chapter_03

# chapter_03 - 11:43 L24-ch03 Sheet number 1 Page number 75...

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January 27, 2005 11:43 L24-ch03 Sheet number 1 Page number 75 black 75 CHAPTER 3 The Derivative EXERCISE SET 3.1 1. (a) m tan = (50 10) / (15 5) =40 / 10 = 4 m/s (b) t (s) 4 10 20 v (m/s) 2. (a) m tan (90 0) / (10 2) =90 / 8 =11 . 25 m/s (b) m tan (140 0) / (10 4) = 140 / 6 23 . 33 m/s 3. (a) m tan = (600 0) / (20 2 . 2) = 600 / 17 . 8 33 . 71 m/s (b) m tan (820 600) / (20 16) = 220 / 4 =55m/s The speed is increasing with time. 4. (a) (10 10) / (3 0) = 0 cm/s (b) t =0, t = 2, and t =4 . 2 (horizontal tangent line) (c) maximum: t = 1 (slope > 0) minimum: t = 3 (slope < 0) (d) (3 18) / (4 2) = 7 . 5 cm/s (slope of estimated tangent line to curve at t =3) 5. From the ±gure: t s t 0 t 1 t 2 (a) The particle is moving faster at time t 0 because the slope of the tangent to the curve at t 0 is greater than that at t 2 . (b) The initial velocity is 0 because the slope of a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases from t 0 to t 1 . (d) The particle is slowing down because the slope decreases as t increases from t 1 to t 2 . 6. t s t 0 t 1 7. It is a straight line with slope equal to the velocity. 8. (a) decreasing (slope of tangent line decreases with increasing time) (b) increasing (slope of tangent line increases with increasing time) (c) increasing (slope of tangent line increases with increasing time) (d) decreasing (slope of tangent line decreases with increasing time)

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January 27, 2005 11:43 L24-ch03 Sheet number 2 Page number 76 black 76 Chapter 3 9. (a) m sec = f (1) f (0) 1 0 = 2 1 =2 (b) m tan = lim x 1 0 f ( x 1 ) f (0) x 1 0 = lim x 1 0 2 x 2 1 0 x 1 0 = lim x 1 0 2 x 1 =0 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 2 x 2 1 2 x 2 0 x 1 x 0 = lim x 1 x 0 (2 x 1 +2 x 0 ) =4 x 0 (d) –2 –1 1 2 1 2 3 x y Secant Tangent 10. (a) m sec = f (2) f (1) 2 1 = 2 3 1 3 1 =7 (b) m tan = lim x 1 1 f ( x 1 ) f (1) x 1 1 = lim x 1 1 x 3 1 1 x 1 1 = lim x 1 1 ( x 1 1)( x 2 1 + x 1 +1) x 1 1 = lim x 1 1 ( x 2 1 + x 1 +1)=3 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 x 3 1 x 3 0 x 1 x 0 = lim x 1 x 0 ( x 2 1 + x 1 x 0 + x 2 0 ) =3 x 2 0 (d) x y Secant Tangent 5 9 11. (a) m sec = f (3) f (2) 3 2 = 1 / 3 1 / 2 1 = 1 6 (b) m tan = lim x 1 2 f ( x 1 ) f (2) x 1 2 = lim x 1 2 1 /x 1 1 / 2 x 1 2 = lim x 1 2 2 x 1 2 x 1 ( x 1 2) = lim x 1 2 1 2 x 1 = 1 4 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 1 /x 1 1 /x 0 x 1 x 0 = lim x 1 x 0 x 0 x 1 x 0 x 1 ( x 1 x 0 ) = lim x 1 x 0 1 x 0 x 1 = 1 x 2 0 (d) x y Secant Tangent 1 4
January 27, 2005 11:43 L24-ch03 Sheet number 3 Page number 77 black Exercise Set 3.1 77 12. (a) m sec = f (2) f (1) 2 1 = 1 / 4 1 1 = 3 4 (b) m tan = lim x 1 1 f ( x 1 ) f (1) x 1 1 = lim x 1 1 1 /x 2 1 1 x 1 1 = lim x 1 1 1 x 2 1 x 2 1 ( x 1 1) = lim x 1 1 ( x 1 +1) x 2 1 = 2 (c) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 1 /x 2 1 1 /x 2 0 x 1 x 0 = lim x 1 x 0 x 2 0 x 2 1 x 2 0 x 2 1 ( x 1 x 0 ) = lim x 1 x 0 ( x 1 + x 0 ) x 2 0 x 2 1 = 2 x 3 0 (d) x y Tangent Secant 1 2 13. (a) m tan = lim x 1 x 0 f ( x 1 ) f ( x 0 ) x 1 x 0 = lim x 1 x 0 ( x 2 1 1) ( x 2 0 1) x 1 x 0 = lim x 1 x 0 ( x 2 1 x

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## This note was uploaded on 06/10/2010 for the course MATH 200-177 taught by Professor Richardwhite during the Spring '10 term at Drexel.

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chapter_03 - 11:43 L24-ch03 Sheet number 1 Page number 75...

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