query13

#$&*

course Mth 173

3/18 10:51 PM

013. `query 13

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Question: `q query problem 2.3.37 . Which graph matches the graph of the bus and why?

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Your solution:

I do not have a question about a graph matching a bus, I have f(t) from its starting point at time t, which of the graphs from figure 2.35 coule be f'(t) for the following scenarios, and (a) is a bus on a popular route, with no traffic.

Assuming this bus is a public bus, and picking up and dropping off passengers, its still going to stop a few times, so its graph should look somewhat ""rollercoaster"" . but I guess for the most part it would be increasing., and dropping back down to 0 for stosp.

confidence rating #$&*: 1

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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. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q describe the graph of the car with no traffic and no lights

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Your solution:

I think this could have two answers, the one that would seem picked most often would be the constant rise straight line graph. But I could see this also being a power graph, in which the car has to pick up speed and getting faster (to a point) before it would be more linear

confidence rating #$&*: 2

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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. *&*&

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q describe the graph of the car with heavy traffic

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Your solution:

I'd say it look closest to the graph that is shaped somewhat like a ""w"" because the car would be somewhat, erratic with its speeds

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. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

problem number 6.

An economist is interested in how the price of a certain item affects its sales. At a price of $p, a wuantity, q of the item is sold. If q=f(p) explain the meaning of each of the following statements:

(a) f(15) = 2000

if price is p in dollars and q=f(p) that means at price 150, the quantity would be 2000.

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. *&*&

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q what is the meaning of f'(150) = -25?

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Your solution:

assuming its still from the same question

the rate is dropping hear, making an assumption that the sales where better at a different price. and if it is a straightforward relationship, as p increases q increases, then it would mean that more where sold at a cheaper price.

confidence rating #$&*:

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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. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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.

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Your solution:

problem number 26

(a) If you jump out of an airplane without a parachute,you fall faster and faster until air resistance causes you to approach a steady velocity, called the terminal belocity. Sketch a graph of your velocity against time.

(b) Explain the concavity of your graph

(c) Assuming air resistance to be negligible at t= 0, what natural phenomenon is represented by the slope of the graph at t=0

(c) of course at t=0, you can view that almost as a ""stasis"" nothing has happened or past, so velocity is also at 0.

(b) the graph startsd of increasing, and slows down, so increasing at a decreasing rate, so concavity is downwards.

????what would the name of this type of graph be?

@&

This would depend on how the drag force changes with speed.

If the drag force is directly proportional to speed, then the graph would approach terminal velocity exponentially. The function would be of the form v(t) = A e^(-k t) + v_term.

If the drag force is proportionaly to v^2 then you get a solution involving hyperbolic tangents (you don't want to know about those right now).

In practice, the rule governing drag force is more complicated than either of the two simplified cases I mention above, and there is no specific name for the sort of velocity function that results.

*@

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 **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q What does the t = 0 acceleration indicate?

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Your solution:

nothing has started to move yet, velocity has yet to start as well, as being 0. or if you are considering what it means to have 0 acceleration, it means your velocity is constant, which means two different things when t = 0, and when it is farther on, t=0 you just havent started traveling, later on you will be at terminal velocity.

confidence rating #$&*: 2

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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.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Self-critique (if necessary): OK

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Self-critique Rating: OK"

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#*&!

&#Your work looks good. See my notes. Let me know if you have any questions. &#