Change in temperature and avg. temperature

[dbag]

Problem statement: Object cools exponentially so that temperature T in Celsius after t-minutes is . Calculate the change in temperature during the first 10 minutes and also the avg. temperature.

Solution:
a) change in temperature would be acquired by deriving the function and i get -15e^-0,1t then plugging T(10)= -5,5C/min as change
b) Avg temperature is calculated using and after the operations it equals to 1/10(30x-1500e^-0,1t) and taking definite integral between 0 and 10 i get an aswer of 124,8 Celsius. Am i missing anything or is it wrong?

 

[ksdhart2]

The wording on part (a) is not 100% clear to me. I can think of a few different ways to interpret what they're asking. The first thought I had of how to interpret it was to find how much the object's temperature has changed over the first 10 minutes. However, this would require no calculus whatsoever, being given by \(\displaystyle T(10) - T(0) = \frac{150(e - 1)}{e} \approx 94.8181\).

What you've found is another way of reasonably interpreting the problem - you found the instantaneous rate of change in temperature at T = 10 minutes. If possible, I'd recommend consulting with your professor and/or whoever wrote the problem, for some clarification on the intent here.

That aside, I agree with your answer for part (b), though the workings you wrote here has a minor mistake - the function is \(T(t)\), so where does \(x\) enter the picture? Second, if I were grading this, I'd like to see it cleaned up a bit and stated more explicitly. Perhaps:

\(\displaystyle \frac{1}{10} \cdot \int\limits_{0}^{10} 30 + 150e^{-0.1t} \: dt = \frac{1}{10} \cdot \bigg[ 30t - 1500e^{-0.1t} \bigg]_{0}^{10}\)

 

[dbag]

Thank you very much

 

[HallsofIvy]

Problem statement: Object cools exponentially so that temperature T in Celsius after t-minutes is View attachment 12454. Calculate the change in temperature during the first 10 minutes and also the avg. temperature.

Solution:
a) change in temperature would be acquired by deriving the function and i get -15e^-0,1t then plugging T(10)= -5,5C/min as change

No, the derivative gives the instantaneous rate of change and that is not asked here. To find the "change in temperature during the first 10 minutes" calculate the temperature at t= 10, \(\displaystyle 30+ 150e^{-1}= 30+ 55.18= 85.18, the temperature at t= 0, 180, and subtract.

b) Avg temperature is calculated using View attachment 12455and after the operations it equals to 1/10(30x-1500e^-0,1t) and taking definite integral between 0 and 10 i get an aswer of 124,8 Celsius. Am i missing anything or is it wrong?

\)

 

[dbag]

No, the derivative gives the instantaneous rate of change and that is not asked here. To find the "change in temperature during the first 10 minutes" calculate the temperature at t= 10, \(\displaystyle 30+ 150e^{-1}= 30+ 55.18= 85.18, the temperature at t= 0, 180, and subtract.

\)

makes sense, afterall the problem asks for a total amount of temp changed during the first 10 minutes. Thanks very much