Question on derivatives [Solved]

Aumni

Hi guys,

I dove into some online courses from MIT and my basic math skills are a little rusty. In the basics physics course the instructor writes an example on the board:

x = 8 -6t + 2t²

Simple enough, next he says to pull the velocity out of that formula he does:

"All I'm using is"

x = t^n

"Which as all you all know is"
dx/dt = nt^(n-1)

"Therefore"

v = -6 + 2t

I'm having trouble following this a little bit. I understand why the velocity is what it is, but taking x = t^n and applying it to x = 8 -6t + 2t² is where I'm missing something.

I looked up some laws on derivatives but still can't piece this together. Thanks for the help guys.

Infidel

Your confusion may stem from the (instructor's?) mistake that 2t^2 should be 4t, not 2t.

v = -6 + 4t

By x = t^n he is talking about the power rule.

Jebus314

Infidel wrote: »

Your confusion may stem from the (instructor's?) mistake that 2t^2 should be 4t, not 2t.

v = -6 + 4t

By x = t^n he is talking about the power rule.

To clarify, the whole part about x=t^n is just an example of how to use the power rule. He then takes that knowledge and uses the power rule to determine what dx/dt is for the original equation of x=8-6t+2t^2.

Al_wat

it just means what you do with the exponent

x = 8 -6t + 2t²

x = t^n

applying the first to the second looks like this:

x = (8t^0) -(6t^1) +(2t^2)

dx/dt = nt^(n-1)

applying the third equation to the fourth gives you:

dx/dt = 0*(8t^-1) -1*(6t^0) +2*(2t^1)

which reduces to:

v = -6 + 4t

remember that anything to the power of zero equals one; ie t^0 = 1

edit: right, there is a mistake, correcting my answer.

Aumni

That clears it up, thanks guys. Figured it was something simple.