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Ahsome

  • one year ago

Question on exponents

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  1. ahsome
    • one year ago
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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

  2. anonymous
    • one year ago
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    is it the temperature inside the oven that is increasing or its film surface outside temperature?

  3. ahsome
    • one year ago
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    I believe the temperature inside the oven that is increasing @chris00

  4. anonymous
    • one year ago
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    what have you been studying in this subject? have you learnt first order equation? differential equation? etc..or not?

  5. ahsome
    • one year ago
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    We've done differential, this is on the "indices and indices laws" section, with exponential growth and decay

  6. anonymous
    • one year ago
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    yeah right

  7. anonymous
    • one year ago
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    its been ages since ive done this um..

  8. anonymous
    • one year ago
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    have you learnt logistic functions?

  9. ahsome
    • one year ago
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    Yes

  10. anonymous
    • one year ago
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    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

  11. phi
    • one year ago
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    It is just saying the rate is dropping , not the max temp.

  12. anonymous
    • one year ago
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    perhaps solve \[\frac{ dT }{ dt }=0.1(500-T)\]

  13. anonymous
    • one year ago
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    i'm only throwing ideas here

  14. ahsome
    • one year ago
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    The question is really confusingly worded. Will giving the equation help us understand what we need to do?

  15. phi
    • one year ago
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    yes, solving the differential equation sounds good.

  16. anonymous
    • one year ago
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    do you know how to solve the differential equation @Ahsome

  17. anonymous
    • one year ago
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    i think thats a logarithm function you learn in final year in high school...

  18. phi
    • one year ago
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    Here is a plot

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  19. anonymous
    • one year ago
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    i was so interested in this question because i was like first order system! thats what i'm learning now in process control

  20. ahsome
    • one year ago
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    Sorry! Had to have dinne. This is the answer \[T=500-480\times0.9^m\]

  21. anonymous
    • one year ago
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    correct, although we tend to use the exponential function in the equation.

  22. anonymous
    • one year ago
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    \[T(t)=500-480e ^{-0.1t}\]

  23. anonymous
    • one year ago
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    so by simply solving the differential equation, we can get to this solution

  24. anonymous
    • one year ago
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    can you do this?

  25. ahsome
    • one year ago
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    We haven't used Euler's number yet @chris00

  26. anonymous
    • one year ago
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    thats how you solve the differential function of \[\frac{ dT }{ dt }=0.1(500-T)\]

  27. anonymous
    • one year ago
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    have you learnt how to solve differential equations?

  28. ahsome
    • one year ago
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    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

  29. anonymous
    • one year ago
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    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.

  30. anonymous
    • one year ago
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    and since the question says that the temperature increases by 10% of (500-T) per MINUTE, then this is a rate

  31. anonymous
    • one year ago
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    we express rates as a differential function, i.e\[\frac{ dT }{ dt}\]

  32. anonymous
    • one year ago
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    this simply means, the change in temperature with respect to the change in time.

  33. anonymous
    • one year ago
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    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

  34. anonymous
    • one year ago
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    i.e \[\frac{ dT }{ dt}=0.1(500-T)\]

  35. anonymous
    • one year ago
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    0.1 is simply a decimal form which represents 10% right?

  36. ahsome
    • one year ago
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    Yup

  37. anonymous
    • one year ago
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    have you sort of grasped this concept yet?

  38. ahsome
    • one year ago
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    I get that. But I am confused where the 480 comes into play :/

  39. anonymous
    • one year ago
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    its comes out when you solve it...

  40. anonymous
    • one year ago
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    \[\frac{ dT }{ dt }=0.1(500-T)\] \[\frac{ dT }{ 500-T }=0.1dt\]

  41. anonymous
    • one year ago
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    \[\int\limits_{}^{}\frac{ dT }{ 500-T }=\int\limits_{}^{}0.1dt\]

  42. anonymous
    • one year ago
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    remember separable differential equations?

  43. anonymous
    • one year ago
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    @Ahsome

  44. ahsome
    • one year ago
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    We haven't done that. This is meant to be WAY simpler than th

  45. ahsome
    • one year ago
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    Than that* @chris00

  46. anonymous
    • one year ago
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    yea interesting :/

  47. anonymous
    • one year ago
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    im not sure if i can help you then :/

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