrom (1),Next we have


Example 2 2
Example 2 (2)

Hence (7) gives N = – cos ti – sin tj

Now, Finally using and


Example 2 3
Example 2 (3)

From (4) we have


Formula for a t a n and curvature
Formula for aT, aN and Curvature

  • Observethen (10) On the other hand


Since ||B|| = 1, it follows that (11)then (12)


Example 32
Example 3

The position vector r(t) = ti + ½t2j + (1/3)t3k is said to be a “twisted” cube”. Find the tangential and normal components of the acceleration at t. Find the curvature.

SolutionSince v  a = t + 2t3 and From (10),


Example 3 21
Example 3 (2)

NowandFrom (11) From (12)


Radius of curvatures
Radius of Curvatures

  •  = 1/ is called the radius of curvature.

  • See Fig 9.20.


9 4 partial derivatives
9.4 Partial Derivatives

  • Functions of Two VariablesSee Fig 9.21. The graph of a function z = f(x, y) is a surface in 3-space.



Level curves
Level Curves

  • The curves defined by f(x, y) = c are called the level curves of f. See Fig 9.22.


Example 12
Example 1

The level curves of f(x, y) = y2 – x2 are defined by y2 – x2 = c. See Fig 9.23. For c = 0, we obtain the lines y = x, y = −x.


Level surfaces
Level Surfaces

  • The level surfaces of w =F(x, y, z) are defined by F(x, y, z) = c.


Example 23
Example 2

Describe the level curves of F(x, y, z) = (x2 + y2)/z.

SolutionFor c  0, (x2 + y2)/z = c, or x2 + y2 = cz. See Fig 9.24. ,



Partial derivatives
Partial Derivatives

  • For y = f(x),For z = f(x, y), (1) (2)


Example 33
Example 3

If z = 4x3y2 – 4x2 + y6 + 1, find ,

Solution


Alternative symbols
Alternative Symbols

  • If z = f(x, y), we have


Higher order and mixed derivatives
Higher-Order and Mixed Derivatives

  • If z = f(x, y), we have:

  • Second-order partial derivatives:

  • Third-order partial derivatives:

  • Mixed second-order partial derivatives:


Alternative symbols1
Alternative Symbols

  • If f has continuous second partial derivatives, thenfxy = fyx(3)


Example 41
Example 4

Ifthen


THEOREM 9.5

If z = f(u, v) is differentiable, and u = g(x, y)and

v = h(x, y) have continuous first partial derivatives,

then(5)

Chain Rule


Example 51
Example 5

If z = u2 – v3, u = e2x – 3y, v = sin(x2 – y2), find and

SolutionSincethen (6) (7)


Special case
Special Case

  • If z = f(u, v) is differentiable, and u = g(t) and v = h(t) are differentiable, then (8)If z = f(u1, u2,…, un) and each variable u1, u2,…, un are functions of x1, x2,…, xk, we have (9)


Similarly, if u1, u2,…, un are functions of a single variable t, then (10)These results can be memorized in terms of a tree diagram. See next page.


Example 6
Example 6

If r = x2 + y5z3and x = uve2s, y = u2 – v2s, z = sin(uvs2), find r/s.

SolutionAccording to the tree diagram,


Example 71
Example 7

If z = u2v3w4and u = t2, v = 5t – 8, w = t3 + t,find dz/dt.

SolutionAnother approach: differentiate z = t4(5t – 8)3(t3 + t)4


Thank You !

Zill工程數學(上)茆政吉譯


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