Now, I know we didn't really cover a lot of material today but were presented with a rather new type of problem and also a quick quiz at the beginning of class. For the problem, we have in fact encountered somewhat similar problems, such as the oil slick problem, but the main difference is that we are now dealing with a numeric representation of the density function, rather than a symbolic one. Remember that on the exam, we could be given a varied amount of different representations, ideally numeric, symbolic and graphical ones. So try to practice not only questions that convey a specified density function in the question, but also practice those that have a numerical representation (say a table of values such as today's question) or an accompanying graph. I believe that this idea was the basis of today's problem and overall lesson.

Before I continue onward with the detailed progression of today's problem, I'll first digress and explain the quiz that preceded the challenging question. Though we went over the answers to the quiz afterwards, we did not detail each questions solutions. I felt obligated (not entirely sure why =p) to do so now for each question.

*f*, in order to determine where

*f*has a local maximum or a local minimum in this case is to analyze the roots of

*f'*.

Knowing the behaviour of the derivative of a function and it's relation to the parent function helps incredibly in this analysis. You must know that not all roots of

*f'*are maximums or minimums, such at x = 2. This is because since there is no sign change, what happens at x = 2 is that there exists a plateau or flat portion in the parent function, thus the horizontal slope at x = 2. Also, looking at x = 4 as another candidate for a local minimum or maximum yields a sign change from positive to negative. This change indicates that x = 4 yields a maximum on the parent function, because since

*f'*is positive to the left of x = 4 and negative to the right of x = 4 the parent function is increasing until it hits x = 4 then turns downwards, therefore it is a local maximum. Now there's only one thing left to analyze since the zero's are now out of the question.

Don't forget about the endpoints of the function. At the endpoints of

*f*', since

*f*is increasing on (0, 4) and then

*f*starts to decrease once again on (4, 5), they turn out to be LOCAL minima on the parent function. This is because since the function start out low at x = 0 and then increases from that point on until x = 4, it is a local minimum. After the function reaches x = 4, the function begins to decrease and therefore reaches another low point at the end of the function. Don't confuse the question with finding the global minimum, which is actually the mistake that I made when I chose my answer. You must recognize that local minima are only considered minima on a small interval central to that point, not over the entire function. Thus both endpoints qualify as LOCAL minima.

*f*again. But this time we are asked to determine the points of inflection of the parent function

*f*. Though the previous question relied on the endpoints of the function, we cannot do that this time since it's impossible to determine the derivative of the endpoints of a function since there's no one specific derivative at that point. If you imagine trying to draw a tangent line at x = 0 or x = 5 on

*f*', there's no way you can draw just one definite tangent line, thus the illegal procedure of differentiating this function. You might be asking as to why I'm mentioning differentiating this function when it already is the derivative. Recall that we're looking for the inflection point of

*f*, which is a change in the concavity of the function, and that

*f''*represents the concavity of the parent function, thus the reason for differentiating

*f'*.

We basically are now looking for the zero's of

*f'',*which occur at the local maxima and minima of

*f'*. We can quickly see that there is a maximum present at x = 3 and a minimum at x =2, therefore these are also the points of inflection of

*f*. Don't forget, however, to analyze the shape of these points to determine whether they are in fact points of inflection or not.

Answer: There are points of inflection at x = 2 and x = 3.

In this question, the level of comprehension required for this question does not really bypass the simple rules of antidifferentiating polynomial functions and knowing the connection between acceleration, velocity, and position. This required connection involves the fundamental theorem of calculus, which implicates the process of integration as the reverse of differentiation. Knowing this, and seeing that velocity is defined as a change in position over a given time interval, it can also be perceived as the time derivative of position, which is also followed by the relationship between acceleration and velocity. This indicates that antidifferentiating the acceleration function given gives us the velocity function. But don't forget that the arbitrary constant of integration, C, must be determined using the fact that v(0) = 1. Using this, we can easily solve for C by substituting 0 in for t and setting this equal to 1. Now, since velocity and position are also related similarly, we can antidifferentiate this new velocity function to determine the position function. Don't forget once again, however, to use the fact that s(0) = 3 in order to determine the constants value and finish off the final function.

Answer: s(t) = t^{3} + t + 3

Question #4:

Since we're finding the area under two separate function on the interval x = 0 to x = k, we know that we're going to be finding the value of two separate integrals with the intervals lower and upper limits. Also, just looking at these functions yields should yield a sigh of relief since these functions aren't complicated at all to antidifferentiate each function, so it would be quite easy to continue on that path now. First, as it shows in Mr. K's work, we must set these two integrals equal to each other on the same interval, 0 to k. Then, all you must do is antidifferentiate each function, giving you an equality as shown in the first line in black print in Mr. K's solution above. Don't forget that the antiderivative of sin(x) is -cos(x) and not just cos(x). Remember this is so because the derivative of cos(x) is -sin(x), so just sin(x) is just missing that negative sign when going backwards. Also, don't neglect substituting the 0 into the trigonometric function -cos(x) when integrating the left side, since cos(0) yields a 1. Watch the negative signs though, and try not to get mixed up by the negatives in the function.

Now once you have determined the final equality after antidifferentiating and inserting k and 0, you could solve this in two different ways. You could either leave the two functions equal to each other, and find points of intersection between the two graphs. Or you could find move everything to one side and find the zero's of the function. Either way, it'll require the use of your calculator but yet still yield the same answer.

Answer: 1.300

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And now, on to the problem for today's class, the Boston By The Sea" problem. First off, Mr. K quickly told us the origin of Boston as a harbour that slowly grew outwards into the ever-growing city it is now. He really only mentioned that since the question modeled the city using a semi-circle. Here is the question.

In class, we were only able to cover question (a), though our solution never came to fruition and is therefore homework for Friday's class. Now on to our discussion concerning the first portion of this question.

Now you must first identify what your first task is while approaching this question. If you analyze the question thoroughly, as you should do for long answer questions similar to this, you should notice that the density function gives a value in units of population (in thousands)/miles^{2}.

You now must be able to try and find a way to multiply the thousands of people by a unit that will reduce with the miles^{2} to only give the answer a final unit of population (in thousands of people). This can be done by multiplying the density function by the areas of various semi-circles with varying radii. All of our work on this problem can be seen on the following slide

Now, follow the yellow section of the figure to the left, and the yellow highlighted function just to the right of the figure as they correspond to each other. Looking at the data, we can see that there is no change in density in the very first semi-circle of radius 1. Thus, we can create a simple expression to show the amount of people in this section by multiplying the density function by the area of this semi-circle. This very expression is what is highlighted in yellow.

But now we must try and model the rest of the semi-circles, and we can do this by breaking each semi-circle into rings between each circumference. One of these rings is represented by the green section in the diagram. This ring can be taken out and straightened out to then be represented by a rectangle, which is also shown by the green highlighted section. The length of this rectangle will be equivalent to the circumference of the semi-circle, and will have a width equivalent to the change in the radius, which is consistently one in this case. Therefore, we now have a model to determine the actual amount of people in the entire section between r = 0 and r = 8. Adding the successive parts of each interval to the expression highlighted in yellow will yield us our final answer for part (a).

Now, our homework for today's class was to finish up this problem and calculate the amount of people in the 8 mile radius. Don't forget to complete your homework for Friday's class everyone.

I think that's all for my scribe post today. As I mentioned in the beginning of my scribe post, the scribe for tomorrow's class will be ETHAN, that is unless he can't make it to class again, in which case the scribe will be VAN. Goodnight everyone and have a pleasant tomorrow! ( I can't remember where that's from, but oh well! =D) I hope my scribe post helped anyone who was in seek of some elaboration upon today's class, or was yearning for some explanation on the subject of applications of integrals. Have a good one everyone!

## 2 comments:

Wow, what an extensive post! and interesting too. When I read the question about the Reimann Sum, I immediately expected an area under a curve approximation, however the semi-circular representation makes more sense in the context of the problem.

I agree an extensive post, and well written. As I read through it I got a sense that you understand the material very, but most importantly I think anyone who read it understands it better. As I understand the purpose of the scribe post, it is to organize your thinking and convey to the teacher a basic understanding of what the class understands about the topic. Your post indicates great understanding, I'm very impressed!!

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