1. Dec 24, 2005

### Cyrus

I'm not getting this. Can someone show me how they got from this:

$$- T cos ( \theta ) + (T + \Delta T ) cos ( \theta + \Delta \theta ) =0$$

to below by dividing by $$\Delta x$$

$$\frac { d (T cos\theta)}{dx} = 0$$

Im just not getting it for some reason.... The first term would be Tcos(theta)/ dx, which should turn into 1/0, which blows up to infinity?

Thank you,

Cyrus

2. Dec 24, 2005

### Tide

Cyrus,

Clearly, you are treating $\Delta T$ and $\Delta \theta$ as "small" numbers. The lowest order term, $T\cos \theta$, cancels out. Just expand everything in terms of small quantities and your result will be evident.

3. Dec 24, 2005

### Cyrus

like this?

$$-Tcos(\theta) + Tcos(\theta + \Delta \theta ) + \Delta T cos (\theta + \Delta \theta) = 0$$

so I can neglect the delta theta, giving me:

$$-Tcos(\theta) + Tcos(\theta ) + \Delta T cos (\theta) =0$$

which cancels out those two terms, but then im left with:

$$\Delta T cos (\theta) = 0$$

Now, im not where I should be....oops?

I'm somehow supposed to get:

$$\Delta T cos (\theta) + T sin(\theta) \Delta \theta = 0$$

4. Dec 24, 2005

### Cyrus

Got it Tide! I'm sorry, but I disagree, or am not seeing, what you said, heres my approach:

$$-T cos (\theta) + (T + \Delta T)cos(\theta + \Delta \theta) = 0$$

which is equivalent to:

$$T cos( \theta + \Delta \theta) - T cos(\theta) + \Delta T cos (\theta + \Delta \theta) = 0$$

which is equivalent to:

$$\frac {[T cos( \theta + \Delta \theta) - T cos(\theta)]\Delta \theta}{\Delta \theta} + \Delta T cos (\theta + \Delta \theta) = 0$$

NOW divide by delta X:

$$\frac {[T cos( \theta + \Delta \theta) - T cos(\theta)]\Delta \theta}{\Delta \theta \Delta x} + \frac {\Delta T cos (\theta + \Delta \theta)}{\Delta x} = 0$$

and take the limit as: $$\Delta x \rightarrow 0; \Delta \theta \rightarrow 0;\Delta T \rightarrow 0$$

So, the $$\frac {cos(\theta + \Delta \theta)-cos(\theta)}{\Delta \theta}$$ term goes to $$sin(\theta)$$, and $$\frac {\Delta \theta}{\Delta x}$$ goes to $$\frac{d \theta } {dx}$$ and $$\frac {\Delta T}{\Delta x}$$ goes to $$\frac {dT}{dx}$$ and $$cos(\theta + \Delta \theta )$$ goes to just good ole, $$cos (\theta)$$

$$T sin(\theta) \frac{ d \theta}{ dx} + \frac {dT}{dx} cos (\theta)}$$

which clearly equals:

$$\frac {d (Tcos\theta)}{dx}$$

Good god, its 4:00 am. Im glad I got this done or I wouldent have fallen asleep. Goodnight.

Last edited: Dec 24, 2005
5. Dec 24, 2005

### HallsofIvy

You can use the fact that $cos(\theta+\Delta\theta)= cos(\theta)cos(\Delta\theta)- sin(\theta)sin(\Delta\theta)$.

Multiply the whole thing out then use the fact that $cos(\Delta\theta)$ goest to 1 and $sin(\Delta\theta)$ goes to 0 as $\Delta\theta$ goes to 0.

6. Dec 24, 2005

### maverick6664

Is it so complicated?
I just think:
$$\theta$$ and $$T$$ are functions of $$x$$ (it means they are not independent to each other. otherwise $$x$$ wouldn't appear here in this context), so let $$S(x)=T\cos(\theta)$$, then the first equation can read $$- S(x) + S(x + \Delta x) = 0$$. Then divide this equation by $$\Delta x$$ and you'll have $$\frac{\Delta S}{\Delta x} = \frac{dS}{dx} = 0$$.

Last edited: Dec 24, 2005
7. Dec 24, 2005

### Cyrus

Hmm,YES, these all seem to be valid methods. Thats very cool. Does my approach seem wrong?

8. Dec 24, 2005

### maverick6664

Your argument #4 looks fine to me. But I just thought it's a bit complicated and not generic. What you wrote in #4 is $$\Delta S = ( \frac{\partial S}{\partial T} \frac{\partial T}{\partial x} + \frac{\partial S}{\partial \theta} \frac{\partial \theta}{\partial x} ) \Delta x$$
It holds even if it isn't $$\cos{\theta}$$, and now that intermediate variables $$\theta$$ and $$T$$ are affected by $$x$$, so we can simply put $$\frac{dS}{dx} = \frac{\partial S}{\partial T} \frac{\partial T}{\partial x} + \frac{\partial S}{\partial \theta} \frac{\partial \theta}{\partial x}$$

and you can ignore these intermediate variables. If you see $$S$$ simply as a function of x, this can be explained without partial differential as in #6.

hope this helps.. and Merry Christmas! (here in Japan it's 2:30am)

Last edited: Dec 24, 2005
9. Dec 25, 2005

### benorin

Are you working the (one dimensional) wave equation for a vibrating string by chance (where T is tension)? As I recall seeing something similar during that proof.

10. Dec 25, 2005

### Cyrus

Its not a wave equation for a string, but close. Its a derviation to find the equation of a string/rope/wire/whatever under a distributed load in the x direction, OR if you replace delta x by delta s, it is the weight distribution of a catenary wire. No vibration, sorry.

11. Dec 26, 2005

### kant

If ( T+ delT)cos(@+ del@) =o, then Tcos@=0.

The change of Tcos@ is d (Tcos@).

The change of 0 is 0.

divdie d( Tcos@) =0 by delX, and allow delX = dx.

12. Dec 26, 2005

### maverick6664

Then there should be other equations like

$$(T + \Delta T) \sin(\theta + \Delta \theta) = T \sin(\theta) + \varrho \frac {\Delta x} { \cos(\theta) } g$$

and

$$\Delta y = \Delta x \tan(\theta)$$

along Y axis where $$\varrho$$ is the mass density of the rope.
and

$$\frac {d(T \sin(\theta))} {dx} = \frac {\varrho g} {\cos(\theta)}$$ or just $$\varrho g$$

$$\frac {dy}{dx} = \tan(\theta)$$ or just $$sin(\theta)$$

or something...

Last edited: Dec 26, 2005
13. Dec 26, 2005

### Cyrus

.....what?

14. Dec 26, 2005

### Cyrus

You're close. No cos in the denominator. Its just the denisty, times gravity, time delta x. Actually its really, w(x)deltaX. w(x) is the more general weight distribution. Your right about the tangent, no sin there.

15. Dec 26, 2005

### maverick6664

oops. you are right..cosine shouldn't be in the denominator

16. Dec 26, 2005

### lalbatros

It seems to me that this is obvious.
T is possibly dependent on $$\theta$$ and therefore:

$$\Delta (T cos(\theta)) = T(\theta + \Delta \theta) * cos(\theta + \Delta \theta) - T(\theta) * cos(\theta)$$

and of course the definition

$$T(\theta + \Delta \theta) = T(\theta) + \Delta T$$