Would this method still work in an expanding universe?

[nrqed]

But as long as he got that far, we might as well nudge him the rest of the way!

tony: I think your problem is just that you haven't simplified your fraction. What is:

$$\frac{a}{b} \div c$$

?

I agree with you, he is almost there... Except that I am worried about him having a test and starting to try to solve things this way and running out of time

 

[tony873004]

$$\begin{array}{l}<br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)}}}}{h} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)h}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{hx^2 + h^2 x}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {x^3 + x^3 + 2hx^4 + x^2 h + h^2 x - 4} \right)}}{{h\left( {x^2 + hx} \right)}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x^3 + x^3 + 2hx^4 + x^2 h + h^2 x - 4} \right)}}{{\left( {x^2 + hx} \right)}} \\ <br /> \\ <br /> {\rm{Remove all terms with h}} \\ <br /> f'(x) = \frac{{2x^3 - 4}}{{x^2 }} \\ <br /> \\ <br /> f'(x) = 2x - 4 \\ <br /> \\ <br /> \end{array}$$
But by the power rule
$$f(x) = x^2 + \frac{4}{x} \\$$
$$f(x) = x^2 + 4x^{ - 1} \\$$
\\
$$f'(x) = 2x - 4^{ - 2} \\ <br />$$
My algebra way and my power rule way are similar, but not the same.

 

[nrqed]

$$\begin{array}{l}<br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)}}}}{h} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)h}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{hx^2 + h^2 x}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {x^3 + x^3 + 2hx^4 + x^2 h + h^2 x - 4} \right)}}{{h\left( {x^2 + hx} \right)}} \\ <br /> \\ <br /> f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x^3 + x^3 + 2hx^4 + x^2 h + h^2 x - 4} \right)}}{{\left( {x^2 + hx} \right)}} \\ <br /> \\ <br /> {\rm{Remove all terms with h}} \\ <br /> f'(x) = \frac{{2x^3 - 4}}{{x^2 }} \\ <br /> \\ <br /> f'(x) = 2x - 4 \\ <br /> \\ <br /> \end{array}$$

OHHHHHH! Almost!
You only made a tiny mistake in the next to last line! it's -4 /x^2 !

 

[tony873004]

Thanks. Curious3141 just posted this answer too in the other thread. Why does this answer differ from the one I get using the power rule (post #62)?

$$\begin{array}{l}<br /> f'(x) = \frac{{2x^3 - 4}}{{x^2 }} \\ <br /> \\ <br /> f'(x) = 2x - \frac{4}{{x^2 }} \\ <br /> \end{array}$$

** edit. I just verified it by graphing it. My power rule answer is wrong.

 

[Curious3141]

Thanks. Curious3141 just posted this answer too in the other thread. Why does this answer differ from the one I get using the power rule (post #62)?$$\begin{array}{l}<br /> f'(x) = \frac{{2x^3 - 4}}{{x^2 }} \\ <br /> \\ <br /> f'(x) = 2x - \frac{4}{{x^2 }} \\ <br /> \end{array}$$

Separate the terms and cancel out the x's where possible.

 

[nrqed]

Thanks. Curious3141 just posted this answer too in the other thread. Why does this answer differ from the one I get using the power rule (post #62)?$$\begin{array}{l}<br /> f'(x) = \frac{{2x^3 - 4}}{{x^2 }} \\ <br /> \\ <br /> f'(x) = 2x - \frac{4}{{x^2 }} \\ <br /> \end{array}$$

But ${2 x^3 - 4 \over x^2 }$ is equal to $2 x - { 4 \over x^2}$ !

Patrick

 

[tony873004]

Never mind. In my power rule solution, my x mysteriously disappeared. Now they agree.

Thanks so much for you help. This problem was definitely harder than the class lecture examples and book examples and homework problems.

5 pages and 60 posts. You people are awesome.

 

[nrqed]

Never mind. In my power rule solution, my x mysteriously disappeared. Now they agree.

Thanks so much for you help. This problem was definitely harder than the class lecture examples and book examples and homework problems.

5 pages and 60 posts. You people are awesome.

You did all th ework

But if you encountered a limit problem like this in a test, it is much better to treat separately the terms...It's easy to get the limit for the x^2 term and for the 1/x term separately... Combining everything makes things more messy in terms of algebra (an it would be much worse with x^3 - 1/x^2, say). So better to separate things.

Best luck

 

[matt grime]

Here's a simple test for you to consider what a 'definition' is.

Look in your notes at the bit where the teacher states:

definition: the derivative of f at x is...

he or she might explain what that definition is good for, but that is not the definition. And if I were to ask you to define an apple, and you were to say 'a round red fruit' I would point out that that is not even close to being a/the definition of an apple (if such a thing were to make sense), and only a description of certain kinds of apple. A granny smith is not red, and a round red fruit is not in general going to be an apple.

In any case, that still doesn't account for the fact that the sentence you wrote doesn't make sense so even if wordy descriptions were acceptable you still would not have got full/any marks.

 

[tony873004]

Granny Smiths are my favorites. How'd I miss that? Thanks, Matt.

 

[AKG]

I realize what you're saying AKG, but since his formula s(t) defined this as a constant acceleration problem, I felt free to use another method that works fine for constant acceleration.

Okay, first realize that it is HIS FORMULA that "defines" it as a constant acceleration problem, and not the physics that go along with the background story. So if you want to use theorems which only apply in the case of constant acceleration, you have to indicate that you know it is a constant acceleration problem (which you didn't), and that you know why it is a constant acceleration problem (not because physics says so, but how he happened to define s - even if he had defined s in total contradiction to physical laws, all that would matter in this math problem is how he defined s - in particular, it is because s'' is constant). If you don't indicate these things, then what sets you apart from a student who thinks the formula [s'(t₂) - s'(t₁)]/2 will always work? As far as the teacher can tell, nothing. And if nothing sets you apart from such a student, then why should your marks be different? They shouldn't. And should such a student get full marks, if they don't really know what's going on?

 

[Curious3141]

Okay, first realize that it is HIS FORMULA that "defines" it as a constant acceleration problem, and not the physics that go along with the background story. So if you want to use theorems which only apply in the case of constant acceleration, you have to indicate that you know it is a constant acceleration problem (which you didn't), and that you know why it is a constant acceleration problem (not because physics says so, but how he happened to define s - even if he had defined s in total contradiction to physical laws, all that would matter in this math problem is how he defined s - in particular, it is because s'' is constant). If you don't indicate these things, then what sets you apart from a student who thinks the formula [s'(t₂) - s'(t₁)]/2 will always work? As far as the teacher can tell, nothing. And if nothing sets you apart from such a student, then why should your marks be different? They shouldn't. And should such a student get full marks, if they don't really know what's going on?

I think you meant : [s'(t₂) + s'(t₁)]/2

 

[matt grime]

I've got the mindset of a businessman who wants to argue that I've provided the work as dictated by the contract. Methods of project completion were not discussed. The job is finished. It works, now PAY.

And when the person who's payed you looks at the job and realizes the only reason why it was finished to spec is apparently a complete fluke, a fortuitous coincidence that in one particular case just happens to be the correct method, I doubt they'll rehire you, so it's bad business practice on your part.

Say I hire you to do some consulting; I want to know how many more people I need to hire, you go away, come back and say 5, and it turns out to be correct. Now, I come back to you with a different question, you say 5 again and it turns out to be a disaster because your policy is just to say 5 all the time.

Am I not entitled to know what methodology you used, or to see some evidence that your work is correct? You surely don't just hand over a piece of paper with the number 5 on it, but write a report, justifying your conclusions.

 

[tony873004]

I'm a low-voltage electrician. When installing an intercom system that included an electric door release, I said to the customer that the specs of the door release require Power Supply A, and the specs of the intercom require Power Supply B. But since they're the same voltage (usually they aren't), I can save you money by purchasing Power Supply C, which is cheaper than A+B, and has more than enough current to power both, and Power Supply C comes with battery backup so both your systems continue to work through power failures.

So although the specs didn't call for power supply C, and it worked only in this special circumstance (intercoms are usually 12vdc, and doorstrikes 12vac), the customer was quite happy with me, and did re-hire me. Unlike your example of the consultant who answers 5 to everything, this wasn't a case where I just blindly substituted power supply C for every job I performed.

And it wasn't a fluke that I got the right answer on the average velocity question. I fully understand that my method works only under constant acceleration. And I fully understood that this was a constant acceleration problem. Unlike your example of the consultant who answers 5 to everything, I don't just blindly apply that formula to every problem I encounter, hoping it works. I use it only when I know it works. And I knew it would work in this situation.

I had a physics test once following the chapter on the Work/Energy theorum. I looked at a problem and decided that although Work/Energy theorum often saves a lot of time over kinmatics solutions, the basic kinmatics formulas would be easier on one particular problem. Although the teacher didn't specifically say to use Work/Energy, he took away a few points. Several other people in the class also did what I did, so he commented when he passed back the tests that several people did it one way, but since this was the Work Energy theorum test, we should have known that's the method he wanted.

Same thing here. This was our derivative test. I didn't just blindly use this formula, hoping it would work. It made sense to me that he wanted us to compute the answer using what we've learned about derivatives.

 

[matt grime]

I'm a low-voltage electrician. When installing an intercom system that included an electric door release, I said to the customer that the specs of the door release require Power Supply A, and the specs of the intercom require Power Supply B. But since they're the same voltage (usually they aren't), I can save you money by purchasing Power Supply C, which is cheaper than A+B, and has more than enough current to power both, and Power Supply C comes with battery backup so both your systems continue to work through power failures.

So although the specs didn't call for power supply C, and it worked only in this special circumstance (intercoms are usually 12vdc, and doorstrikes 12vac), the customer was quite happy with me, and did re-hire me. Unlike your example of the consultant who answers 5 to everything, this wasn't a case where I just blindly substituted power supply C for every job I performed.

but presumably you explained what you were doing to the person involved, otherwise they might be worried by the fact that you installed a seemingly inappropriate power supply. You didn't explain why you made the substitution in the maths, did you?

And it wasn't a fluke that I got the right answer on the average velocity question. I fully understand that my method works only under constant acceleration. And I fully understood that this was a constant acceleration problem. Unlike your example of the consultant who answers 5 to everything, I don't just blindly apply that formula to every problem I encounter, hoping it works. I use it only when I know it works. And I knew it would work in this situation.

but how were we to know that from what you did?

Suppose, in that power supply situation, that you didn't tell me what you were doing or why, and when I looked at it after you'd finished the job I saw you'd apparently fitted the wrong thing, what am I suppoesd to think? If you don't explain why you're doing something then you can't get the credit for it working.

Of course, as with every analogy, it is a spurious likening really, it might perhaps be better to remember that you're back to being the trainee now, not the trained, you've got to prove to someone that you canjustify why you can switch power supplies and not have any problems. I mean, suppose that in order to pass your electrical engineer practical exam you had to demonstrate installing that system, and you just installed the 'wrong' parts in front of your supervisor without telling them what you were doing. Sure it works, sure the supervisor knows it works, but if they don't hear you say it, they don't know that you actually know it works. It could just be a lucky guess. Switch it round, imagine you're training up an apprentice, and you ask them to demonstrate that they know what they're doing and they do it all against the strict instructions but in a way that works without issue but they don't tell you that they know this is correct. How do you distinguish between the intelligent and the merely lucky?

It is one of the main things to realize in maths that it absolutely does not suffice to just write out some numbers. You need to explain what you're doing, and why. Imagine your teacher were to just write numbers on the board and not explain what he or she was doing with them. You'd be very unhappy with that class I imagine.

 

[tony873004]

That's a good point you make. I've taken 6 physics classes in my life, so when the math questions start to deal with physics, I often can see other ways that students who have never taken physics might see.

 

[tony873004]

Let me re-visit the question I asked in post #8 in this thread

Calculate the average rate of change for the indicated value of x.
f(x)=x²-5; x=3

Since we were not prohibited from using the power rule, my answer was:

f'(x)=2x
f'(3)=2*3=6

My score: 1.25/2

What the teacher wanted is:

6+h.

My reasoning as to why he's wrong:

6+h would be the average rate of change for [x, x+h], but not for x.
x is a single point. It's like asking what's the batting average of a player who has 1 hit in 1 at bat. It's 1/1. Here too, with only a single point, the denominator of a fraction designed to compute an average must contain a number n=number of data points.

So the average rate of change for a single point should be the instantaneous value of that point.

A couple of you have already said that the question makes no sense the way it is worded. Nor does my response that the average rate of change for a single point should be the instantaneous value of that point.

Looking at the the teacher's red ink, I think I can see how he wanted it solved.

scanned copy here:
http://orbitsimulator.com/BA/test3.GIF

$$\begin{array}{l}<br /> {\rm{average rate of change = }}\frac{{f(x + h) - f(x)}}{h} = \\ <br /> \\ <br /> \frac{{\left( {x + h} \right)^2 - 5 - \left( {x^2 - 5} \right)}}{h} = \\ <br /> \\ <br /> \frac{{x^2 + 2xh + h^2 - 5 - x^2 + 5}}{h} = \\ <br /> \\ <br /> \frac{{2xh + h^2 }}{h} = \\ <br /> \\ <br /> \frac{{h\left( {2x + h} \right)}}{h} = \\ <br /> \\ <br /> 2x + h \\ <br /> \\ <br /> {\rm{Average rate of change = }}2\left( 3 \right) + h = 6 + h \\ <br /> \end{array}$$
Does this solution make sense considering the way the question was phrased? Or is it still reasonable to assume that the question made no sense?

 

[Curious3141]

Let me re-visit the question I asked in post #8 in this thread

A couple of you have already said that the question makes no sense the way it is worded. Nor does my response that the average rate of change for a single point should be the instantaneous value of that point.

Looking at the the teacher's red ink, I think I can see how he wanted it solved.

scanned copy here:
http://orbitsimulator.com/BA/test3.GIF

$$\begin{array}{l}<br /> {\rm{average rate of change = }}\frac{{f(x + h) - f(x)}}{h} = \\ <br /> \\ <br /> \frac{{\left( {x + h} \right)^2 - 5 - \left( {x^2 - 5} \right)}}{h} = \\ <br /> \\ <br /> \frac{{x^2 + 2xh + h^2 - 5 - x^2 + 5}}{h} = \\ <br /> \\ <br /> \frac{{2xh + h^2 }}{h} = \\ <br /> \\ <br /> \frac{{h\left( {2x + h} \right)}}{h} = \\ <br /> \\ <br /> 2x + h \\ <br /> \\ <br /> {\rm{Average rate of change = }}2\left( 3 \right) + h = 6 + h \\ <br /> \end{array}$$
Does this solution make sense considering the way the question was phrased? Or is it still reasonable to assume that the question made no sense?

The question would ONLY make sense if the question asked "Calculate the average rate of change over the interval [3, 3 + h].

 

[tony873004]

Thanks, Curious. Sorry for asking the same question twice, but I thought after seeing what he expected as the solution, it might change your mind.

Considering the solution he expected us to use, would it have made sense if the question read: "Estimate the instanteanous rate of change for the indicated value of x."?

 

[Curious3141]

Thanks, Curious. Sorry for asking the same question twice, but I thought after seeing what he expected as the solution, it might change your mind.

Considering the solution he expected us to use, would it have made sense if the question read: "Estimate the instanteanous rate of change for the indicated value of x."?

You don't have to estimate, that would be exactly six.

 

[tony873004]

I just thought of an interesting twist.

If the universe can expand faster than the speed of light, as current theories suggest, but no object can travel faster than the speed of light, if we do ∆p / ∆t between 2 objects far enough apart that one is in a part of the universe which is expanding at a rate > c compared to the other, we'll get an answer that implies it did travel faster that light, while in reality, that's impossible.

So using the definition of average velocity, ∆p / ∆t, wouldn't an explanation that this method only works in a non-expanding universe be required?

I know, I'm grasping for straws here . And I don't know much about relativity, so I'm probably wrong. Any comments?