A group of students heated a thermal sensor with a blow dryer and then let it cool down. The temperatures $T$ Celsius degrees for each $t$ minutes are shown below:

(It's a long list of values, so I am going to just show 5)

$$T(0) = 69.5$$ $$T(5) = 66.9$$ $$T(10) = 63.8$$ $$T(100) = 35.7$$ $$T(175) = 26.4$$

The adequate model to relate the temperature of the sensor, in Celsius degrees, $t$ minutes after being cooled down, is:

$$T(t) = a + be^{-kt}$$

a. Show that $a = 26$ and $b = 43.5$

The problem here is that I am not sure exactly how to solve this problem. In my view, there are 3 things that I can try:

  • replace $a$,$b$, $T$ and $t$ with the given values and test it for a couple values from the list;

  • put these values in lists in my calculator and use the regression options from the calculator to generate the proper values for the model for the graph

  • calculate this analitically

The problem doesn't specify which method to use, I am just supposed to "assume" the method because I have already solved more than 10 variations of this same exercise only with different numbers and wording in my schoolbook and because that's how it is explained in the examples.

However, whenever possible I like to solve problems analitically whenever possible.

$$\\$$ Anyway, I have tried to solve this analitically the following way:

Isolate $a$ (I used $t = 0$ for this one): $$69.5 = a+be^{-k0} \Leftrightarrow 69.5 = a+b \Leftrightarrow a = 69.5 - b$$

Then I tried to isolate $k$ and replaced $a$ by $69.5 - b$ (I used $t = 10$):

$$63.8 = (69.5 - b) + be^{-k10} \Leftrightarrow 63.8 - 69.5 +b = be^{-k10} \Leftrightarrow \frac{-5.7+b}{b} = (e^{-10})^k \Leftrightarrow \\ k = \log_{e^{-10}}(\frac{-5.7+b}{b})$$

Then I replaced $k$ for $k = \log_{e^{-10}}(\frac{-5.7+b}{b})$ and used $t=100$:

$$35.7 = (69.5 -b)+be^{-(\log_{(e^{-10})}(\frac{-57+b}{b}))100}$$

I wanted to make sure I was solving this right so I put these in the calculator and tried to find the intersection, however I couldnt' get a graph, probably because the value are too disperse.

Solving this analitically I got as far as:

$$-33.8 = \frac{3249b-57b^2+57b^{11}}{-57b^{10}+b^{11}}$$

My question is: Can this problem be solved analitically? If yes and if I have been solving it correctly, could you continue? If not, could you just show me how to solve this?

  • $\begingroup$ What is the last value of the list? It is important because after a long time (you can consider it as infinite) the thermal sensor reaches the thermal equilibrium.. $\endgroup$ – MattG88 Nov 29 '16 at 1:25
  • $\begingroup$ @MattG88 $T(190) = 26.4$ $\endgroup$ – Mark Read Nov 29 '16 at 1:28
  • $\begingroup$ Hint: $e^{-10k}=\left(e^{-5k}\right)^2$. Eliminate $e^{-5k}$ between the equations at $t=5$ and $t=10$, then you get a second relation in $a,b$ alone. $\endgroup$ – dxiv Nov 29 '16 at 1:31

Assuming the model $$T(t) = a + be^{-kt}$$ and admitting that htere is no error in the measurements, using the first points, yo have $$a+b=69.5\tag 1$$ $$a+b e^{-5k}=66.9\tag 2$$ $$a+b e^{-10k}=63.8\tag 3$$ Subtracting $(1)$ from $(2)$ and $(3)$ gives $$b(1-e^{-5k})=2.6 \tag 4$$ $$b(1-e^{-10k})=5.7 \tag 5$$ Making the ratio $$\frac{1-e^{-10k}} {1-e^{-5k} }=\frac{57}{26}$$ Let $x=e^{-5k}$ gives $$\frac{1-x^2} {1-x }=\frac{57}{26}\implies 26 x^2-57 x+31=26(x-1)(x-\frac {31}{26})=0$$ So $$e^{-5k}=\frac {31}{26}\implies k=-\frac{1}{5} \log \left(\frac{31}{26}\right)\tag 6$$ Back to $(4)$ $$b(1-\frac{31}{26})=2.6\implies b=-\frac{338}{25}=-13.52$$ and back to $(1)$, $a=83.02$.

So, based on the first points $$T(t)=83.02-15.52\,e^{-\frac{1}{5} \log \left(\frac{31}{26}\right)t}$$ The problem is that this model would give just ridiculous values for the other values of time.

This shows that some regression works needs to be done; the model being nonlinear with respect to its parameters, "reasonable" estimates are required. So in a preliminary step, let us consider assume that $a=25$ and consider $$T(t)-25=b e^{-kt}\implies \log(T(t)-25)=\log(b)-kt=c-k t$$ which is a linear model.

The linear regression then gives $c=3.88149$ and $k=0.0190052$. So, using (as starting guesses $a=25$, $b=e^{3.88149}=48.50$, $k=0.019$) we can start the nonlinear regression which would lead to $$T(t)=19.6208 +49.8278 e^{-0.0113629 t}$$ which is a very good fit $(R^2=0.999991 )$ and which would lead to the following results. $$\left( \begin{array}{ccc} t & T & T_{calc} \\ 0 & 69.5 & 69.45 \\ 5 & 66.9 & 66.70 \\ 10 & 63.8 & 64.10 \\ 100 & 35.7 & 35.62 \\ 175 & 26.4 & 26.44 \end{array} \right)$$

I really do not understand "Show that $a=26$ and $b=43.5$"


Hint: $\;T(t) = a +be^{-kt}\;$ has $3$ free parameters $a,b,k$. Assuming what you are given are exact values, those parameters can be determined analytically from the first $3$ of the given equations. With the substitution $x=e^{-5k}$ (so that $e^{-10k}=x^2$):

$$ \begin{alignat}{2} T(0) & \;=\; 69.5 && \;=\; a+b \tag{1} \\ T(5) & \;=\; 66.9 && \;=\; a + b \,x \tag{2} \\ T(10) & \;=\; 63.8 && \;=\; a + b \,x^2 \tag{3} \\ \end{alignat} $$

The above solves to $a = 4151/50 = 83.02$, $b = -338/25 = -13.52$, and $x = 31/26 \approx 1.19$.

The values for $a,b$ don't match the ones you posted. If you meant to find a "best fit" function, instead, then the question should state so, and also provide the full set of experimental numbers.

  • $\begingroup$ The problem doesn't state anything, but considering the following exercises refer to a "cloud of points" and "regression models" and I am given a bit list of values then I assume I am supposed to use the calculator. What a mess. Thanks for answering $\endgroup$ – Mark Read Nov 29 '16 at 14:50

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