# Line integral to calculate work

1. Apr 11, 2012

### Hernaner28

Hi. I have a concrete doubt with this problem. Here's the pic.

It asks me to calculate the work done by force P (the ball moves with constant speed). So the solution is in the book and I understood everything, but the problem comes here,

the force P in axis y is zero so the work of P should be:

$$\int\limits_{{x_o}}^{{x_f}} {{P_x}dx}$$

And we know that Px is equal to the tension in axis x so:

$$\int\limits_{}^{} {T\sin \theta dx}$$

But we need to convert the variable dx into theta variable. And the books states that as:
$$x = L\sin \theta$$
then:
$$dx = L\cos \theta d\theta$$

But shouldn't it be:
$$dx = L\sin\theta d\theta$$??

Why did it take the derivative of sine and not of x on the other side? Thank you!

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2. Apr 11, 2012

### Staff: Mentor

Given:
$~~~x = L sin(\theta)$

Differentiate w.r.t. θ:
$~~~\frac{dx}{d \theta} = L cos(\theta)$

Rearrange:
$~~~dx = L cos(\theta) d\theta$

3. Apr 11, 2012

### Hernaner28

Sorry but what did you do in the second step (wrt?)? I think I'm understaning now but I still don't get it.

Thanks!

4. Apr 11, 2012

### Whovian

Differentiate both sides with respect to theta.

5. Apr 11, 2012

### Hernaner28

Aham! So it would be an intermediate step there:

$$\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}L\sin \theta$$

I didn't know that you could take dO to the other side as a product. Thanks!

6. Apr 11, 2012

### Whovian

Think about it. If two functions are constantly the same, shouldn't their derivatives also be equal?

7. Apr 13, 2012

### Hernaner28

And could this exercise be solved easily using the concepts of energy and conservative forces? Then, how could I begin?

Thanks!

Edit. I'll create a new thread.