I'm not quite sure my process is right or not.
---Here is the question---
Let $f(t)$ be a given function for $0 \le t \le T$. Assume $g(\tau ) = f(t)$,where $\tau \equiv T-t$ for $0 \le \tau \le T$. Also define $ F(t) = \int_0^t f(\sigma) d \sigma$
(a.) Show that $\displaystyle\frac {dg(\tau)} {d\tau} = -\frac {df(t)} {dt}$
My method:
$\frac {dg(\tau)} {d\tau} = \frac {dg(\tau)} {dt} * \frac {dt} {d\tau} =\frac {df(t)} {dt} * \frac {d(T-\tau)} {d\tau} = - \frac {df(t)} {dt}$
(b.) Show that $\displaystyle F\left( t \right) = \int\limits_{T - t}^T {g\left( \sigma \right)d\sigma } $
My method:
$\because g\left( \sigma \right) = g\left( {T - t} \right) = f\left( t \right)$
let $a = T - \sigma$
$\Rightarrow da = - d\sigma$
$\begin{gathered} \Rightarrow F\left( t \right) = \int\limits_0^t {g\left( {T - \sigma } \right)d\sigma } = - \int\limits_T^{T - t} {g\left( a \right)da} = \int\limits_{T - t}^T {g\left( a \right)da} \\ \end{gathered} $
(c.) Let $f(t)=t^2+2$ for $0 \le t \le 6$
Find $g(\tau)$,and $\displaystyle\frac {dg(\tau)} {d\tau}$
I have two methods, which one is right? or both?
My method 1:
$g\left( \tau \right) = g\left( {T - t} \right) = f\left( t \right) = {t^2} + 2$
$\frac{{dg\left( \tau \right)}}{{d\tau }} = - \frac{{df\left( t \right)}}{{dt}} = - 2t$
My method 2:
$\begin{align*} g\left( \tau \right) & = {\left. {f\left( t \right)} \right|_{t = T - \tau }} \\ & = {\left. {\left( {{t^2} + 2} \right)} \right|_{t = 6 - \tau }} \\ & = {\left( {6 - \tau } \right)^2} + 2 \\ & = 36 - 12\tau + {\tau ^2} + 2 \\ & = {\tau ^2} - 12\tau + 38 \\ \frac{{dg\left( \tau \right)}}{{d\tau }} & = 2\tau - 12 \end{align*} $
Is anything wrong?