- #1
Devil Moo
- 44
- 1
Is Differentiation exact or just an approximation?
I am wonder whether this question is meaningful or not. Slope is expressed as "it is approaching to a value as x is approaching 0" so it is inappropriate to ask such question. But when I deal with uniform circular motion, it is very confusing.
Suppose ##A## is constant for vector ##\mathbf A##. And the angle between ##\mathbf A(t+\Delta t)## and ##\mathbf A(t)## is ##\Delta \theta##.
##\begin{align}
\Delta \mathbf A & = \mathbf A (t + \Delta t) - \mathbf A(t) \nonumber \\
| \Delta \mathbf A | & = 2A \sin (\Delta \theta / 2) \nonumber
\end{align}##
if ##\Delta \theta \ll 1##, ##\sin (\Delta \theta / 2) \approx \Delta \theta / 2##
##\begin{align} | \Delta \mathbf A | & \approx 2A (\Delta \theta / 2 \nonumber \\
& =A \Delta \theta \nonumber \\
| \Delta \mathbf A / \Delta t | & \approx A (\Delta \theta / \Delta t) \nonumber
\end{align}##
if ##\Delta t \rightarrow 0##,
##| d \mathbf A / dt | = A (d \theta / dt)##
But isn't it ##| d \mathbf A / dt | = 2A (d \sin (\Delta \theta / 2) / dt)##?
Is ##v = r \omega## not accurate compared with ##v = 2r (d \sin (\Delta \theta / 2) / dt)##?
I am wonder whether this question is meaningful or not. Slope is expressed as "it is approaching to a value as x is approaching 0" so it is inappropriate to ask such question. But when I deal with uniform circular motion, it is very confusing.
Suppose ##A## is constant for vector ##\mathbf A##. And the angle between ##\mathbf A(t+\Delta t)## and ##\mathbf A(t)## is ##\Delta \theta##.
##\begin{align}
\Delta \mathbf A & = \mathbf A (t + \Delta t) - \mathbf A(t) \nonumber \\
| \Delta \mathbf A | & = 2A \sin (\Delta \theta / 2) \nonumber
\end{align}##
if ##\Delta \theta \ll 1##, ##\sin (\Delta \theta / 2) \approx \Delta \theta / 2##
##\begin{align} | \Delta \mathbf A | & \approx 2A (\Delta \theta / 2 \nonumber \\
& =A \Delta \theta \nonumber \\
| \Delta \mathbf A / \Delta t | & \approx A (\Delta \theta / \Delta t) \nonumber
\end{align}##
if ##\Delta t \rightarrow 0##,
##| d \mathbf A / dt | = A (d \theta / dt)##
But isn't it ##| d \mathbf A / dt | = 2A (d \sin (\Delta \theta / 2) / dt)##?
Is ##v = r \omega## not accurate compared with ##v = 2r (d \sin (\Delta \theta / 2) / dt)##?
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