#$&*
course Phy 231
1:25 pm 9-7-14
My questions for the chapters and the videos are:When plotting a velocity function the velocity is usually plotted on the y-axis and the time is on the x-axis. That means that at any point the acceleration at that time is = to the slope of the line at that point, or the derivative of the velocity function. What would happen if you flipped the time to the y-axis and the velocity to the x-axis? Would the relationship stay the same? Or be inverted? I made this mistake in one of the activities and ended up scrapping all of my problem to start new.
@&
If y vs. x is the function f(x), then x vs. y is the inverse function, which we will denote f^-1 (x).
The derivative of f^-1(x) is related to the derivative of f(x) as follows:
Since
f(f^-1(x)) = x,
it follows that
f(f^-1(x)) ' = x '
where ' indicates derivative with respect to x.
x ' = 1
f(f^-1(x)) ' = (f^-1(x)) ' * f ' ( f^-1(x))
the latter being so by the chain rule.
It follows that
(f^-1(x)) ' * f ' ( f^-1(x)) = 1
so
(f^-1(x)) ' = 1 / f ' ( f^-1(x))
For example, if f(x) = e^x, then f^-1(x) = ln(x).
(f^-1(x)) ' is the derivative of ln(x), and f ' (u) = e^u so that f ' (f^-1(x)) = e^(ln(x)) = x.
Thus
(f^-1(x)) ' = 1 / f ' ( f^-1(x))
gives us
(ln(x)) ' = 1 / e^(ln(x)) = 1 / x.
*@
Also, I am not exactly sure how to put it into one question, but in video 231 018 when you were talking about partial derivatives and percent uncertainty I just couldn’t grasp the concept. If you wouldn’t mind touching on that again in class I would appreciate it.
@&
I addressed your second question in class today.
See my notes on the first question (where the simple answer is that you would get the derivative of the inverse function).
*@
#$&*
course Phy 231
1:25 pm 9-7-14
My questions for the chapters and the videos are:When plotting a velocity function the velocity is usually plotted on the y-axis and the time is on the x-axis. That means that at any point the acceleration at that time is = to the slope of the line at that point, or the derivative of the velocity function. What would happen if you flipped the time to the y-axis and the velocity to the x-axis? Would the relationship stay the same? Or be inverted? I made this mistake in one of the activities and ended up scrapping all of my problem to start new.
@&
If y vs. x is the function f(x), then x vs. y is the inverse function, which we will denote f^-1 (x).
The derivative of f^-1(x) is related to the derivative of f(x) as follows:
Since
f(f^-1(x)) = x,
it follows that
f(f^-1(x)) ' = x '
where ' indicates derivative with respect to x.
x ' = 1
f(f^-1(x)) ' = (f^-1(x)) ' * f ' ( f^-1(x))
the latter being so by the chain rule.
It follows that
(f^-1(x)) ' * f ' ( f^-1(x)) = 1
so
(f^-1(x)) ' = 1 / f ' ( f^-1(x))
For example, if f(x) = e^x, then f^-1(x) = ln(x).
(f^-1(x)) ' is the derivative of ln(x), and f ' (u) = e^u so that f ' (f^-1(x)) = e^(ln(x)) = x.
Thus
(f^-1(x)) ' = 1 / f ' ( f^-1(x))
gives us
(ln(x)) ' = 1 / e^(ln(x)) = 1 / x.
*@
Also, I am not exactly sure how to put it into one question, but in video 231 018 when you were talking about partial derivatives and percent uncertainty I just couldn’t grasp the concept. If you wouldn’t mind touching on that again in class I would appreciate it.
@&
I addressed your second question in class today.
See my notes on the first question (where the simple answer is that you would get the derivative of the inverse function).
*@