course Mth 173 Calculus I10-22-2007
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22:39:38 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 problem is different from the book problem but with those zeros we would get y= -(3+x)(x-1)(x-3.5) which gives us a graph similiar to cubic function. The derivative goes from a negative slope to 0 to a positive slope to 0 to an negative slope and continuing at a negative slope.
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22:46:57 ** 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 --> The derivative of the function is a graph of the slope so the the points in between the zeros are a point on the derivative function that is a zero and since there are two of these there are two zeros and this would indicate a parabola, also since y is negative for a +x then the parabola would be concave down.
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22:48:31 query problem 2.4.37 . Which graph matches the graph of the bus and why?
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RESPONSE --> The highest problem in my book is 22, I will look ahead and see if I can make since of the solution.
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22:48:49 ** 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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RESPONSE --> okay
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22:54:34 describe the graph of the car with no traffic and no lights
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RESPONSE --> The car would most likely accelerate at a near steady rate untill it reaches a certain speed and then has no acceleration. There for the graph of speed vs time, would climb steadily then plateu off at a certain speed. The graph of the derivative would show the steady acceleration as a straight line then the graph would drop off to zero or near zero as the speed plateued off.
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22:55:43 ** 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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RESPONSE --> I assumed the car was starting from a stop, but without traffic lights of traffic there would be no reason to stop so a straight line makes perfect sense.
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22:57:10 describe the graph of the car with heavy traffic
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RESPONSE --> the car would be a stop and go with a very low average velocity. It would make a very ""roller-coaster-y"" (up and down) graph showing the starts and stops.
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22:57:22 ** 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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RESPONSE --> okay
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22:58:52 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 --> The price for 2000 units is $150.
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23:00:25 *&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*&
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RESPONSE --> Sorry I misunderstood But p=150 and q=2000
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23:08:07 what is the meaning of f'(150) = -25?
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RESPONSE --> Since q is on the Y axis and p is one the X axis and the rate is Y/x/x so it would be -25 units/ dollar / dollar
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23:08:25 ** 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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RESPONSE --> okay
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23:16:47 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 --> Since you have no chute your going to acclerate untill you hit terminal velocity so a v vs t graph would look like start at 0 because you are not moving verticly and velocity takes into account direction, at the point at which you start to fall the graph will curve upward sharply then plateau out at terminal velocity. The derivative of this would show a constant acceleration of 9.8 m/s/s untill terminal velocity is reached at which point you are still acted on by gravity but you are not accelerating anymore so the derivative would form a straight line then drop to 0
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23:19:06 ** 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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RESPONSE --> The air resistance has the same force as gravity at terminal velocity so it cancells out the force of gravity. I like physics based questions!
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23:21:52 What does the t = 0 acceleration indicate?
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RESPONSE --> The t=0 acceleration indicates that you have reached terminal velocity and the air resistance is equal to the force of gravity and you go no faster and no slower, without an outside system acting upon it.
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23:22:02 ** 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 --> okay
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23:23:25 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I think I pretty much understand the graphical representation of the derivative when related to the function from which it is derived.
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