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# Lesson 8: Curves, Arc Length, Acceleration

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The velocity of a vector function is the absolute value of its tangent vector. The speed of a vector function is the length of its velocity vector, and the arc length (distance traveled) is the integral of speed.

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### Lesson 8: Curves, Arc Length, Acceleration

1. 1. Sections 10.3–4 Curves, Arc Length, and Acceleration Math 21a February 22, 2008 Announcements Problem Sessions: Monday, 8:30, SC 103b (Sophie) Thursday, 7:30, SC 103b (Jeremy) Oﬃce hours Tuesday, Wednesday 2–4pm SC 323.
2. 2. Outline Arc length Velocity
3. 3. Pythagorean length of a line segment Given two points P1 (x1 , y1 ) and P2 (x2 , y2 ), we can use Pythagoras to ﬁnd the distance between them: P2 y y2 − y1 P1 x2 − x1 x |P1 P2 | = (x2 − x1 )2 + (y2 − y1 )2 = (∆x)2 + (∆y )2
4. 4. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y x
5. 5. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y x
6. 6. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y x
7. 7. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y x
8. 8. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y x
9. 9. Length of a curve Break up the curve into pieces, and approximate the arc length with the sum of the lengths of the pieces: y n L≈ (∆xi )2 + (∆yi )2 i=1 x
10. 10. Sum goes to integral If x, y is given by a vector-valued function r(t) = f (t), g (t), with domain [a, b], we can approximate: ∆xi = f (ti )∆ti ∆xi = g (ti )∆ti So n n L≈ (∆xi )2 + (∆yi )2 ≈ [f (ti )∆ti ]2 + [g (ti )∆ti ]2 i=1 i=1 n = [f (ti )]2 + [g (ti )]2 ∆ti i=1 As n → ∞, this converges to b L= [f (t)]2 + [g (t)]2 dt a
11. 11. Sum goes to integral If x, y is given by a vector-valued function r(t) = f (t), g (t), with domain [a, b], we can approximate: ∆xi = f (ti )∆ti ∆xi = g (ti )∆ti So n n L≈ (∆xi )2 + (∆yi )2 ≈ [f (ti )∆ti ]2 + [g (ti )∆ti ]2 i=1 i=1 n = [f (ti )]2 + [g (ti )]2 ∆ti i=1 As n → ∞, this converges to b L= [f (t)]2 + [g (t)]2 dt a In 3D, r(t) = f (t), g (t), h(t) , and b L= [f (t)]2 + [g (t)]2 + [h (t)]2 dt a
12. 12. Example Example Find the length of the parabola y = x 2 from x = 0 to x = 1.
13. 13. Example Example Find the length of the parabola y = x 2 from x = 0 to x = 1. Solution Let r(t) = t, t 2 . Then √ 1 5 1 √ L= 1+ (2t)2 = + ln 2 + 5 0 2 4
14. 14. Example Find the length of the curve r(t) = 2 sin t, 5t, 2 cos t for −10 ≤ t ≤ 10.
15. 15. Example Find the length of the curve r(t) = 2 sin t, 5t, 2 cos t for −10 ≤ t ≤ 10. Answer √ L = 20 29
16. 16. Outline Arc length Velocity
17. 17. Velocity and Acceleration Deﬁnition Let r(t) be a vector-valued function. The velocity v(t) is the derivative r (t) The speed is the length of the derivative |r (t)| The acceleration is the second derivative r (t).
18. 18. Example Find the velocity, acceleration, and speed of a particle with position function r(t) = 2 sin t, 5t, 2 cos t
19. 19. Example Find the velocity, acceleration, and speed of a particle with position function r(t) = 2 sin t, 5t, 2 cos t Answer r (t) = 2 cos(t), 5, −2 sin(t) √ r (t) = 29 r (t) = −2 sin(t), 0, −2 cos(t)
20. 20. Example The position function of a particle is given by r(t) = t 2 , 5t, t 2 − 16t When is the speed a minimum?
21. 21. Example The position function of a particle is given by r(t) = t 2 , 5t, t 2 − 16t When is the speed a minimum? Solution The square of the speed is (2t)2 + 52 + (2t − 16)2 which is minimized when 0 = 8t + 4(2t − 16) =⇒ t = 4
22. 22. Example A batter hits a baseball 3 ft above the ground towards the Green Monster in Fenway Park, which is 37 ft high and 310 ft from home plate (down the left ﬁeld foul line). The ball leaves with a speed of 115 ft/s and at an angle of 50◦ above the horizontal. Is the ball a home run, a wallball double, or is it caught by the left ﬁelder?
23. 23. Solution The position function is given by r(t) = 115 cos(50◦ )t, 3 + 115 sin(50◦ )t − 16t 2 The question is: what is g (t) when f (t) = 310? The equation 310 f (t) = 310 gives t ∗ = , so 115 cos(50◦ ) 2 2 310 310 g (t ∗ ) = 3 + 115 sin(50◦ ) − 16 115 cos(50◦ ) 115 cos(50◦ ) ≈ 91.051 ft
24. 24. Solution The position function is given by r(t) = 115 cos(50◦ )t, 3 + 115 sin(50◦ )t − 16t 2 The question is: what is g (t) when f (t) = 310? The equation 310 f (t) = 310 gives t ∗ = , so 115 cos(50◦ ) 2 2 310 310 g (t ∗ ) = 3 + 115 sin(50◦ ) − 16 115 cos(50◦ ) 115 cos(50◦ ) ≈ 91.051 ft Home run!