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- Reading Assignment: pp. 325-top 329
- Reading Assignment: pp. 210-211 (Examples 1-2), p. 213 (Example 4), p. 219 (Examples 2-5)
- Reading Assignment: bottom of p. 177 and top of p. 178 (proof of Sum/Difference Rule), pp. 186-187 (expls. 1-3), p. 188 (expls. 4-5)
- Reading Assignment: pp. 125-127 (start at bottom of p. 125), 133-134
- Reading Assignment: pp. 27-29, 36-38

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- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329
- Jonathan E Lopez on Reading Assignment: pp. 325-top 329

I was really confused on how they did the square root of 500/3.14. Why did they do it and how? on page 326-328. And on page 329 example 4 I wish they would show them using Pythagoram theorem, and where are they getting the roots from? Same question for example 3 on 328 where are they getting the roots from?

In example 2 on page 327, the critical number is the solution to pi*r^3=500. To solve for r, divide by pi so that r^3=500/pi. Then take the cube root of both sides, so r=(500/pi)^{1/3}. This is the only critical number, so it is the radius that will minimize cost.

In Example 4, 7x^2=81. Divide both sides by 7 to get x^2=81/7. Then take the square root of both sides (and keep in mind x can’t be negative) to get x= sqrt(81/7). This simplifies to 9/sqrt(7).

In Example 3, we need to solve y^3-8=0. Add 8 to both sides to get y^3=8, then take cube root of both sides, so y=8^{1/3}=2 is the only critical number.

When would it be advantageous to use implicit differentiation rater than substituting to get a single variable? Also, what if we get two critical values rather than just 1?

If one of the equations is “complicated” (like involving a square root as in Example 4), then using implicit differentiation can be easier (assuming a high comfort level with implicit differentiation). Generally, implicit differentiation itself can be tricky for those just learning it, so coupling it with optimization often isn’t done in the textbooks.

If you get two critical values, you would have to be a bit more careful in answering the question. If you were looking for maximum volume, for instance, and you had two critical values (assuming both give you local max), you could plug both into the volume function to see which produces a larger volume. That often doesn’t come up in the types of problems we do in section 4.7.

Example 1 was easy to follow and understand for me, however I got a little lost during Example 2 when substituting for A and also during the steps of finding the critical numbers and differentiating. Also I am confused on where the critical numbers are coming from. I also find that using implicit differentiation is a bit easier to follow.

To find the critical numbers, differentiate whatever quantity you want to max/min, and solve for what makes it equal to 0.

In example 2, they solve the volume formula for h and substitute into the surface area equation. They simplify a bit and then differentiate (remember that pi is a constant). The derivative of 2pi*r^2 is 2pi*2r=4pi*r. The derivative of 2000/r = 2000r^{-1} is -2000r^{-2}=-2000/r^2.

Next they get a common denominator (which is r^2) and factor a 4 out of the numerator. Since the derivative A'(r) is a fraction, it is equal to 0 when the numerator is 0, i.e., when pi*r^3-500=0. Solving for r gives r=cuberoot(500/pi).

The steps from 325-326 were easy to comprehend and the example 1 on 326 was easy to follow. Example 2 on 327 was a bit confusing. Although it was explained I didn’t quite understand how a whole half of the equation was able to be eliminated therefore causing the rest of the problem to give me trouble. The first derivative test for extreme vales is not complicated but I would like some clarification on Example 4.

There is some explanation of the steps for finding the critical number in Example 2 above.

In example 4, there is one critical number. Back in 4.1, we learned that the abs. max/min occurs at a critical number or an endpoint of the interval. In this case, x can be as small as 0 (this means the boat is rowed directly from A to C, and then the full 8km is run), or x could be as large as 8 (this would be rowing from A to B, and not running at all). So theoretically, the abs. max/min could happen at x=0, x=8, or the critical number x=9/sqrt(7). Generally, in the problems we do in 4.7, the abs. max/min occurs at the critical number.

The process/steps to follow for optimization are pretty straightforward and easy to understand, and the first few examples are similar to what we did in class with the fence and can. However, example 4 gave me some trouble. Without looking at the answer, I was not sure how to finish the problem after labeling what we know so far.

Yes, Example 4 is tricky, mainly because the thing we want to minimize is time. To come up with a formula for time, we need to think of time as distance/rate. Since the problem specifies the rate at which this man can run and row, we really just need to determine the distances for each. Once we have this, we differentiate and find the critical number.

I liked how the book gave us steps in solving optimization problems and explained each step. After reading it, I understand it a little bit more. Optimization problems are like related rates in that you have to come up with your own formula by using what they give you. To me, example 3 didn’t make much sense because it’s hard to imagine. With the other problems, you get a real life scenario, like the one with the fence, so it’s easier to imagine. This one is confusing.

Example 3 is certainly a bit more abstract. Note that the point (x,y) that’s closest to (1,4) can also be thought of as the point at which the tangent line is perpendicular to the line through (x,y) and (1,4). This means the slopes need to be negative reciprocals of each other.

The slope of the tangent line is found by using implicit differentiation:

y^2=2x

2yy’=2

so y’=1/y.

The slope of the line through (x,y) and (1,4) is

m=(4-y)/(1-x).

If these two slopes are negative reciprocals of each other, then

(4-y)/(1-x) = -y, which gives

4-y=xy-y

so 4=xy.

Using 4=xy along with y^2=2x, the only solution is the point (2,2) (which agrees with their answer).

I am having trouble following the steps in example 3. It would be helpful if there was more visualization than just the graph of the parabola.

See the comment above. Draw the scenario where the line trough (1,4) and (x,y) is perpendicular to the tangent line at (x,y). That is the point we need.

Example 1 and 2 were easy to follow, especially since we had done examples just like them in class. It was interesting to see that implicit differentiation came up in example 2. I feel like it makes the process a bit simpler or easier to understand. Example 3 was very different compared to the other examples since it was not a “real world” example or something you would be able to visualize easily in your mind. This example confused me a bit, mainly because it was so different, but after reading over everything I understood how the process worked.

See the previous two comments above for another way to think about visualizing Example 3.

Example 1 made a lot of sense, and I had no issues. Example 2 was very similar to a problem we completed in class, so it made sense. But once the textbook states “the first derivative test for absolute extreme values” I became very confused. After reading this definition about 5 times, I just gave up. What we learned in class make sense so this just seemed to over complicate things. This seems to be a consistent theme with the book. Example 3 was a bit confusing as well. How are the minimum of d and d^2 the same thing?

The minimum of d and d^2 are not the same, but the input that produces the minimums is the same.

For example, if the minimum of d is 2, then the minimum of d^2 is 4. If you take the input that gives you d=2, then that same input gives you d^2=4.

The”First Derivative Test for Absolute Extreme Values” is just saying that if there is only one “change” (either inc/dec or dec/inc), then the max/min must be absolute.

For example, if a function is increasing and then decreasing (and that’s the only change), then the place where the change happens has to be the highest point (abs. max).

If a function is decreasing and then increasing (and that’s the only change), then the place where the change happens has to be the lowest point (abs. min).

Generally, you can think of it like this. If there is only one critical number, then local max = abs. max and local min = abs. min.

The optimization problems I sometimes find rather confusing however they provided good steps as a way to solve them right from the start as discussed in class. The examples that they provided were like the ones that we did in class so it helped to read over them again. I thought all the examples that they provided along with showing that the first example you could have drawn it out different ways but showed how to get the right drawing for the problem made the optimization problems a lot easier to understand.

I’ve never done an optimization problem before. All of the information in the reading seems very confusing to me at the moment.

Yes, optimization problems are one of the most difficult types encountered in first-semester calculus. The only way to become proficient at them is do practice a lot. Then you begin to see that you are performing the same steps in every problem, even though all problems seem totally different at first.

Im very thankful that the book gave steps in solving the problems pertaining to optimization. The steps that are provided help but to me its still difficult to follow the steps through the examples, especially example 3. Also since this is my first time learning and reading about optimization, that could also be a reason for my confusion as well.

Yes, optimization problems are tricky and definitely require a lot of practice. Some of the previous comments above may help you visualize and understand how to think about Example 3.

The goal of and procedure for optimization problems makes sense. Making the equations and establishing the relationships between them might be difficult. I was a little confused about the closed interval method and also how it was determined that x has to be less than or equal to 1200, in example 1. I also don’t understand how implicit differentiation can be used instead, is it just another way to solve for a variable like we did for the last test?

The closed interval method goes back to section 4.1. The abs. max/min has to occur at a critical number or at an endpoint of an interval. In most of these problems, there is a interval of possible values for the variable. In example 1, for instance, since there are only 2400 feet of fencing available, the largest x can be is 1200. This isn’t a realistic value of course, because in this case, we would have two vertical section of fence (1200 feet each), and then nothing left for the horizontal side. So the area enclosed would be 0. At the other extreme, x could be 0, which means y=2400. But again, this is not realistic, since then there would just be a 2400 ft length of fence up against the river. Since there are no vertical sections, the area enclosed is again 0.

Generally, the endpoints in these types of problems produce unrealistic situations (but are “theoretically possible”). So for the most part, you are safe with assuming that the number you’re looking for is the critical number.

I thought i was very helpful with the idea of doing it in steps to get the quantity that is to be maximized or minimized to get expressed as a function of one of the variables in the function and that it has to be expressed in terms of some other. I wanted to know about the alternative method of implicit differentiation and when is it the best case to use it for optimization problems

Yes, implicit differentiation is another technique that can be used. Sometimes it can be easier, but it requires you to be very proficient with implicit differentiation.

The steps to solving optimization problems was very helpful because in class I didn’t feel like I fully understood how to do it by myself but now I am confident. The examples were also helpful because the illustrations and the break down. But the steps are still a bit confusing especially example 4 because I don’t understand how they go from the time equation to inputting in the numbers. But besides that I understood the rest of the examples and it made a bit more easier to grasp the concept.

In Example 4, they find an equation for time, T(x). They differentiate and solve for when T'(x)=0. This gives the one critical number x=9/sqrt(7).

The smallest possible value of x is 0 (this means running all 8km in the picture) and the largest possible value of x is 8 (this means rowing all the way, and not running at all). From section 4.1, the abs. minimum time occurs at a critical number (x=9/sqrt(7)) or at an endpoint (x=0,8). At the bottom of page 329, they plug these three values in T(x) to see which one produces the minimum time.

I always have trouble with optimization so these examples were a little difficult to follow to begin with. Example 2 is similar to the one you did in class but i think I got it. What happened to the h’ at the end of the problem? In example 3, how was d=sqrt ((x-1^2)+(y-4)^2) calculated.? It looks as if it just appeared. He rest of the example was easy to follow.

At the top of page 328, they calculate h based on the value they found for r. They note that when they simplify, the value they get for h is actually twice the value of r, so they write h=2r (and r=cuberoot(500/pi)).

Example 3 uses the distance formula. To find the distance between two points, find the distances between the x- and y-coordinates, square them, and add them; then take the square root.

In general, distance between (x1,y1) and (x2,y2) is

d = sqrt((x2-x1)^2 + (y2-y1)^2)

This is really just the pythagorean theorem if you draw a picture. d is the hypotenuse of a right triangle with base x2-x1 and height y2-y1.

Example one was fairly simple on page 326 but Im confused with the substituting for example 2 and most of the equation disappeared with out explanation. Very confused. And is it possible to get multiple critical values or no??

Do you mean at the top of page 328, when substituting to find h?

If so, here is the detail, which basically involves playing around with exponents:

r = (500/pi)^{1/3}

r^2 = ((500/pi)^{1/3})^2

= (500/pi)^{2/3}

= 500^{2/3} / pi^{2/3}

Substitute into h = 1000/(pi*r^2) to get

h = 1000 / (pi * (500^{2/3}/pi^{2/3}))

= 2*500 / pi^{1/3}*500^{2/3}

= 2*500^{1/3} / pi^{1/3}

= 2*(500/pi)^{1/3}

= 2*cuberoot(500/pi)

= 2*r.

Multiple critical values are possible, although don’t generally occur in the types of problems we do in section 4.7.

I understand the first 6 steps on page 325, they are very clear and actually make it easy to understand. For example 1, I follow the math and the steps pretty well. As for drawing the picture from the description, that will definitely be the most complicated part of the problem for me and the part that I need to practice on the most. I was wondering if there was a way you could do the problem without drawing a picture? Because if that is the only way to start the problem, I may struggle to even get it started. For example 3 on page 328, I think starting the problem is also confusing because it is set up so differently than the other two problems. But, again, all of the mathematical steps in it make sense to me.

You can do these problems without a picture, but it makes it harder to determine what variables to use and how to get the equations.

There is some discussion in the comments above about another way to think of the scenario in Example 2.

Think of the parabola as a curved wall in a building, and imagine there is an electrical pole at position (1,4). The question is basically saying, if we want to use the least amount of wire, and need to run a wire from the electrical pole to the building, where should the wire go?