If t = 1 and x = 1.00; If t = 2 and x = 0.45; If t = 3 and x = 0.29;
How do you find dx/dt = ... if the function is assumed exponential?
Thanks!
If t = 1 and x = 1.00; If t = 2 and x = 0.45; If t = 3 and x = 0.29;
How do you find dx/dt = ... if the function is assumed exponential?
Thanks!
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Assume that $x=f(t)=A\cdot B^t+C$.
Then, $\displaystyle \frac{dx}{dt} = f'(t)=A \cdot B^t \cdot \ln B$
First, observe that $A \cdot B^1 +C=1, \quad A\cdot B^2+C=0.45, \quad A\cdot B^3+C=0.29$
This will give,
$A\cdot B= 1-C, \quad A\cdot B^2 =0.45-C, \quad A\cdot B^3 =0.29-C$.
But we know $\displaystyle \frac{A \cdot B^2}{A\cdot B} = \frac{A \cdot B^3}{A\cdot B^2}=B$.
So, $\displaystyle \frac{0.45-C}{1-C} = \frac{0.29-C}{0.45-C}=B$
From here, you can find the value of $C$.
Then you can find $B$ and $A$. Hence you can calculate $f'(t)$.