course Mth 173 ???????????~?y?assignment #013??????[?H??????Calculus I
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09:17:58 query problem 2.4.6 derivative of fn (poly zeros at -3,1,3.5, neg for pos x)Describe the graph of your function, including all intercepts, asymptotes, intervals of increasing behavior, behavior for large |x| and concavity
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RESPONSE --> I cannot locate this problem in my textbook.
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09:18:02 ** The x intercepts of the derivative function will occur when the given function levels off, which occurs at approximately x = -1.5 and at x = 2.5. Between these x values the function is increasing so the derivative will be positive. Every where else the function is decreasing so the derivative will be negative. The derivative will take its greatest positive value where the original graph has its steepest upward slope, which probably occurs around x = .5. As x approaches +infinity the steepness of the original graph approaches -infinity so the value of the derivative function approaches -infinity. As x approaches -infinity the steepness of the original graph approaches -infinity so the value of the derivative function approaches -infinity. This description would be satisfied, for example, by a parabola opening downward, with vertex above the x axis around x = .5. **
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09:19:24 query problem 2.4.37 . Which graph matches the graph of the bus and why?
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RESPONSE --> I think this problem is 2.3.35 If so, I choose II for the bus because it stops and starts so often but since it is on a road with no traffic the height of the levels should be consistent.
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09:19:27 ** The bus only makes periodic stops, whereas the graph for III only comes to a stop once. I would matche the bus with II. **
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09:20:37 describe the graph of the car with no traffic and no lights
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RESPONSE --> I. should be this because there is no lights so there is no stops and the car level is constant because there is also no traffic to slow the car down.
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09:20:39 ** The car matches up with (I), which is a continuous, straight horizontal line representing the constant velocity of a car with no traffic and no lights. *&*&
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09:21:11 describe the graph of the car with heavy traffic
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RESPONSE --> III. is the answer because with heavy traffic the car is constantly changes the levels of speed and stops.
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09:21:13 ** The car in heavy traffic would do a lot of speeding up and slowing down at irregular intervals, which would match the graph in III with its frequent increases and decreases. **
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09:22:18 query 2.5.10 (was 2.4.8) q = f(p) (price and quantity sold)what is the meaning of f(150) = 2000?
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RESPONSE --> p = 150 which is the price q = 2000 which is the amount sold
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09:22:20 *&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*&
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09:23:20 what is the meaning of f'(150) = -25?
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RESPONSE --> p = 150 is the price q = -25 which means there will be a drop of 25 in the amount sold.
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09:23:22 ** f' is the derivative, the limiting value of `df / `dp, giving the rate at which the quantity q changes with respect to price p. If f'(150) = -25, this means that when the price is $150 the price will be changing at a rate of -25 units per dollar of price increase. Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. **
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09:34:28 query problem 2.5.18 graph of v vs. t for no parachute. Describe your graph, including all intercepts, asymptotes, intervals of increasing behavior, behavior for large |t| and concavity, and tell why the graph's concavity is as you indicate.
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RESPONSE --> 2.4.18 The graph will be increasing at a decreasing rate, which means it will concave down. When you fall you go rapidly and then slow down the farther you fall.
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09:34:29 ** When you fall without a parachute v will inrease, most rapidly at first, then less and less rapidly as air resistance increases. When t = 0 we presume that v = 0. The graph of v vs. t is therefore characterized as an increasing graph beginning out at the origin, starting out nearly linear (the initial slope is equal to the acceleration of gravity) but with a decreasing slope. The graph is therefore concave downward. At a certain velocity the force of air resistance is equal and opposite to that of gravity and you stop accelerating; velocity will approach that 'terminal velocity' as a horizontal asymptote. The reason for the concavity is that velocity increases less and less quickly as air resistance increases; the approach of the velocity to terminal velocity is more and more gradual **
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09:35:38 What does the t = 0 acceleration indicate?
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RESPONSE --> t = 0 means the acceleration in gravity.
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09:35:40 ** t = 0 acceleration is acceleration under the force of gravity, before you build velocity and start encountering significant air resistance. Acceleration is rate of velocity change, indicated by the slope of the v vs. t graph. **
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09:36:41 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I could not locate the first problem in order to make this query complete. If you will give me another problem I will gladly try to do it. However, I could not complete the problem because I needed to refer to the text and could not because I could not find the problem.
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