Question on exponents

- Ahsome

Question on exponents

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- Ahsome

An oven has a theoretical maximum temperature of 500°C. The temperature of the room is 20°C. Because of radiation losses, when the oven is switched on its temperature T increases each minute by 10% of the difference between its temperature and the theoretical maximum; that is, (500 − T) decreases by 10% each minute.
Set up a model of the oven temperature m minutes after it is switched on.
Question is confusing me so much

- anonymous

is it the temperature inside the oven that is increasing or its film surface outside temperature?

- Ahsome

I believe the temperature inside the oven that is increasing @chris00

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## More answers

- anonymous

what have you been studying in this subject?
have you learnt first order equation? differential equation? etc..or not?

- Ahsome

We've done differential, this is on the "indices and indices laws" section, with exponential growth and decay

- anonymous

yeah right

- anonymous

its been ages since ive done this um..

- anonymous

have you learnt logistic functions?

- Ahsome

Yes

- anonymous

im unsure of the quesiton, you say there is a theortical maximum of 500 degrees right? now, the temperature of the oven increases by 0.1(500-T) each minute, yet the theortical maximum drops by 10% of (500-T)?
yes, theortical max temperature would drop because of radiation heat transfer, convection heat transfer etc but i'm unsure how we could model this with such info

- phi

It is just saying the rate is dropping , not the max temp.

- anonymous

perhaps solve \[\frac{ dT }{ dt }=0.1(500-T)\]

- anonymous

i'm only throwing ideas here

- Ahsome

The question is really confusingly worded. Will giving the equation help us understand what we need to do?

- phi

yes, solving the differential equation sounds good.

- anonymous

do you know how to solve the differential equation @Ahsome

- anonymous

i think thats a logarithm function you learn in final year in high school...

- phi

Here is a plot

##### 1 Attachment

- anonymous

i was so interested in this question because i was like first order system! thats what i'm learning now in process control

- Ahsome

Sorry! Had to have dinne. This is the answer
\[T=500-480\times0.9^m\]

- anonymous

correct, although we tend to use the exponential function in the equation.

- anonymous

\[T(t)=500-480e ^{-0.1t}\]

- anonymous

so by simply solving the differential equation, we can get to this solution

- anonymous

can you do this?

- Ahsome

We haven't used Euler's number yet @chris00

- anonymous

thats how you solve the differential function of \[\frac{ dT }{ dt }=0.1(500-T)\]

- anonymous

have you learnt how to solve differential equations?

- Ahsome

We don't need to differentiate the equation, we just need to come up with it @chris00 :)
I just have no idea how to make that equation from the question itself

- anonymous

well simply temerpature is affected by the time right? We say that the temerpature changes with resepct to time. Hence, temperature being the rate of change with resepect to time.

- anonymous

and since the question says that the temperature increases by 10% of (500-T) per MINUTE, then this is a rate

- anonymous

we express rates as a differential function, i.e\[\frac{ dT }{ dt}\]

- anonymous

this simply means, the change in temperature with respect to the change in time.

- anonymous

and since we know that the the temperature changes by 10% of (500-T) with respect to time (in minute basis) then, we simply equal this to the differential term

- anonymous

i.e \[\frac{ dT }{ dt}=0.1(500-T)\]

- anonymous

0.1 is simply a decimal form which represents 10% right?

- Ahsome

Yup

- anonymous

have you sort of grasped this concept yet?

- Ahsome

I get that. But I am confused where the 480 comes into play :/

- anonymous

its comes out when you solve it...

- anonymous

\[\frac{ dT }{ dt }=0.1(500-T)\]
\[\frac{ dT }{ 500-T }=0.1dt\]

- anonymous

\[\int\limits_{}^{}\frac{ dT }{ 500-T }=\int\limits_{}^{}0.1dt\]

- anonymous

remember separable differential equations?

- anonymous

@Ahsome

- Ahsome

We haven't done that. This is meant to be WAY simpler than th

- Ahsome

Than that* @chris00

- anonymous

yea interesting :/

- anonymous

im not sure if i can help you then :/

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