Calculus: Early Transcendentals, by Anton, 7th Edition,ch03

Calculus - Early Transcendentals

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70 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) (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) 3. From the figure: 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 . 4. t s t 0 t 1 5. It is a straight line with slope equal to the velocity. 6. (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) 7. (a) m sec = f (4) f (3) 4 3 = (4) 2 / 2 (3) 2 / 2 1 = 7 2 (b) m tan = lim x 1 3 f ( x 1 ) f (3) x 1 3 = lim x 1 3 x 2 1 / 2 9 / 2 x 1 3 = lim x 1 3 x 2 1 9 2( x 1 3) = lim x 1 3 ( x 1 + 3)( x 1 3) 2( x 1 3) = lim x 1 3 x 1 + 3 2 = 3
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Exercise Set 3.1 71 (c) 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 / 2 x 2 0 / 2 x 1 x 0 = lim x 1 x 0 x 2 1 x 2 0 2( x 1 x 0 ) = lim x 1 x 0 x 1 + x 0 2 = x 0 (d) x y Tangent Secant 5 10 8. (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 9. (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 10. (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
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72 Chapter 3 (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 11. (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 2 0 x 1 x 0 = lim x 1 x 0 ( x 1 + x 0 ) = 2 x 0 (b) m tan = 2(2) = 4 12. (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 + 3 x 1 + 2) ( x 2 0 + 3 x 0 + 2) x 1 x 0 = lim x 1 x 0 ( x 2 1 x 2 0 ) + 3( x 1 x 0 ) x 1 x 0 = lim x 1 x 0 ( x 1 + x 0 + 3) = 2 x 0 + 3 (b) m tan = 2(2) + 3 = 7 13. (a) m
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