Question

PLEASE TELL ME THE CONCEPT:

We can say that Instantaneous acceleration is The derivative of velocity w.r.t time
= \(\frac{dv}{dt}\)

But what does it mean when we say that instantaneous acceleration can also mean
The second derivative of x w.r.t time
= \(\frac{d^2x}{dt^2}\) ?

I am new to calculus!!
Please tell me what this means PRACTICALLY! :)

Answer 1

velocity is the derivatve of x w.r.t time

Answer 2

Okay.

Answer 3

I know that but then what? :D

Answer 4

\(v = \frac{dx}{dt}\)

\(a= \frac{dv}{dt}\)

combine them both

Answer 5

Like this?

\[a = \frac{ dv }{ dt } = \frac{ d \frac{ dx }{ dt } }{ dt }\]

So we can just multiply the d and the dt like that directly?

Answer 6

So would I get

\[a = \frac{ d^2x }{ d^2t^2} ?\]

Answer 7

\(a= \frac{dv}{dt} = \frac{d}{dt} (\frac{dx}{dt})\)

think of d/dt as an operation. taking derivative of some quantity w.r.t. time

Answer 8

But but what DOES IT MEAN?

Answer 9

From what I understand v = dx/dt
Means as time becomes really small
dx = small change in distance
dt = small change in time.

WHat does, what you wrote, mean??

Answer 10

okay you're mixing two different but related questions :-
1) wat does the *notation* \(\frac{d^2x}{dt^2}\) mean
2) wat does it mean wid regards to acceleration/velocity/distance

Answer 11

I have no clue. :/
TEACH ME. :)

Answer 12

answering first question is simple.
that notation is, as the name implies is just a notation which refers to "taking derivative of x two times"

Answer 13

And what does that mean?
Taking the derviative twice? :/

Answer 14

if x is the position function, say a projectile : -x^2+2x
then, dx/dt is the first derivative, dx/dt = -2x+2

Answer 15

fine wid that ?

Answer 16

Yes. Kind of.

Answer 17

d^2x/dt^2 is the second derivative, d^2x/dt^2 = d/dx(-2x+2) = -2

so the second derivative of poistion function(Acceleration) is -2

Answer 18

Okay. Thanks but can you, by any chance, tell me what it means practically?

Answer 19

we can take third derivative also for x :-
d^3x/dt^2 = d/dt(-2) = 0

Answer 20

okay, you comfortable wid notation part right ?
d^2x/dt^2 is just a notation. dont do algebra with dx and dt...

Answer 21

Next, lets try to relate dx/dt and d^x/dt^2 to velocity and acceleration

Answer 22

say the position function x = -t^2 + 2t

Answer 23

function x gives us exactly wat position the object is at, at any given time right ?

Answer 24

this is the beginning

Answer 25

What derivative means? Rate of change.
So 'rate of change' of distance x is velocity v
and 'rate of change' of velocity is acceleration...

Answer 26

What does it MEAN?!

Answer 27

the "derivative of a function" produces another function.

since you can take the "derivative of a function" ... you can also take the (derivative of the "derivative of a function") ...

all a derivative does is relate how the function is moving at a given point in the domain. When 2 function have the same successive derivatives all the way thru ... they move and act as one. if they move an act alike over all their domains, they are equivalent. If there is only a finite region in their domains that they act alike in, then their sameness is defined by the interval in which they converge.

Answer 28

the rate of change in distance with respect to time is defined as velocity, velocity is a function of time

the rate of change in velocity with respect to time is defined as acceleration, acceleration is a function of time

the rate of change in acceleration with respect to time is defined as jerk, jerk is a function of time

the rate of change of jerk with respect to time is to my knowledge not defined by a specific name; but it is still a function of time ...

Answer 29

Thank you! I got quite a bit of that!
But what does d mean in dx/dt
And from that what would be d^2 signify?

Answer 30

d/dt of a function is just an operation on the function. You could just as well have wrote this as D(x) for 1st derivative or D(D(x)) to represent second derivative. If you think of this as just a function, then it might make more sense that D(D(x)) = \(\Large {d\over dt} {d\over dt}x={d^2x\over dt^2}\). The "d" and "dt" portions of this notation are derived from the ratios of "differences" between endpoints taken at smaller and smaller intervals: http://en.wikipedia.org/wiki/Derivative#Differentiation_and_the_derivative