query 23

#$&*

course Mth 173

7/16 7:10 pm

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:

The graph is increasing from (0,0) to (3,3) so it is concave up, therefore the 1st and 2nd derivatives are positive on this interval

The then after the point (3,3) the graph is decreasing at an increasing rate, therefore the 1st and 2nd derivatives are opposite in sign

I would think that the local max and global max are the same (3,3)

I don’t think there are local min

confidence rating #$&*: 1

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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):

????I am having trouble visualizing this graph. Would it be sort of like a weird parabola?

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Self-critique Rating:

@&

Imagine a circle of radius 1, centered at the origin. Among other points, it passes through (1, 0) and (-1, 0).

Now cut the circle along the x and y axes. Slide the part in the third quadrant one unit to the right, so that the point at (-1, 0) shifts to the origin. Slide the part in the fourth quadant one unit to the left, so that the point at (1, 0) also shifts to the origin.

You will now have a shape that starts at (-1, -1), increases at an increasing rate to the point (0, 0), then decreases at a decreasing rate to (1, -1). The shape will have a pointed peak at the origin.

The graph described here will have this sort of peak at the point (3, 3).

*@

query 23

#$&*

course Mth 173

7/16 7:10 pm

023. `query 23

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph is increasing from (0,0) to (3,3) so it is concave up, therefore the 1st and 2nd derivatives are positive on this interval

The then after the point (3,3) the graph is decreasing at an increasing rate, therefore the 1st and 2nd derivatives are opposite in sign

I would think that the local max and global max are the same (3,3)

I don’t think there are local min

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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):

????I am having trouble visualizing this graph. Would it be sort of like a weird parabola?

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Self-critique Rating:

07-17-2014

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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:

(rate of energy usage vs. velocity)

The graph shows that for higher velocities the rate of energy usage increases. Which makes sense because the faster the bird goes the more energy it will need

confidence rating #$&*: 1

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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):

I did not understand why the graph showed that at lower velocities more energy usage is required than for higher velocities. After reading your explanation this makes sense

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Self-critique Rating: 3

07-17-2014

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

It is hard for me to visualize graphs when given certain conditions. I struggled with this in the material for the last test.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

07-17-2014

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

It is hard for me to visualize graphs when given certain conditions. I struggled with this in the material for the last test.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good responses. Let me know if you have questions. &#