linear regression on calculator

Question

I have the equation T(t)=T_0 e^kt . and some coordinates (t,T). t: (10,20,30,40,50,60). (T: 45,40,37,33,27,24)

I transform it to y = ae^bx = lna+bx.

I then change the coordinates to the form (x, lny) and use the stat function of my calc to plot in the coordinates.

I get a= 3.981, b= -0.0003636

how do I find the value for a and k? I thought since y = ae^bx is the same form as T(t)= T_0e^kt , a would qual to T_0, and b would equal to k, but in the solutions manual it says that k = -0.127 with no further explanation. Am I supposed to do something with the a and b value to get the right answer?

Im using a casio fx9860GII

Answer 1

You wrote : I transform it to y = ae^bx = lna+bx.

This is not correct. You should have written : y = ln(ae^bx) = lna+bx.

In order to avoid confusion, use clear notations : $$T=T_0e^{kt}$$ which is transformed to $$\ln(T)=kt+\ln(T_0)$$ $$y=ax+b\quad\text{with}\quad \begin{cases} y=\ln(T)\\ x=t\\ a=k\\ b=\ln(T_0) \end{cases}$$ Now you proceed to the linear regression according to the standard form $y=ax+b$

Once you got $a$ and $b$ there is no difficulty at all to find $k$ and $T_0$ from the above equations : $$\begin{cases} k=a\\ T_0=e^b \end{cases} $$

NOTE :

I wonder why you use an exponential function. If you plot the data $(t,T)$ you see that the points are quite lined. Thus, no need for $T=T_0e^{kt}$. More simply : $$T=At+B$$

Answer 2

$$T=T_0e^{kt}$$

becomes

$$\ln T=kt+\ln T_0$$ which is a linear relation of the form $$y=mx+p$$

with the following correspondences:

$$\begin{align}y&\leftrightarrow \ln T,\\x&\leftrightarrow t,\\m&\leftrightarrow k,\\p&\leftrightarrow \ln T_0.\end{align}$$

So after performing the linear regression with the input pairs $(t, \ln T)$, $k$ is given by the slope and $T_0$ by the antilogarithm of the intercept.