course Mth 173
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16:33:15 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 --> The graph of the derivative function has zeros at about x = -1.4 and x = 2.3, is negative before and after these zeros, and is positive between them. It is a concave down parabola. It increases at a decreasing from a very large negative value to about 2.5, then decreases at an increasing rate as x gets arbitrarily large.
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16:34:28 ** 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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RESPONSE --> When I mentioned the function was greatest at 2.4, I meant when f(x) was 2.4, which does indeed occur around x = .5
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16:35:33 query problem 2.4.37 . Which graph matches the graph of the bus and why?
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RESPONSE --> Graph II matches the bus, as it represents an increase in velocity up to traveling speed, then a decrease as it makes its stop, then 0 velocity as it loads and unloads, then repeating the process
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16:35:37 ** 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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16:36:38 describe the graph of the car with no traffic and no lights
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RESPONSE --> The graph of the car with no traffic or lights is represented by graph I - the car accelerates from rest and eventually achieves a constant velocity, unhindered by traffic or obstacles.
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16:36:43 ** 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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16:37:44 describe the graph of the car with heavy traffic
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RESPONSE --> The erratic graph III representing the velocity of the car in traffic exhibits seemingly random increases and decreases, and even one stop, in velocity, which goes along with unpredictable patterns characteristic of heavy traffic.
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16:37:47 ** 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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16:38:11 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 --> At the $150 price point, 2000 units will be sold.
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16:38:15 *&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*&
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16:38:50 what is the meaning of f'(150) = -25?
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RESPONSE --> When the price is going up from $150 per unit, we expect to be selling about 25 less units per dollar of price increase (for prices near $150).
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16:38:57 ** 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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16:39:56 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 --> The graph increases at a gradually decreasing rate, eventually becoming constant at the point of terminal velocity. The increase is at a decreasing rate due to the wind resistance. Concave down, then constant.
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16:40:35 ** 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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16:40:56 What does the t = 0 acceleration indicate?
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RESPONSE --> The rate at which the velocity is changing at t = 0.
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16:41:29 ** 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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RESPONSE --> Yes, you could say that, since at that precise point wind resistance has not come into play yet.
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16:41:40 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> All good so far.
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