1. ## Time derivative

Hi,
I have the following problem: I have a function to evaluate the obliquity of the ecliptic (astronomy) in the form:
Code:
f(t) = a + bt + ct^2 + dt^3
Its derivative (rate) is:
Code:
derivative of f(t) = b + 2ct + 3dt^2
Everything works out until here. But I've improved the formula; but now the t factor is equal to t/100;let's call it t', and with the new coefficients a', b', c' and d':
Code:
f_2(t') = a' + b't' + c't'^2 + d't'^3
If I calculate f_2 for a given t the result is the same, but improved. But what about the derivative of f_2? It's:
Code:
derivative of f_2(t') = b' + 2c't' + 3d't'^2
The rate, or derivative is referred to t' = t / 100; to referrer it to t, must I divide de derivative of f_2 by 100?
Kind regards,
Kepler

2. Originally Posted by kepler
Hi,
I have the following problem: I have a function to evaluate the obliquity of the ecliptic (astronomy) in the form:
Code:
f(t) = a + bt + ct^2 + dt^3
Its derivative (rate) is:
Code:
derivative of f(t) = b + 2ct + 3dt^2
Everything works out until here. But I've improved the formula; but now the t factor is equal to t/100;let's call it t', and with the new coefficients a', b', c' and d':
Code:
f_2(t') = a' + b't' + c't'^2 + d't'^3
If I calculate f_2 for a given t the result is the same, but improved. But what about the derivative of f_2? It's:
Code:
derivative of f_2(t') = b' + 2c't' + 3d't'^2
The rate, or derivative is referred to t' = t / 100; to referrer it to t, must I divide de derivative of f_2 by 100?
Kind regards,
Kepler
Yes, the derivative with respect to t' is $b'+ 2c't+ 3d't'^2$. Now, by the chain rule, to get the derivative with respect to t, multiply by the derivative of t' with respect to t. Since t'= t/100, that is 1/100.