query_23

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course MTH 173

12/8 17:00

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

023. `query 23

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Question: `qQuery 4.3.34 (3d edition extra problem): Sketch a possible graph for a function which is positive, continuous, with a global maximum at (3,3); the 1st and 2d derivatives have the same sign for x<3, opposite signs for x> 3.

Describe your graph, telling where it is increasing in decreasing, where it is concave up where it is concave down, and where (if anywhere) it has local maxima and minima.

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Your solution:

It is given that the sign of the first and the second derivative should be same for x < 3, the derivative should be equal to 0 for x = 3, and the sign should be opposite for x > 3.

We can assume a graph that is increasing for x < 3, making the sign of the derivative of the function positive for x>3. According to the function the sign of the second derivative function should now also be positive. For this the function should be concave up for x < 3. We can graph such a function by allowing the function to tend to 0 as x tends to - infinity. That is the negative x axis will be asymptote to the curve of the function. The function will then be increasing until it reaches x = 3, where it achieves its maximum value, has a zero slope and then starts to decrease. For the graph the point (3,f(3)) appears to be a sharp point, but is actually a curve which allows the derivative to be zero. The function then decreases beyond x = 3, decreasing further tending 0 as x tends to + infinity. Thus the positive x axis also behaves as an asymptote to the curve f(x). Since the curve is decreasing and will be shown decreasing at a decreasing rate, the sign of the derivative will be negative and since the shape is concave up, the sign of the second derivative will remain positive. This makes the sign of the first and the second derivative opposite for x >3.

Thus the point (3,f(3)) will be a local maxima, and also the global maxima, the function whereas does not have a global or a local minima.

confidence rating #$&*: 3

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Given Solution:

`a** The function would have to be increasing for x < 3, which would make the first derivative positive. The second derivative could also be positive, with the function starting out with an asymptote to the negative x axis and gradually curving upward to reach (3,3). It would then have to start decreasing, which would make the first derivative negative, so the second derivative would have to be positive. The function would have be sort of 'pointed' at (3,3). The graph, which would have to remain positive, could then approach the positive x axis as an asymptote, always decreasing and always concave up.

The horizontal asymptotes would not have to be at the x axis and could in fact by at any y < 3. The asymptote to the right also need not equal the asymptote to the left. **

STUDENT QUESTION

Wouldn’t the graph be concave down?

INSTRUCTOR RESPONSE

If the graph is concave down the slope is decreasing, so the second derivative is negative.

If that was the case then since the second and first derivatives have the same sign for x < 3, the first derivative would also be negative on that interval. This would contradict the condition that the point (3, 3) is a global maximum.

There would also be a contradiction of the given conditions for x > 3. If the second derivative was negative, the first derivative would have to be positive, which again contradicts the global maximum.

The correct graph turns out not to be a smooth curve, but rather one with a 'point' at (3, 3):

The graph must be increasing to the left and decreasing to the right of the global maximum. So the first derivative must

be positive to the left of the global maximum and negative to the right.

So according to the given conditions the second derivative must be positive to the left and positive to the right.

Thus the graph must be concave up on both sides of the point (3, 3).

The only way for that to happen is for the graph to come to some sort of a 'point' at (3, 3).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `qQuery problem 4.3.31 (3d edition 4.3.29) f(v) power of flying bird vs. v; concave up, slightly decreasing for small v; a(v) energy per meter.

Why do you think the graph has the shape it does?

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Your solution:

The graph of the function represents the rate at which the energy consumed vs. velocity. We can see that the graph of the function is a concave graph indicating that the rate at which energy is consumed is higher for low velocities and high velocities.

It is very practical that the starting rate of energy consumption will be huge. This is because, taking the flight involves taking flight, lifting the body in the air, much more over just simple gliding. Thus the bird would demand greater use of energy for reaching the height it wants to, then just simply hovering in the air.

The second thing about the graph is high rate of energy consumption for high velocities. When the bird has high velocities, it has reached the height required to be reached and is thus hovering in the air at that height. The factor responsible is the air resistance. This air resistance is calculated by a formula, according to which the air resistance is directly proportional to the square of the velocity. Thus as the velocity of the bird increases as it glides higher in the air, the air resistance will thus increase. The bird will have to use its energy to fight this increasing resistance. Thus as the velocity increases, the resistance increases and thus the rate at which energy is consumed increases considerably. Thus we have the shape of the graph as a concave curve.

confidence rating #$&*: 3

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Given Solution:

`a** the graph actually doesn't give energy vs. velocity -- the authors messed up when they said that -- it gives the rate of energy usage vs. velocity. They say this in the problem, but the graph is mislabeled.

The graph says that for high velocities the rate of energy usage, in Joules / second, increases with increasing velocity. That makes sense because the bird will be fighting air resistance for a greater distance per second, which will require more energy usage. To make matters worse for the bird, as velocity increases the resistance is not only fought a greater distance every second but the resistance itself increases. So the increase in energy usage for high velocities isn't too hard to understand.

However the graph also shows that for very low velocities energy is used at a greater rate than for slightly higher velocities. This is because low velocities imply hovering, or near-hovering, which requires more energy than the gliding action the bird achieves at somewhat higher velocities. **

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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

I had some difficulty with the graphical interpretations, but I think going over more notes can give me a better understanding

STUDENT COMMENT: I actually think I understand the general concept of the graphical interpretation and the first and second derivative but it helps me to think of this in relation to the trappazoid approximation graphs that we worked on earlier in the year is that a bad way fo thinking about this concept????

INSTRUCTOR RESPONSE The characteristics of trapezoidal approximation graphs seem to provide a good basis for interpreting graphs in general.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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