Question about integration modeling

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Orsa
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Question about integration modeling

Postby Orsa » Mon Jan 03, 2011 4:54 pm UTC

So today in calc I got this question.

The rate at which your home consumes electricity is measured in kilowatts. If your home consumes electrucity at the rate of one killowatt per hour, you will be charged for one "kilowatt-hour" of electricity. Suppose that the average consumption rate for a certain household is modeled by the function C(t) = 3.9 - 2.4sin(pi*t/12), where C(t) is measured in kilowatts and t is the number of hours past midnight. Find the average daily consumption for this home in kilowatt-hours.


So at first I figured just to integrate C(t) from 0 to 24. When I did that, I got 93.6 kilowatts. However, the problem wants the units in kilowatt-hours. So I'm not really sure how to convert the units without changing the answer to average hourly consumption. Any suggestions?

gorcee
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Re: Question about integration modeling

Postby gorcee » Mon Jan 03, 2011 5:49 pm UTC

Orsa wrote:So today in calc I got this question.

The rate at which your home consumes electricity is measured in kilowatts. If your home consumes electrucity at the rate of one killowatt per hour, you will be charged for one "kilowatt-hour" of electricity. Suppose that the average consumption rate for a certain household is modeled by the function C(t) = 3.9 - 2.4sin(pi*t/12), where C(t) is measured in kilowatts and t is the number of hours past midnight. Find the average daily consumption for this home in kilowatt-hours.


So at first I figured just to integrate C(t) from 0 to 24. When I did that, I got 93.6 kilowatts. However, the problem wants the units in kilowatt-hours. So I'm not really sure how to convert the units without changing the answer to average hourly consumption. Any suggestions?


Integration with respect to time is similar to adding one "dimension" of time to the units.

Think about it like this: If you have a fence, measured in feet, and you integrate it, you get an area: square feet.

When you differentiate distance with respect to time, you get speed as a unit: feet per second, for instance.

So when you integrate (with respect to time) something measured in kilowatts, you get... kilowatt hours.

So in general, here are some handy-dandy rules:

[math]\int f dx = \mbox{units of f} \cdot \mbox{units of x}[/math]
[math]\frac{d}{dx}\left(f\right) = \frac{\mbox{units of f}}{\mbox{units of x}}[/math]

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lamemaar
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Re: Question about integration modeling

Postby lamemaar » Mon Jan 03, 2011 6:40 pm UTC

You wrote
... I got this question:
"If your home consumes electrucity at the rate of one killowatt per hour..."

I cannot imagine that you really got that question. Not only does it contain two spelling errors, but 'kilowatt per hour' is nonsense. You should not confuse units of power (kilowatt) and of energy (kilowatt-hour). You could say that a kilowatt is a kilowatt-hour per hour.

gorcee
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Re: Question about integration modeling

Postby gorcee » Mon Jan 03, 2011 7:22 pm UTC

lamemaar wrote:You wrote
... I got this question:
"If your home consumes electrucity at the rate of one killowatt per hour..."

I cannot imagine that you really got that question. Not only does it contain two spelling errors, but 'kilowatt per hour' is nonsense. You should not confuse units of power (kilowatt) and of energy (kilowatt-hour). You could say that a kilowatt is a kilowatt-hour per hour.


The spelling errors could be typos. And it could be a paraphrased example.

People often have trouble with the idea that a kilowatt, by itself, is defined as a thing per unit time. I wouldn't be terribly surprised, even, if the textbook got it wrong.

Notwithstanding, you do emphasize a good point: kilowatts measure comsumption [imath]rate[/imath], so kilowatt-hours measure consumption.

Dark Avorian
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Re: Question about integration modeling

Postby Dark Avorian » Mon Jan 03, 2011 7:58 pm UTC

C(t)is consumption rate, measured in KILOWATTS

x is time, measure in HOURS

An integral is an area, the units of which are the product of the units of the function, and the units of the variable

therefore the integral is in kilowatt hours

(Further Explanation: If we go back to the most basic integral, it's the sum of an infinite number of rectangles... (limit of the sum actually) So look at each rectangle, it is a number of kilowatts times a number of hours, so it's a number of kilowatt-hours, and adding them together doesn't change the units.
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Re: Question about integration modeling

Postby skeptical scientist » Mon Jan 03, 2011 8:08 pm UTC

You can think of dt as having the same units as t. So if you measure t in seconds (and C(t) in kilowatts), and you integrate \(\int_0^{3600} C(t) dt\), you are integrating something (C(t)dt) that has units of kilowatt-seconds, so the result is in kilowatt-seconds (or kilojoules). If you measure t in hours, and you integrate \(\int_0^1 C(t) dt\), you are integrating something (C(t)dt) that has units of kilowatt-hours, so the result is in kilowatt-hours.
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Orsa
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Re: Question about integration modeling

Postby Orsa » Mon Jan 03, 2011 8:28 pm UTC

Thank you guys. I did know about the units tricks; I guess the wording of the question was just what was confusing me. :)


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