1. Nov 23, 2004

### Fritz

Uniform motion:

dv/dt = 0

ds/dt = v (a constant)

integral(dx) = v. integral(dt) => x = x(subsript 0) +vt ???

Uniform accelerated motion: dv/dt = constant

Equations of motion are:

dv/dt = a => v = v(subscript 0) + a.t ???

dx/dt = v => x = x(subscript 0) + v(subscript 0).t + 1/2.a.t^2 ???

v.dv/dt = a => v^2 = v(subscript 0)^2 + 2.a.(x-x(subcript 0)) ???

I don't understand the lines with question marks next to them.

2. Nov 23, 2004

### matt grime

They've just got an integral equation, from a differential equation, and solved it, albeit with the standard abuse of notation.

3. Nov 23, 2004

### Fritz

I still don't understand.

How do you get from ds/dt = v (a constant) to integral(dx) = v. integral(dt) to x = x(subsript 0) +vt?

4. Nov 23, 2004

### Fritz

Can you explain it step by step?

5. Nov 23, 2004

### Fritz

6. Nov 23, 2004

### matt grime

Have you not done basic differential equations (linear first order)? Sorry, I'm not the best person to spoon feed answers to anyone, least of all on a subject like this. Good luck finding yourself a teacher, but it ain't going to be me.

7. Nov 23, 2004

### Fritz

^ I've done normal differentiation and partial differentiation. Is that what you mean?

8. Nov 23, 2004

### Fritz

ds/dt = v,

=> dx = vdt,

=> integral between x and x(subscript 0) of dx =v. integral of dt between x and x(subscript 0)

=> x = x(subscript 0) + vt

Is this correct?

9. Nov 23, 2004

### matt grime

No, I mean differential equations. Linear first order. Look it up on the web, get an A-level text book out of the library, if you're not from the UK, then I don't know what level you meet it at or what you'd call it - calc 101 perhaps. Latexing up notes for you is more effort than it merits, sorry, but _you_ should put the effort in on this one.

10. Nov 23, 2004

### matt grime

And no that last post of yours isn't correct.

11. Nov 23, 2004

### Fritz

does the integral of dx = x?

12. Nov 23, 2004

### matt grime

Seeing as you're making the effort:

the *indefinite* integral of 1 dx (with respect to x) is x+k where k is a constant.

13. Nov 23, 2004

### Fritz

I'm a bit confused. Here's our syllabus:

FUNCTIONS OF ONE REAL VARIABLE (five lectures)
Definition, domain, range; odd, even, periodic; polynomials,
factorisation; rational functions, partial fractions; exponential and
logarithmic functions; circular and hyperbolic functions; inverse function;
limits and continuity; composite functions.

DIFFERENTIATION (six lectures)
Definition, chain rule, implicit, parametric, product rule; stationary
values, points of inflection, curve sketching; Maclaurin and Taylor
expansions, polynomial approximation; l’Hopital’s rule.

FUNCTIONS OF TWO REAL VARIABLES (three lectures)
Definition, partial differentiation, total derivative.

COMPLEX NUMBERS (three lectures)
Definition, conjugate, representation; roots of complex numbers, de
Moivre’s theorem.

How do they expect us to understand this stuff when we haven't even done differential equations?

14. Nov 23, 2004

### Fritz

We don't do differential equations until the next symester. I'll be back tomorrow to check this thread (I have to go home now).

15. Nov 23, 2004

### matt grime

differential equations come *after* differentiation.

16. Nov 23, 2004

### arildno

Okay, I'll do a bit of spoon-feeding.
1)Suppose the velocity of an object is constant, that is:
$$\frac{dx}{dt}=v$$
Here, v is a constant, wheras the function x(t) is the position of the object as a function of time.
2. If we integrate both sides of an equality in the same way, we still have an equality:
$$\int_{0}^{T}\frac{dx}{dt}dt=\int_{0}^{T}vdt$$
Here, we perform the integration over the interval $$0\leq{t}\leq{T}$$
where T is some arbitrarily chosen instant.
3. Left-hand side:
$$\int_{0}^{T}\frac{dx}{dt}dt=x(T)-x(0)$$
(Why is this true?)
4. Right-hand side:
$$\int_{0}^{T}vdt=vT-v*0=vT$$
(Why is this true?)
5. Equating the results from 3. and 4., we gain:
$$x(T)-x(0)=vT$$
or, since T is arbitrary, we might as well call it "t":
$$x(t)=x(0)+vt$$
The last step could also have been reached by introducing a dummy variable $$\tau$$ and performed the integration:
$$\int_{0}^{t}\frac{dx}{d\tau}d\tau=\int_{0}^{t}vd\tau$$

17. Nov 24, 2004

### Fritz

Thank you my friend. I get it now.