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Business Calc Homework w answers_Part_18

# Business Calc Homework w answers_Part_18 - 86 Section 3.4...

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26. (a) To graph the velocity, we estimate the slopes at several points as follows, then connect the points to create a smooth curve. To graph the acceleration, we estimate the slope of the velocity graph at several points as follows, and then connect the points to create a smooth curve. (b) } d d s t } 5 30 t 2 3 t 2 [0, 15] by [ 2 300, 100] } d dt 2 2 s } 5 30 2 6 t [0, 15] by [ 2 100, 50] The graphs are very similar. 27. (a) Solving 160 5 490 t 2 gives t 5 6 } 4 7 } . It took } 4 7 } of a second. The average velocity was sec = 280 cm/sec. (b) v 5 s 9 ( t ) 5 980 t a 5 s 0 ( t ) 5 980 The velocity was s 9 1 } 4 7 } 2 5 560 cm/sec. The acceleration was s 0 1 } 4 7 } 2 5 980 cm/sec 2 . (c) Since there were about 16 flashes during } 4 7 } of a second, the light was flashing at a rate of about 28 flashes per second. 28. Graph C is position, graph A is velocity, and graph B is acceleration. A is the derivative of C because it is positive, negative, and zero where C is increasing, decreasing, and has horizontal tangents, respectively. The relationship between B and A is similar. 29. Graph C is position, graph B is velocity, and graph A is acceleration. B is the derivative of C because it is negative and zero where C is decreasing and has horizontal tangents, respectively. A is the derivative of B because it is positive, negative, and zero where B is increasing, decreasing, and has horizontal tangents, respectively. 30. (a) } d d y t } 5 } d d t } 3 6 1 1 2 } 1 t 2 } 2 2 4 5 } d d t } 3 6 1 1 2 } 6 t } 1 } 1 t 4 2 4 } 24 5 } d d t } 1 6 2 t 1 } 2 1 4 } t 2 2 5 0 2 1 1 } 1 t 2 } 5 } 1 t 2 } 2 1 (b) The fluid level is falling fastest when } d d y t } is the most negative, at t 5 0, when } d d y t } 5 2 1. The fluid level is falling slowest at t 5 12, when } d d y t } 5 0. (c) [0, 12] by [ 2 2, 6] y is decreasing and } d d y t } is negative over the entire interval. y decreases more rapidly early in the interval, and the magnitude of } d d y t } is larger then. } d d y t } is 0 at t 5 12, where the graph of y seems to have a horizontal tangent. 160 cm } 1 } 4 7 } 2 a (km/hr 2 ) 50 25 –25 –50 t (hours) 15 10 v (km/hr) 100 –100 –200 –300 t (hours) 15 5 86 Section 3.4 t (hours) 0 2.5 5 7.5 10 12.5 15 v (km/hour 2 ) 30 15 0 2 15 2 30 2 45 2 60 t (hours) 0 2.5 5 7.5 10 12.5 15 v (km/hour) 0 56 75 56 0 2 94 2 225

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31. (a) } d d V r } 5 } d d r } 1 } 4 3 } p r 3 2 5 4 p r 2 When r 5 2, } d d V r } 5 4 p (2) 2 5 16 p cubic feet of volume per foot of radius. (b) The increase in the volume is } 4 3 } p (2.2) 3 2 } 4 3 } p (2) 3 < 11.092 cubic feet. 32. For t . 0, the speed of the aircraft in meters per second after t seconds is } 2 9 0 } t . Multiplying by } 360 1 0 h sec } ? } 1 1 00 k 0 m m } , we find that this is equivalent to 8 t kilometers per hour. Solving 8 t 5 200 gives t 5 25 seconds. The aircraft takes 25 seconds to become airborne, and the distance it travels during this time is D (25) < 694.444 meters. 33. Let v 0 be the exit velocity of a particle of lava. Then s ( t ) 5 v 0 t 2 16 t 2 feet, so the velocity is } d d s t } 5 v 0 2 32 t . Solving } d d s t } 5 0 gives t 5 } 3 v 2 0 } . Then the maximum height, in feet, is s 1 } 3 v 2 0 } 2 5 v 0 1 } 3 v 2 0 } 2 2 16 1 } 3 v 2 0 } 2 2 5 } v 6 0 4 2 } . Solving } v 6 0 4 2 } 5 1900 gives v 0 < 6 348.712. The exit velocity was about 348.712 ft/sec. Multiplying by } 360 1 0 h sec } ?
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