AP Calculus AB 11/7/2003

Name: ________________________

Notes on §4-6, #35 and 38

Instructions
         Problems 31, 33, 35, and 38, or others of equivalent difficulty, will be included on the quiz on Monday, 11/10/2003. If you cannot understand anything below, or if you cannot achieve the book’s answers for #33 or #35, I will expect to receive a voice mail message from you.

35.	(Note: Your book uses the phrase “increases uniformly with x” to mean “has rate of change proportional to x, with a positive constant of proportionality.” If you do not recall what the terms proportional or constant of proportionality mean, please review your Algebra II textbook.)
38.(a)	Because d(t) has left- and right-hand derivatives that are not numerically equal at t = 0.5, we conclude that d ¢(0.5) does not exist. By the quotient rule, d ¢(t) = 60.5[(0.5 + t)(–1) – (0.5 – t)(1)]/(0.5 + t)^2 = –60.5/(0.5 + t)^2, if t < 0.5 d ¢(t) = d/dt [300 – 150/t] = 0 – (–150t^–2) = 150/t^2, if t > 0.5 Note the use of the strict less-than and greater-than signs.
(b)	d ¢(1) = 150/1^2 = 150 Þ d(t) is continuous at t = 1 Reason: differentiability Þ continuity.
(c)	Since d(t) is continuous from the left, limx®0.5– d ¢(t) = –60.5/(0.5 + 0.5)^2 = –60.5. Real-world meaning: As the ball was about to hit the bat, it was approaching the plate at 60.5 ft/sec. Since d(t) is continuous from the right, limx®0.5+ d ¢(t) = 150/(0.5)2 = 150/(¼) = 600. Real-world meaning: Immediately after the ball hit the bat, it was leaving the plate at 600 ft/sec.
(d)	continuous since lim d(t) = 0 from both left and right, and this limit agrees with both function definitions at t = 0.5 not differentiable since left-hand derivative disagrees with right-hand derivative at t = 0.5
(e)	If we plug in t = 0 into the first equation for d(t), we get d(0) = 60.5 ft. That makes sense, since the ball is released from the pitcher’s mound at time 0, and on a professional baseball diamond, the distance from the pitcher’s mound to home plate is 60 feet, 6 inches.