Question on exponents

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Question on exponents

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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
is it the temperature inside the oven that is increasing or its film surface outside temperature?
I believe the temperature inside the oven that is increasing @chris00

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what have you been studying in this subject? have you learnt first order equation? differential equation? etc..or not?
We've done differential, this is on the "indices and indices laws" section, with exponential growth and decay
yeah right
its been ages since ive done this um..
have you learnt logistic functions?
Yes
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.
perhaps solve \[\frac{ dT }{ dt }=0.1(500-T)\]
i'm only throwing ideas here
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.
do you know how to solve the differential equation @Ahsome
i think thats a logarithm function you learn in final year in high school...
  • phi
Here is a plot
1 Attachment
i was so interested in this question because i was like first order system! thats what i'm learning now in process control
Sorry! Had to have dinne. This is the answer \[T=500-480\times0.9^m\]
correct, although we tend to use the exponential function in the equation.
\[T(t)=500-480e ^{-0.1t}\]
so by simply solving the differential equation, we can get to this solution
can you do this?
We haven't used Euler's number yet @chris00
thats how you solve the differential function of \[\frac{ dT }{ dt }=0.1(500-T)\]
have you learnt how to solve differential equations?
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
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.
and since the question says that the temperature increases by 10% of (500-T) per MINUTE, then this is a rate
we express rates as a differential function, i.e\[\frac{ dT }{ dt}\]
this simply means, the change in temperature with respect to the change in time.
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
i.e \[\frac{ dT }{ dt}=0.1(500-T)\]
0.1 is simply a decimal form which represents 10% right?
Yup
have you sort of grasped this concept yet?
I get that. But I am confused where the 480 comes into play :/
its comes out when you solve it...
\[\frac{ dT }{ dt }=0.1(500-T)\] \[\frac{ dT }{ 500-T }=0.1dt\]
\[\int\limits_{}^{}\frac{ dT }{ 500-T }=\int\limits_{}^{}0.1dt\]
remember separable differential equations?
We haven't done that. This is meant to be WAY simpler than th
Than that* @chris00
yea interesting :/
im not sure if i can help you then :/

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