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# Contents - PowerPoint PPT Presentation

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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• 9.1 Vector Functions

• 9.2 Motion in a Curve

• 9.3 Curvature and Components of Acceleration

• 9.4 Partial Derivatives

• 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

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.

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

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

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.

Limit of a Vector Function

If exist, then

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

Properties of Limits

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.

Proof

If , where f, g, and h are

Differentiable, then

Differentiation of Components

• 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.

See Fig 9.5.

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) =

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

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

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

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

is a differentiable scalar function.

(i)

(ii)

(iii)

(iv)

Chain Rule

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

• 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)

• 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).

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

Using then (4)Thus

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

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.

‖v(t)‖2 = c2or v‧v = c2a(t)‧v(t) = 0

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

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

• Fig 9.9

• 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

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)

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

(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

• 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)

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)

• 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.

• 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.

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

SolutionSincefrom (1),Next we have

Hence (7) gives N = – cos ti – sin tj

Now, Finally using and

From (4) we have

Formula for aT, aN and Curvature

• Observethen (10) On the other hand

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

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),

NowandFrom (11) From (12)

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

• See Fig 9.20.

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

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

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.

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

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. ，

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

If z = 4x3y2 – 4x2 + y6 + 1, find ,

Solution

• If z = f(x, y), we have

• If z = f(x, y), we have:

• Second-order partial derivatives:

• Third-order partial derivatives:

• Mixed second-order partial derivatives:

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

Ifthen

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

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

SolutionSincethen (6) (7)

• 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.

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

SolutionAccording to the tree diagram,

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

Zill工程數學(上)茆政吉譯