Questions 

#$&*

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).

*@

Questions 

#$&*

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).

*@