2
$\begingroup$

Let $f(t)$ and $g(t)$ both be smooth functions of time and $dh(t)/dt = af(t) - bh(t)$. At the turning point in $h(t)$ we have

$$ af(t) - bh(t) = 0 $$

Can I re-arrange this and take the time derivative of both sides as follows?

$$ \frac{a}{b}\frac{df(t)}{dt} = \frac{dh(t)}{dt} = 0 $$

thus generating a condition for the turning point of $h$ to be

$$ \frac{df(t)}{dt} $$

$\endgroup$
1
  • $\begingroup$ That condition is only true for a critical (turning) point of $h$, so the derivative of the condition may not be $0$, $\endgroup$
    – Keshav
    May 3, 2020 at 15:39

1 Answer 1

2
$\begingroup$

There is an inconsistency in how you are treating your symbols between these two equations:

$$h'(t) = af(t) - bh(t)$$ $$0 = af(t) - bh(t)$$

Note that in the first, you establish that the image of a particular value $t$ under the function $h'$, is equal to some algebraic combination of the images of this particular $t$ under two different functions, $f$ and $h$.

In the second equation, you continue to treat $f$ and $h$ as functions. However, since you have asserted that $h'(t) = 0$, we are now saying that for some particular value of $t$ in the domain, call it $t_0$, that $$h'(t_0)= 0 = af(t_0) - bh(t_0)$$

Rewriting it this way we can see that the RHS is reduced to a value. It is the image of $t_0$ under the function $h'$.

In other words, $\frac{a}{b} \frac{d}{dt} f(t_0) = \frac{d}{dt} h(t_0) = 0$. This is not the same as the general conclusion $\frac{a}{b} \frac{d}{dt} f(t) = \frac{d}{dt} h(t) = 0$. In the former we are taking the derivative with respect to a value, in the latter it is the derivative with respect to a function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.