I use it from time to time to check answers and compare my derivative to what it actually should be, but that won't do me any good on my final.Quote:
Originally Posted by 2faststealth4u
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I use it from time to time to check answers and compare my derivative to what it actually should be, but that won't do me any good on my final.Quote:
Originally Posted by 2faststealth4u
This would be better explained with a pencil and paper in person, or perhaps over Skype/Facetime.Quote:
Originally Posted by Vantage
I'll do the best I can with only a keyboard at my disposal.
Part A) Not sure what they're on about with the C^-1(t) business. Are they trying to say the inverse of the answer? Basically, the temperature of the soda is C, and the temperature varies with time in minutes, t. This means that C is a function of t or C(t). After ten minutes, C(10), the temperature of the soda is 62F. C some function of t (could be linear, could be polynomial or anything - we don't know yet) after 10 minutes is 62 Fahrenheit.
C^-1(45 minutes) = 40. If I am understanding this then the inverse of the inverse will give you C(45 min) = 1/40 of a degree Fahrenheit. Seems quite cold for a refrigerator so I may be missing something here. Pretty shaky on this Part A, and unfortunately this will screw us later on Part D.. Moving on.
Part B) Ah! C prime. The first derivative of our unknown function "C(t)". So what is C'(t)? It is the rate of change of temperature over time. Given our units, it is Degrees F-per-minute. So how quickly is the soda cooling down at any given moment. Given what we know in Part B, after exactly 10 minutes of sitting in the fridge the soda is cooling down at a rate of -.4F/minute. Because the number is negative, temperature is decreasing as time increases.
Part C) Integral of a derivative! Integrating C'(t) (Degrees F-per-minute) puts you back to C(t) or Degrees F. The bounds of this integral are 0 to 10 minutes, so from the point of time that you put the soda in the fridge (0 minutes) to now (10 minutes later), integrate that C'(t) function. Confused? Let's think of this in terms of a car moving down the road for 10 minutes. If C(t) is position of the car in miles, then the derivative C'(t) is the velocity of the car in miles-per-minute. If you integrate the velocity C'(t) from 0 minutes to 10 minutes then you will have the total distance that the car traveled in miles. Getting back to the soda, integrating C'(t) from 0 to 10 minutes will give you the total temperature in Fahrenheit that the soda has cooled down. The number you get will then need to be subtracted from the initial temperature to determine how much the soda has cooled down.
In this example it says that integral of C'(t) from 0 to 10 minutes is -5F, so the soda has cooled down 5F since you put it in the fridge 10 minutes ago.
Part D) Find the initial temperature at t=0 minutes. Alright! So we know a few things here:
- The temperature of the soda after 10 minutes is 62F
- The temperature after 45 minutes is 1/40th of a degree F (Shaky on this one, I must be misinterpreting C^-1(45))
- The soda's temperature is decreasing at a rate of -.4 F/min after sitting in the fridge for 10 minutes
- The soda has lost 5F in temperature after 10 minutes.
First we need to figure out what our equation is, then we need to put "0 minutes" into our equation and solve. There should be some standard form of differential equation in your book that knowing the above things you can back out the equation. Do you have a section on Newton's Law of Cooling?
Part E) Once you integrate C'(t) you are done. Integrating C(t) is meaningless. You would end up with Fahrenheit-minutes as the units, and what does that mean?
Here's the part of the interpretation worksheet that includes the inverse if it helps:
http://s18.postimg.org/6gu3zhodl/Section_2_4.jpg
So then C^-1(45)=40 just means that C^-1(45) is changing by approximately C^-1(40)' (but then can this be expressed more in words?)? Or is it so simple as to say C(40) = 45, so that after 40 minutes in the fridge, the temp is now 45 *F?
It looks to really be as simple as C(40min)=45F. Makes me wonder why even have this magical C(t)^-1 terminology.Quote:
Originally Posted by Vantage
Anyway, now that you have your boundary conditions, you can solve your differential equation for the constants. Then plug in 0 and find the initial temperature.
Do you know what the generic form of your equation is? Should be something like C(t)=At^2+Bt+C or C(t)=A*ln(t)+B and you just solve for A, B and C.
42. Duh.
Just wow lol.. Its a gift to understand this kinda stuff. I would struggle to the ends of the earth with math like this.
Haha. Win.Quote:
Originally Posted by eljefefx
OP, good luck. Lol
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this is why I use to fight with teachers in school. Where in life would I ever need this?? Good luck to you. You are a better man than me lol
http://www.youtube.com/watch?v=FzCsDVfPQqkQuote:
Originally Posted by hunt.chad
High Quality - Apollo 8 Saturn V rocket launch - YouTube
Please, have a seat and allow me to explain.
Just like everything else guys, it's hard right up until the moment it all 'clicks,' and then you'll wonder how it never made sense before. Thanks again for your help Alex, I'm sure I'll have a more questions over the next few days if you don't mind.