Lesson 8: Curves, Arc Length, Acceleration

Slide 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.

Slide 2: Outline Arc length Velocity

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 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

Slide 12: Example Example Find the length of the parabola y = x 2 from x = 0 to x = 1.

Slide 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

Slide 14: Example Find the length of the curve r(t) = 2 sin t, 5t, 2 cos t for −10 ≤ t ≤ 10.

Slide 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

Slide 16: Outline Arc length Velocity

Slide 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).

Slide 18: Example Find the velocity, acceleration, and speed of a particle with position function r(t) = 2 sin t, 5t, 2 cos t

Slide 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)

Slide 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?

Slide 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

Slide 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?

Slide 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

Slide 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!