# Intrinsic Equations

1. Nov 3, 2005

### Focus

Hello, I was doing some past papers and I couldn't solve one.
I don't even know where to begin with this question.
$$s=asec^{3}\psi - a$$
Show that
$$y=atan^{3}\psi$$

If you can tell me where to start from or how to solve the question that would be great.
Thanks

Last edited: Nov 3, 2005
2. Nov 3, 2005

### Fermat

I think the [ TEX ] should be lower case [ tex ]

$$s=asec^{3}\psi - a$$
Show that
$$y=atan^{3}\psi$$

3. Nov 3, 2005

### Fermat

try using,

$$\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}$$

4. Nov 3, 2005

### Focus

Is there no way of getting to the second one, just using the first one, the second part of the question asks for an expression for x

5. Nov 3, 2005

### Fermat

I haven't found it yet. I've only been able to show that differential coeficients are the same. But integration, to give you the original expression, y = a.tan³psi, involves a constant of integration. I've not been able to get rid of that. I was hoping you might manage it yourself !

6. Nov 3, 2005

### Fermat

I suppose that, in effect, that was what I was doing.

$$\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}$$

$$\mbox{We know that\ }\frac{dy}{ds} = sin\psi$$

So,

$$y = \int sin\psi\cdot \frac{ds}{d\psi}\ d\psi$$

You should be able to do similar to find an expression for x.

7. Nov 3, 2005

### Focus

Ah ok thanks a lot, I can do the x myself, I just didn't know how to start

8. Nov 3, 2005

### Fermat

Let me know how you get rid of the constant of integration. Ta.

9. Nov 4, 2005

### Focus

Oh I forgot to write this bit...
$$When y=0 x=0 \psi=0$$
But I still can't solve it...

10. Nov 4, 2005

### Fermat

You can insert spaces when using latex with backslash-space "\ "

$$When\ y=0\ x=0\ \psi=0$$

11. Nov 4, 2005

### Fermat

When you say you can't solve it, do you mean you can't do the integral for the x-function ?

$$x = \int cos\psi\cdot \frac{ds}{d\psi}\ d\psi$$

If it's that one, do you have any working to show ?

12. Nov 4, 2005

### Focus

$$y= \int 3sin^{2}\psi sec^{4}\psi d\psi$$
$$u=sec \psi$$
$$du/(tan\psi sec\psi) = d\psi)$$
some canceling and stuff
$$y= \int 3sin\psi u^{2} du$$
I can't get rid of the sin psi

I think its to do with y=0 x=0 and psi=0

Last edited: Nov 4, 2005
13. Nov 4, 2005

### Fermat

It doesn't have anything to do wiht the initial values: y=0 x=0 and psi=0

You already know what the answer is. Just differentiate that and see how that can be manipulated to give you the expression you have to integrate.

Then work backwards.