Preliminary things required.

[PrudensOptimus]

What are the things required for a physics problem to be solvable by Calculus? like given valueables..

and again, what can calculus solve that normal equations couldn't?

 

[chroot]

Calculus is used when quantities change throughout a problem as a function of distance, or time, or whatever.

For example, the basic definition of work is [itex]W = \mathbf{F} \cdot \mathbf{x}[/itex]. This definition is only valid when the force remains constant over the entire distance. If it is not constant, you will have to use calculus to determine the work:

[tex]W = \mathbf{F} \cdot \mathbf{x} \ \ \Rightarrow\ \ W = \int F\ dx[/tex]

- Warren

 

[PrudensOptimus]

You can determine it with normal equations too...

But can calculus be used to solve a "equation only" problem?


Edit: No wonder those kenematics equations always say, only if Acceleration is uniform(constant).

 

[chroot]

If F changes over the distance, there is no way to find the work except by calculus.

And I have no idea what a "normal equation" or an "equation only problem" are.

- Warren

 

[PrudensOptimus]

Originally posted by chroot
If F changes over the distance, there is no way to find the work except by calculus.

And I have no idea what a "normal equation" or an "equation only problem" are.

- Warren

So far I have only touched problems in Physics that are asked to solve by equations derived from calculus. eg. Kenematics. etc

Which restricted acceleration to be constant.

As you mentioned, calculus and integrals come in the play when the problem has a changing acceleration, or force.

But, I was wondering, can calculus be used to solve those problems with constant acceleration? In other words, can the same calculus approach be used to "calculus" problems, and "non claculus" problems?

Example:

[tex]\int F dx = W[/tex]

Given there was 1 N of Force exerting on an object on a table, find the work.

so

[tex]\int 1 dx = x J?[/tex]

 

[chroot]

When F is a constant, it can be pulled out of the integral, like this:

[tex]W = \int F dx[/tex]
[tex]W = F \int dx[/tex]

The integral of dx is just x -- therefore,

[tex]W = F \cdot x[/tex]

when F is constant.

- Warren

 

[PrudensOptimus]

I see. So basically we are using Calculus even though we don't know it!

So that means calculus can be used to solve both constant and nonconstant problems?

 

[chroot]

Originally posted by PrudensOptimus
So that means calculus can be used to solve both constant and nonconstant problems?

Yes. Calculus is just a generalization from constant problems to non-constant problems. When the quantities are constant, the calculus just reduces to algebra.

- Warren

 

[PrudensOptimus]

that's amazing.

but how do you find the velocity if given the displacement was 3 meters?

d3/dx = 0...

velocity = 0?

 

[chroot]

Originally posted by PrudensOptimus
that's amazing.

but how do you find the velocity if given the displacement was 3 meters?

d3/dx = 0...

velocity = 0?

You didn't provide enough information. Velocity is defined as

[tex]v \equiv \frac{\Delta x}{\Delta t}[/tex]

where [itex]\Delta x[/itex] is the change in distance and [itex]\Delta t[/itex] is the change in time.

If you consider the velocity of a car by measuring the distance is moves in one hour, you can get a rough estimate of how fast it is moving -- [itex]\Delta t = 1 \ hr.[/itex] You're probably more used to thinking about the velocity of a car as measured by the speedometer though. You can get a better estimate of its "speedometer" velocity by considering the distance it moves in a smaller unit of time -- maybe a second? -- [itex]\Delta t = 1\ s[/itex]. In calculus, we continue this on... not just a second, not just a microsecond, not just a femtosecond, but an infinitely small interval of time. We call this quantity the car's instantaneous velocity, and we represent it with the symbols

[tex]v \equiv \frac{dx}{dt}[/tex]

where dx and dt mean "infinitesimal change in distance" over "infinitesimal change in time," respectively. Note we went from using [itex]\Delta[/itex] to represent large changes to d to represent infinitesimal changes. Your car's speedometer gives you a pretty decent indication of the car's instantaneous velocity.

Does this make sense?

- Warren

 

[HallsofIvy]

The two main requirements for using calculus to solve physics problems are:

1. You must know calculus!

2. You must know phyics!

 

[PrudensOptimus]

Originally posted by HallsofIvy
The two main requirements for using calculus to solve physics problems are:

1. You must know calculus!

2. You must know phyics!

I know a little.

So when problems give you displacement = 3 m... find velocity .. you can't just take s' ?

 

[chroot]

Originally posted by PrudensOptimus
I know a little.

So when problems give you displacement = 3 m... find velocity .. you can't just take s' ?

There is not enough information to do this, as I said before. You need both distance and time to find velocity. Your question is like asking "I drove 100 miles. How fast did I drive?" The answer is: "Who the hell knows?"

- Warren

 

[HallsofIvy]

As chroot has already said, that's not enough information.


HOW LONG did it take for the object to move the 3 m?

Are you assuming the velocity was constant throughout the move?

IF the object took T seconds to move 3 m AND you know that the velocity was constant, then the velocity was 3/T m/s.

If you do not know if the velocity was constant, then the best you can say is that the average velocity was 3/T m/s.

If you are given the "position function" as a function of time,
x(t),then velocity= dx/dt.

If you are given that the position (not "displacement") is a constant, say x(t)= 3 (meters from some reference point) then
velocity= dx/dt= 0 (the object doesn't move at all).