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# Lesson 7: Vector-valued functions

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We define vector-valued functions, whose image is a curve in the plane or space. We also show how to differentiate and integrate them

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### Lesson 7: Vector-valued functions

1. 1. Sections 10.1–2 Vector-Valued Functions and Curves in Space Derivatives and Integrals of Vector-Valued Functions Math 21a February 20, 2008 Announcements Problem Sessions: Monday, 8:30, SC 103b (Sophie) Thursday, 7:30, SC 103b (Jeremy) Oﬃce hours Wednesday 2/20 2–4pm SC 323.
2. 2. Outline Vector-valued functions Derivatives of vector-valued functions Integrals of vector-valued functions
3. 3. Recall −→ −→ If P and Q are two points in the plane, u = OP, and v = OQ, then the line through P and Q can be parametrized as r(t) = tv + (1 − t)u
4. 4. Recall −→ −→ If P and Q are two points in the plane, u = OP, and v = OQ, then the line through P and Q can be parametrized as r(t) = tv + (1 − t)u This is a function whose domain is R and whose range is a subset of R3 (the line).
5. 5. Deﬁnition A vector-valued function or vector function is a function r(t) whose domain is a set of real numbers and whose range is a set of vectors. We can split r(t) into its components r(t) = f (t)i + g (t)j + h(t)k Then f , g , and h are called the component functions of r. The range of r is a curve in R2 or R3 .
6. 6. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve.
7. 7. r(t) = r(t) = sin(t)i + 2 cos(t)j y t r(t) 0 2j π/2 i r(π/4) π −2j x 3π/2 −i 2π 2j
8. 8. Curves and functions Example Two particles travel along the space curves r1 (t) = 3t, 7t − 12, t 2 r2 (t) = 4t − 3, t 2 , 5t − 6 Do the particles collide?
9. 9. Curves and functions Example Two particles travel along the space curves r1 (t) = 3t, 7t − 12, t 2 r2 (t) = 4t − 3, t 2 , 5t − 6 Do the particles collide? Answer Yes. r1 (3) = r2 (3).
10. 10. Outline Vector-valued functions Derivatives of vector-valued functions Integrals of vector-valued functions
11. 11. Derivatives of vector-valued functions Deﬁnition Let r be a vector function. The limit of r at a point a is deﬁned componentwise: lim r(t) = lim f (t), lim g (t), lim h(t) t→a t→a t→a t→a The derivative of r is deﬁned in much the same way as it is for real-valued functions: dr r(t + h) − r(t) = r (t) = lim dt h→0 h
12. 12. Example Given r(t) = t, cos 2t, sin 2t , ﬁnd r (t).
13. 13. Example Given r(t) = t, cos 2t, sin 2t , ﬁnd r (t). Answer 1, −2 sin 2t, 2 cos(2t)
14. 14. Fact If r(t) = f (t), g (t), h(t) , then r (t) = f (t), g (t), h (t)
15. 15. Fact If r(t) = f (t), g (t), h(t) , then r (t) = f (t), g (t), h (t) Proof. Follow your nose: r(t + h) − r(t) r (t) = lim h→0 h 1 = lim [ f (t + η), g (t + η), h(t + η) − f (t), g (t), h(t) ] η→0 η 1 = lim [ f (t + η) − f (t), g (t + η) − g (t), h(t + η) − h(t) ] η→0 η f (t + η) − f (t) g (t + η) − g (t) h(t + η) − h(t) = lim , lim , lim η→0 η η→0 η η→0 η = f (t), g (t), h (t)
16. 16. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve. (b) Find r (t)
17. 17. r(t) = r(t) = sin(t)i + 2 cos(t)j r (t) = cos(t)i − 2 sin(t)j y t r(t) 0 2j π/2 i r(π/4) π −2j x 3π/2 −i 2π 2j
18. 18. Example Given the plane curve described by the vector equation r(t) = sin(t)i + 2 cos(t)j (a) Sketch the plane curve. (b) Find r (t) (c) Sketch the position vector r(π/4) and the tangent vector r (π/4).
19. 19. r(t) = r(t) = sin(t)i + 2 cos(t)j r (t) = cos(t)i − 2 sin(t)j y t r(t) 0 2j π/2 i r(π/4) r (π/4) π −2j x 3π/2 −i 2π 2j
20. 20. Rules for diﬀerentiation Theorem Let u and v be diﬀerentiable vector functions, c a scalar, and f a real-valued function. Then: d 1. [u(t) + v(t)] = u (t) + v (t) dt d 2. [cu(t)] = cu (t) dt d 3. [f (t)u(t)] = f (t)u(t) + f (t)u (t) dt d 4. [u(t) · v(t)] = u (t) · v(t) + u(t) · v (t) dt d 5. [u(t) × v(t)] = u (t) × v(t) + u(t) × v (t) dt d 6. [u(f (t))] = f (t)u (f (t)) dt
21. 21. Leibniz rule for cross products Let u = f1 (t), g1 (t), h1 (t) and v = f2 (t), g2 (t), h2 (t) . The ﬁrst component of u(t) × v(t) is (u(t) × v(t)) · i = g1 h2 − g2 h1 Diﬀerentiating gives (u(t) × v(t)) · i = g1 h2 + g1 h2 − g2 h1 − g2 h1 = g1 h2 − g2 h1 + g1 h2 − g2 h1 = (u (t) × v(t)) · i + (u(t) × v (t)) · i = u (t) × v(t) + u(t) × v (t) · i
22. 22. Meet the Mathematician: Isaac Newton English, 1643–1727 Professor at Cambridge (England) Philosophiae Naturalis Principia Mathematica published 1687
23. 23. Meet the Mathematician: Gottfried Leibniz German, 1646–1716 Eminent philosopher as well as mathematician Contemporarily disgraced by the calculus priority dispute
24. 24. Smooth curves Example Which of the following curves are smooth? That is, which curves satisfy the property that r (t) = 0 for all t? (a) r(t) = t 3 , t 4 , t 5 (b) r(t) = t 3 + t, t 4 , t 5 (c) r(t) = cos3 t, sin3 t
25. 25. The ﬁrst curve r(t) = t 3 , t 4 , t 5 has r (t) = 3t 2 , 4t 3 , 5t 4 , and is not smooth at t = 0. z x y Projecting r(t) onto the yz-plane gives y = z 4/5 , which is not diﬀerentiable at 0.
26. 26. If r(t) = t 3 + t, t 4 , t 5 , then r (t) = 3t 2 + 1, 4t 3 , 5t 4 , which is never 0. So this curve is smooth.
27. 27. If r(t) = cos3 t, sin3 t , then r (t) = −3 cos2 (t) sin(t), 3 sin2 (t) cos(t) . This is 0 when cos t = 0 or sin t = 0, i.e., when t = π/2, π, 3π/2, 2π. y x
28. 28. Outline Vector-valued functions Derivatives of vector-valued functions Integrals of vector-valued functions
29. 29. Integrals of vector-valued functions Deﬁnition Let r be a vector function deﬁned on [a, b]. For each whole number n, divide the interval [a, b] into n pieces of equal width ∆t. Choose a point ti∗ on each subinterval and form the Riemann sum n Sn = r(ti∗ ) ∆t i=1 Then deﬁne b n r(t) dt = lim Sn = lim r(ti∗ ) ∆t a n→∞ n→∞ i=1 n n n = lim f (ti∗ ) ∆ti + g (ti∗ ) ∆tj + h(ti∗ ) ∆ n→∞ i=1 i=1 i=1 b b b = f (t) dt i + g (t) dt j + h(t) dt a a a
30. 30. Example Given r(t) = t, cos 2t, sin 2t , ﬁnd π r(t) dt 0
31. 31. Example Given r(t) = t, cos 2t, sin 2t , ﬁnd π r(t) dt 0 Answer π2 , 0, 0 2
32. 32. FTC for vector functions Theorem (Second Fundamental Theorem of Calculus) If r(t) = R (t), then b r(t) dt = R(t) a
33. 33. FTC for vector functions Theorem (Second Fundamental Theorem of Calculus) If r(t) = R (t), then b r(t) dt = R(t) a Proof. Let R(t) = F (t), G (t), H(t) . To say that R (t) = r(t) means that F = f , G = g , and H = h. That and the componentwise b deﬁnition of r(t) dt are all you need. a