Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 235: 2

2a) v ( t ) = $\frac{( 81 – 9t² )}{(t²+9)²}$ 2b) The velocity after 1 second is 0.72 feet per second. 2c) The particle is at rest at 3 seconds. 2d) The particle is moving in the positive direction from zero seconds until 3 seconds. 2e) The particle travels 1.8 feet in the first 6 seconds. 2f) See sketch for 2f 2g) a ( t ) = $\frac{18t( t² - 27 )}{(t²+9)³}$ 2g) a ( 1 ) = 0.468 feet per second per second 2h) see screenshot of 2h) 2i) The particle is speeding up on the interval ( 3 , 3 $\sqrt 3$ ). 2i) The particle is slowing down on the interval [ 0 , 3 ) ∪ ( 3 $\sqrt 3$ , ∞).

a) Find the velocity at time ( t ). Quotient Rule. s ‘ = f ‘( t ) = v ( t ) = $\frac{d/dx[9t](t²+9) - d/dx[t²+9](9t)}{(t²+9)²}$ v ( t ) = $\frac{( 9 )(t²+9) - ( 2t )(9t)}{(t²+9)²}$ v ( t ) = $\frac{9t² + 81 – 18t² }{(t²+9)²}$ v ( t ) = $\frac{( 81 – 9t² )}{(t²+9)²}$ 2a) v ( t ) = $\frac{81 – 9t² }{(t²+9)²}$ b) What is the velocity after 1 second? v ( 1 ) = $\frac{81 – 9( 1 )²}{( ( 1 )²+9)²}$ v ( 1 ) = $\frac{( 81 – 9 )}{( 1 + 9 )²}$ v ( 1 ) = $\frac{72}{( 10 )²}$ v ( 1 ) = $\frac{72}{100}$ v ( 1 ) = $\frac{72}{100}$ = 0.72 feet per second. 2b) The velocity after 1 second is 0.72 feet per second. c) When is the particle at rest? v ( t ) = $\frac{81 – 9t² }{(t²+9)²}$ 0 = $\frac{81 – 9t² }{(t²+9)²}$ The denominator can never equal zero so the domain is all real numbers. If the numerator equals zero, the velocity function also equals zero. 0 = 81 – 9t² 0 = 9( 9 – t² ) 0 = 9( 3 + t ) ( 3 - t ) The velocity function is zero at ( – 3 ) seconds and ( 3 ) seconds. 2c) The particle is at rest at 3 seconds. d) When is the particle moving in the positive direction? When is v ( t ) > 0 ? When the signs of the numerator and denominator are both positive. v ( t ) = $\frac{81 – 9t² }{(t²+9)²}$ The denominator (t²+9)² is always positive. The numerator ( 81 – 9t² ) is positive when 0 ≤ t < 3 . The signs of the numerator and denominator are both positive when 0 ≤ t < 3. 2d) The particle is moving in the positive direction from zero seconds until 3 seconds. e) Find the total distance traveled in the first 6 seconds. s = f ( t ) = $\frac{9t}{t²+9}$ s = f ( 0 ) = $\frac{9( 0 )}{( 0 )²+9}$ s = f ( 0 ) = $\frac{0}{9}$ f ( 0 ) = 0 feet s = f ( 3 ) = $\frac{9( 3 )}{( 3 )²+9}$ s = f ( 3 ) = $\frac{27}{18 }$ f ( 3 ) = $\frac{3}{2}$ = 1.5 feet | f ( 3 ) – f ( 0 ) | = | 1.5 – 0 | = 1.5 feet s = f ( 6 ) = $\frac{9( 6 )}{( 6 )²+9}$ s = f ( 6 ) = $\frac{54}{45}$ f ( 6 ) = $\frac{6}{5}$ = 1.2 feet | f ( 6 ) – f ( 3 ) | = | 1.2 – 1.5 | = 0.3 feet The particle travels 1.5 feet in the positive direction, then 0.3 feet in the negative direction, for a total of 1.8 feet in the first 6 seconds. 2e) The particle travels 1.8 feet in the first 6 seconds. 2f) See sketch for 2f) g) Find the acceleration at time ( t ) and after 1 second. s ‘ = f ‘( t ) = v ( t ) = $\frac{81 – 9t² }{(t²+9)²}$ s ‘’ = f ‘’( t ) = v ‘( t ) = a ( t ) = $\frac{d/dx [81 – 9t²](t²+9)² – d/dx[(t²+9)²]*d/dx[t²+9]( 81 – 9t² )}{(t²+9)^4}$ a ( t ) = $\frac{( - 18t )(t²+9)² – 2( t²+9 )( 2t )( 81 – 9t² )}{(t²+9)^4}$ a ( t ) =$\frac{( t²+9 )( ( - 18t )(t²+9) – 2( 2t )( 81 – 9t² ) )}{(t²+9)^4}$ a ( t ) =$\frac{ ( - 18t )(t²+9) – 2( 2t )( 81 – 9t² ) }{(t²+9)³}$ a ( t ) =$\frac{( - 18t³-162t) – ( 324t – 36t³ )}{(t²+9)³}$ a ( t ) =$\frac{- 18t³-162t – 324t + 36t³}{(t²+9)³}$ a ( t ) =$\frac{18t³ - 486t}{(t²+9)³}$ 2g) a ( t ) =$\frac{18t( t² - 27 )}{(t²+9)³}$ a ( 1 ) =$\frac{18( 1 )( ( 1 )² - 27 )}{( ( 1 )²+9)³}$ a ( 1 ) =$\frac{18( 1 - 27 )}{( 1+9 )³}$ a ( 1 ) =$\frac{18( - 26 )}{( 10 )³}$ a ( 1 ) = $\frac{468}{1000 }$ 2g) a ( 1 ) = 0.468 feet per second per second 2h) see screenshot of 2h) i) When is the particle speeding up? When is the particle slowing down? When velocity [ v ( t ) ] has the same sign as acceleration [ a ( t ) ], the particle is speeding up. a ( t ) =$\frac{18t( t² - 27 )}{(t²+9)³}$ The denominator (t²+9) is positive for all real numbers, so the numerator [ 18t( t² - 27 ) ] needs to be greater than zero for the acceleration function to be positive. ( 18t )( t² - 27 ) > 0 ( 18t ) > 0 18t > 0 t > 0 ( t² - 27 )> 0 t² - 27 > 0 t² > 27 $\sqrt t²$ > $\sqrt 27$ t > $\sqrt 27$ t > $\sqrt9 $ * $\sqrt 3 $ t > 3 $\sqrt 3 $ 2d) The signs of the numerator and denominator ( of velocity ) are both positive when 0 ≤ t < 3. For interval [ 0 , 3 ), velocity is positive and acceleration is negative. The signs are different. For interval ( 3 , 3 $\sqrt 3 $ ), velocity is negative and acceleration is negative. The signs are the same. For interval ( 3 $\sqrt 3 $ , ∞ ) , velocity is negative and acceleration is positive. The signs are different. 2i) The particle is speeding up on the interval ( 3 , 3 $\sqrt 3 $ ). 2i) The particle is slowing down on the interval [ 0 , 3 ) ∪ ( 3 $\sqrt 3 $ , ∞).