Unformatted text preview: Rates of Change Recall that And then and the average rate of change of And the instantaneous rate of change of (of ) with respect to with respect to lim 208-2 A particle moves according to a. Determine the velocity at time Since b. the rate of change of position with respect to time , we know that is is What is the velocity after seconds? ft/s This means calculate graph of c. When is the particle at rest? This means when is ?
MTH 173 section 3.3 notes page 1 . The particle is "at rest" at and seconds. Note that these results match with the zeros of the graph of the velocity function. d. When is the particle moving in the positive direction? This means when is ? Graphically, remembering the horizontal axis represents time and the vertical axis above represents , we are asking in what time intervals is the graph above the time axis. So, from the graph, we have the intervals and . To answer this question without the graph, we consider: and make a sign graph for each factor and the product: t-1 t-5 t= 0 neg neg 0 pos neg 0 5 neg
and pos pos 1 pos 0 3(t-1)(t-5) 0 pos From this sign graph, we also read that the velocity is positive for e. Determine the total distance traveled in the first 8 seconds. It is tempting here to calculate , but that just determines the difference in position (or "net position change") between those two times. Imagine the function gave your location at time during the day. If at 6 AM and 11 PM you are home, then PM AM , even if you were out and had traveled 1000 miles during the day. So Recognize that the particle changed direction at the 1 sec mark and he 5 sec mark. So we determine the distance moved from to , from to , and then from to MTH 173 section 3.3 notes page 2 distance from to distance from to distance from to ft f. Draw a diagram to illustrate the motion: t=5 t=0 t=8 t=1 MTH 173 section 3.3 notes page 3 10. If a ball is thrown vertically upward with a velocity of ft/s, then its height after seconds is What is the maximum height reached by the ball? We can obtain an answer from the graph, but the intent of the question is different. The question asks that we determine the high point on the graph. Note that at this point the slope (think velocity here) is . Thus, we shall determine the time when the velocity is , then determine the height at that time. sec is when the maximum height is reached. Then ft. What is the velocity when it is 96 feet above the ground on the way up? On the way down? This is a request to calculate velocities at some times we don't know - but we can use the position function to determine the times when , then substitute in So: At At seconds , the ball is going up, and its velocity is , the ball is going down, and its velocity is ft/s ft/s MTH 173 section 3.3 notes page 4 18. If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the volume of water remaining in the tank after minutes as Determine the rate at which the water is draining from the tank after a. 5 min b. 10 min c. 20 min d. 40 min This is a request to determine at these times. Since the function gives volume left in the tank and since the water is leaving the tank, we know that getting smaller (decreasing) and thus we should to be negative Since , we calculate is We have: gal/min gal/min gal/min gal/min As the tank empties, it empties more and more slowly. 26. Suppose a bacteria population starts with 500 bacteria and triples every hour. Population after 1 hour 2 hours 3 hours 4 hours hours Estimate the rate of increase after hours: eqn 5 on page 189 says that MTH 173 section 3.3 notes page 5 So and we evaluate this rate of growth expression at bacteria/hr 30. The cost function for a certain commodity is to answer the question: This function gives the total cost of producing items. Calculate Calculate the cost of producing the 101st item: is the cost of producing item 101. It is approximated very well by the marginal cost at MTH 173 section 3.3 notes page 6 0.1303 dollars 1 unit MTH 173 section 3.3 notes page 7 ...
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