What is the Antiderivative of Position in Calculus?

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In summary, the conversation covered the topic of antiderivatives in math class, specifically the relationship between acceleration, velocity, and position. The group discussed integrating the position function with respect to time and the resulting units of length*time. It was acknowledged that while it may not have a practical physical use, it is an interesting concept to explore. The original question of whether the antiderivative of a position has a specific physical meaning was also addressed.
  • #1
JimbozGrapes
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Well sitting in math class today learning about antiderivatives we talked about the following

Accelerations antiderivative is velocity

and,

velocities antiderivative is position.

So I asked my teacher what the antiderivative of a position would be then, and he did not know.

So if someone could tell me what he antiderivative of a position would represent that would be a massive releif to my curiosity.

Thanks.
 
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  • #2
Integrating the position function with respect to time would give you a quantity with units m s (length*time). Do you know of any physical quantity with those units? I don't.
 
  • #3
well i have read this in line integrals that to integrate a continuous function f(x,y,z) which is here position function of an object over curve c integratef(g(t),h(t),k(t))lv(t)ldt where v(t)=dr/dt and coordinates x,y,z are functions of time i guess it has something to do this with this type of problem
 
  • #4
Line integrals are something different. Seeing as the examples he listed were [itex]\int a(t)dt=v(t), \int v(t)dt=x(t)[/itex] it stands to reason he meant [itex]\int x(t)dt=X(t)[/itex]. This is of course not a line integral.
 
  • #5
if position is integrated with respect to time it must be position function and its coordinates can be function of time also.Is it not so?
 
  • #6
I am not quite sure what you mean with that, could you elaborate and perhaps give a concrete example?
 
  • #7
suppose position function is f(x,y,z) of a particle where x,y,z are functions of time t i.e x=h(t) y=g(t) and z=k(t) then it can be integrated i think so
 
  • #8
Of course it can. The question is whether or not there is a useful application for the integral of the position. I cannot think of one, that doesn't mean there isn't one of course. But your example just extends it to three dimensions which doesn't really change the physical interpretation.
 
  • #9
Yes, no one is disputing that the position function can be integrated. What is in dispute is the idea that its integral must have some specific physical meaning. As Cyosis said in the first response to your post, if the position function is integrated with respect to time, then the result must have units of "length times time" or "meters times seconds". There is not regularly defined physics quantity that even has those units!
 
  • #10
Units:
Acceleration has the units of length /time^2. Taking the antiderivative of the acceleration causes you to multiply the acceleration units by time. So you end up with length/time which are the units of velocity. Taking the antiderivative of velocity, multiply by time again, to get units of length, which is ...um..length or distance. If you take the antiderivative of distance, you have to multiply by seconds again to get units of
length*time. Can you think of any physical measurement that you can do that gets you units of length*time?

You can certainly take antiderivatives of a function all day long if you want. Nothing is preventing you. The question though, is does that answer have any physical meaning? I could take cats and divide by the color red. Does that have any physical meaning? (I hope not :)
 
  • #11
Bah...beaten to the punch.
 
  • #12
thanks a lot for the responses everyone, so the antiderivative of a position would give the units length*time is what i gathered. so something like m*s, which is kind of interesting weather it has practical physical use or not haha. Ill have to run that by my teacher and hopefully get a bonus mark or two :) .
 
  • #13
Not a slap on the ears? A very patient teacher!:rofl:
 

1. What is an antiderivative of position?

An antiderivative of position is a mathematical function that represents the original position function of an object. It is the inverse operation of differentiation, which calculates the rate of change of position over time.

2. Why is the antiderivative of position important in physics?

The antiderivative of position is important in physics because it helps us understand the motion of objects. By knowing the position function, we can determine the velocity and acceleration of an object at any given time.

3. How is the antiderivative of position calculated?

The antiderivative of position is calculated by using the reverse power rule, which states that the antiderivative of x^n is (x^(n+1))/(n+1), where n is any real number except -1. This means that if the position function is known, the antiderivative can be found by raising the variable to the power of one more than the original exponent, and then dividing by that new exponent.

4. Can the antiderivative of position be negative?

Yes, the antiderivative of position can be negative. This indicates that the object is moving in the negative direction, or in the opposite direction of the positive direction on a coordinate axis. The sign of the antiderivative is important in determining the direction of an object's motion.

5. How does the antiderivative of position relate to the area under a velocity-time graph?

The antiderivative of position is directly related to the area under a velocity-time graph. This is because the antiderivative represents the displacement of an object, and the area under the velocity-time graph represents the change in position over time. Therefore, the antiderivative of a velocity function is the position function of the object.

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