#$&* course MTH 173 1510 17MAR2013
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Given Solution: ** The bus only makes periodic stops, whereas the graph for III only comes to a stop once. I would matche the bus with II. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q describe the graph of the car with no traffic and no lights YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This would resemble graph I. We can see the car accelerate and stay at a constant speed. This would indicate no red lights and little traffic. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q describe the graph of the car with heavy traffic YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This would be graph III. We see the car accelerate a little but is slowed down and accelerate often and at varying rates. This would indicate a car in heavy traffic perhaps on the interstate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query 2.4.11 5th, 2.4.10 4th; 2.5.10 (was 2.4.8) q = f(p) (price and quantity sold)what is the meaning of f(150) = 2000? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At a price of $150 the company can expect to sell 2000 units of the item. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: *&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q what is the meaning of f'(150) = -25? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At f`(150) we can expect a decline in the number of units sold at a rate of -25. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 $150 the quantity will be changing at a rate of -25 units per dollar of price increase. Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### I did not put that it would be a decrease of 25 units per dollar increase. #### ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q query problem 2.4.23 5th; 2.4.18 4th; 2.4.7 graph of v vs. t for no parachute. Describe your graph, including all intercepts, asymptotes, intervals of increasing behavior, behavior for large |t| and concavity, and tell why the graph's concavity is as you indicate. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My graph lies in quadrant I. The line starts at (0,0) and increases very fast at first but decreases at a decreasing rate. At the point where the person hits the ground, the graph drops almost instantaneously. This makes a vertical asymptote near x = 0. As the line moves to the right, we get a horizontal asymptote at y = (z) where z = a fast but unknown speed. The graph appears to be concave to the right, up until the point where the person hits the ground because the speed dramatically drops to zero. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** When you fall without a parachute v will increase, 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q What does the t = 0 acceleration indicate? Your solution: At this instant the person will fall as if in a vacuum because air resistance will be negligible. If this were the case, and a person did fall while in a vacuum towards earth that person would continue to accelerate without bound and this would create a graph that is concave up. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** STUDENT QUESTION I understand how acceleration and vel. are related, just not the first part of solution INSTRUCTOR RESPONSE If it wasn't for air resistance, acceleration would be equal to that of gravity. When you first jump out you aren't falling very fast, so there isn't much air resistance to counter the acceleration of gravity, so you accelerate pretty much at the acceleration of gravity. You quickly speed up, and air resistance becomes more and more significant. So your acceleration becomes less than the acceleration of gravity. Assuming the ground is far away, this continues until air resistance is effectively equal to the force of gravity, at which point your acceleration will be zero. You will be at terminal velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### This assignment was not too bad. I understood the problems. #### ------------------------------------------------ Self-critique Rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!