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September 8[edit]

The Hawthorne Effect and "improvements due to monitoring" of performance[edit]

Hello,

I am an engineer - working in the area of energy efficiency improvements in industrial processes.

We are attempting to build evidence or provide past performance reports where we can show people the value of measruing watts or power consumption - due to something parallel to the Hawthorne Effect. Essentially, we know that management principals state that "what gets monitored gets improved".

Thus, we would appreciate any reports or resources which document the fact that installing watt meters and monitoring the level of energy consumption - continuously - will change the amount of watts/energy being consumed simply due to the effects on human nature to improve upon what is being monitored.

The Hawthorne effect appears to indicate that a temporary improvement will occur when people are "watched". We are looking for long term improvements due to continuous feedback and an awareness that the watts being consumed will be compared to others - thus, taking advantage of the natural competitive nature of people to be the best/most efficient.

Thank you for your input/help.

Sincerely,

Don Voigt, P.E. —Preceding unsigned comment added by 75.86.190.159 (talk) 12:37, 8 September 2008 (UTC)[reply]

There is a list of books, papers, scholarly works, etc, located in the Further Reading section of the article. Have you investigated these? Nimur (talk) 14:00, 8 September 2008 (UTC)[reply]

Thank you, Nimur, for your suggested readings. I was hopeful that someone had already made a review of the literature and may have had a similar experience - looking for statistical correlation of "what gets monitored and recorded, gets improved upon"....without going through a survey of all papers, I'd like to know where to focus my attention to find the answer.

Don —Preceding unsigned comment added by 75.86.190.159 (talk) 14:21, 8 September 2008 (UTC)[reply]

Electric power is comparatively cheap in Norway which has a small population served by hydroelectric generators. At one time domestic electricity was charged at a fixed cost per household installation (main fuse rating) ignoring actual consumption. Then household meters were introduced with charging by watthour; the electricity was still cheap. Only recently has there been much incentive to reduce household electric consumption, partly due to the national interest in exporting more electric power to foreign markets rather than "wasting" it at home. Household electric consumption typically has strong summer/winter variation due to more power being used for heating in winter. Electricity charging in Oslo is now by quarterly invoices that present each household with a graph of the previous 12 months consumption, thereby focusing attention on how one's latest quarter consumption is relative to "par" (as in golf). My understanding of Hawthorne effect and the foregoing is that to reduce domestic consumption consumers need regular fresh reminders that their micro-choices e.g. whether to bother turning off unnecessary lights, etc. matter, or the old relaxed mindset returns. Cuddlyable3 (talk) 23:35, 8 September 2008 (UTC)[reply]

Differentiation[edit]

Can someone show me, in really small steps, how to get ${\frac {dP}{dx}}$ given that $P=2x+2{\sqrt {d^{2}-x^{2}}}$ . --RMFan1 (talk) 13:38, 8 September 2008 (UTC)[reply]

It's unlikely that you mean

d^{2}

since you're going to wind up with confusion over what 'd' is. You should look to differentiate term by term. Mrh30 (talk) 15:01, 8 September 2008 (UTC)[reply]

And you'll want the chain rule, the power rule and the linearity of differentiation. Algebraist 15:06, 8 September 2008 (UTC)[reply]

Well yop can use another letter if you like but I don't think using a d would be confusing. Anyway, there was a mistake: it should have been d^2 - x^2 not +. Ive gone ahead and changed it. The answer should come to
${\frac {dP}{dx}}=2-2{\frac {x}{\sqrt {d^{2}-x^{2}}}}$ . I just don't know why? I doubt there's anything wrong with the question though--it's an example from a the Cambridge STEPs. You can see it yourself here [1], Question 9 --RMFan1 (talk) 15:54, 8 September 2008 (UTC)[reply]

Well, what do you know? Do you know what differentiation is? Do you know the rules of differentiation I linked to above? What have you done in attempting to answer the question? (the given answer is correct, btw) Algebraist 15:59, 8 September 2008 (UTC)[reply]

And if you don't know the chain rule etc. you could do this from first principles - write down an expression for P(x+ δ), use the binomial theorem to expand the second term in powers of δ, subtract P(x), divide by δ, then use

{\frac {dP}{dx}}=\lim _{\delta \rightarrow 0}{\frac {P(x+\delta )-P(x)}{\delta }}

(but knowing and using the rules makes life easier). Gandalf61 (talk) 16:18, 8 September 2008 (UTC)[reply]

If you're learning to do STEP, you won't be expected to do it from first principles, but you will be expected to apply the various rules linked above. In your answer, you would quote them as you use them. Mrh30 (talk) 07:26, 9 September 2008 (UTC)[reply]

mathematical representation of geography[edit]

hi,,can you help me to understand what a mathematical model mean? i want to know the mathematical representation of : the rotation of earth,its longitudes and latitudes59.95.205.94 (talk) 14:58, 8 September 2008 (UTC)chhaupa@yahoo.com[reply]

You could start by reading our mathematical model article. The "latitude and longitude" model represents the Earth by a sphere, and uses two spherical co-ordinates to model the location of any point on the surface of the Earth. We know it isnt 100% accurate - the Earth isn't a perfect sphere, and the model doesn't represent altitude, for example - but it is good enough for many purposes. Gandalf61 (talk) 22:07, 8 September 2008 (UTC)[reply]

questions about polynomials[edit]

I realise that this falls somewhat under the category of 'begging for homework help', but I have been trying for quite a while with two problems from my homework (IB Higher Maths, if you're interested), and I'd appreciate any help you feel you can ethically give me. I apologise also for my mangling of the correct ways to put maths notation into Wikipedia.

14. A cubic polynomial gives remainders (5x+4) and (12x-1) when divided by x^2-x+2 and x^2+x-1 respectively. Find the polynomial.

24. If the polynomial P(x)=x^2+ax+1 is a factor of T(x)=2x^3-16x+b, find the values of a and b.

For the first, I attempted to put them in the form p(x)=d(x)q(x)+r(x) (with the quotients replaced by (ax+b) and (cx+d)) and to set them equal to one another, in the hope of equating the coefficients, but this didn't work. I have a niggling feeling that the solution has to do with some use of the remainder or factor theorem that I haven't fully understood, but I'm not sure how to proceed. For the second I again tried to put them into the form p(x)=d(x)q(x), infering that q(x)=(2x+b), but this led to answers that were wrong.

So if anyone has any helpful suggestions, I would be very grateful.

Thank you,

D aniel (‽) 20:24, 8 September 2008 (UTC)[reply]

Your method for the first question seems right - it worked for me. What went wrong when you tried it? --Tango (talk) 21:18, 8 September 2008 (UTC)[reply]

I don't have it to hand, but as far as I recall I simply got results to do with relating (ax+b) and (cx+d) to each other (ie c=d and so on). I couldn't see a way from there to working out the starting polynomial. D aniel (‽) 06:08, 9 September 2008 (UTC)[reply]

As Tango said, your method is correct. You might just be making silly errors because you're expanding a fair number of terms by hand (I'll admit I had to do the system of equations twice to get the right answer). But in case you're setting up the initial equation incorrectly, here is what you should be starting with:

p(x) = (ax + b)(x² – x + 2) + (5x + 4)

. . . = (cx + d)(x² + x – 1) + (12x – 1)

Then proceed to expand, simplify, and move all the terms without coefficient variables to one side of the equation. You should get 7x – 5 on that side. Just keep checking your algebra until you get a solution where a, b, c, and d are all integers. You will get some relations like a = c, but if you keep working through all the equations you will find a single unique solution set.

Regarding the second problem, note that there are two solution sets (it is a plus/minus pair of solutions). 75.63.57.93 (talk) 06:41, 9 September 2008 (UTC)[reply]

Indeed. First problem leads to four linear equations in the four unknowns. They are independent, so you have a unique solution for a, b, c and d.

In the second problem you get an equation for 2a+b and another for ab, and they have a pair of conjugate solutions.

Incidentally, don't assume that the solutions are all integers. The unknown coefficients happen to be integers in these two cases, but there is nothing in the problems that tells you that. Different values for the constant coefficients in the problems would lead to non-integer solutions for the missing coefficients (for example, consider what happens if you replace -16 by -18 in the second problem). Gandalf61 (talk) 08:46, 9 September 2008 (UTC)[reply]

Thanks a lot to all of you; in the end I did 14 as mentioned above and got (what I assume is) the right answer. I ran out of time before the lesson to do 24, but I will probably try it later as suggested. D aniel (‽) 11:36, 9 September 2008 (UTC)[reply]