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CHAPTER 9.1 ~ 9.4. Vector Calculus. 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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CHAPTER 9.1 ~ 9.4 Vector Calculus
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 • IntroductionA parametric curve in space or space curve is a set of ordered (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 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 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 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) limta r(t)exists, and (iii) limta r(t) = r(a). DEFINITION 9.3 Derivative of Vector Function The derivative of a vector function r is (2) for all t which the limits exists.
THEOREM 9.2 Proof If , where f, g, and h are Differentiable, then Differentiation of a Components
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,and the corresponding curve is called a smooth curve.
Geometric Interpretation of r’(t) See Fig 9.5.
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 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
Example 6 • If r(t) = (t3 – 2t2)i + 4tj + e-tk, thenr’(t) = (3t2 – 4t)i + 4j−e-tk, andr”(t) = (6t – 4)i + e-tk.
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 If r(s) = cos2si + sin2sj + e–3sk, s = t4, then
THEOREM 9.4 If r1 and r2 be differentiable vector functions and u(t) A differentiable scalar function. (i) (ii) (iii) (iv) Chain Rule
Example 8 If r(t) = 6t2i + 4e–2t j + 8cos 4t k, thenwhere c = c1i + c2j + c3k.
Length of a Space Curve • If r(t) = f(t)i + g(t)j + h(t)k, then the length of this smooth curve is (3)
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 • Velocity and AccelerationConsider the position vector r(t) = f(t)i + g(t)j + h(t)k, then
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.
Note: ‖v(t)‖2 = c2or v‧v = c2a(t)‧v(t) = 0
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 • See Fig 9.9. For circular motion, a(t) is called the centripetal acceleration. • Fig 9.9
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 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) 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) (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 • Unit TangentWe know r’(t) is a tangent vector to the 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.
Rewrite (3) asthat is, (4) DEFINITION 9.4 Curvature From (2) we have T = dr/ds, then the curvature of C at a point is (3)
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 • 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 tangential and normal components.
