Show that y = a tan³ψ given s = a sec³ψ - a

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

If you can tell me where to start from or how to solve the question that would be great.
Thanks
 
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I think the [ TEX ] should be lower case [ tex ]

[tex]s=asec^{3}\psi - a[/tex]
Show that
[tex]y=atan^{3}\psi[/tex]
 
try using,

[tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]
 
Fermat said:
try using,
[tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]
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 :cry:
 
Focus said:
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 :cry:
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 !
 
Focus said:
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 :cry:
I suppose that, in effect, that was what I was doing.

[tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]

[tex]\mbox{We know that\ }\frac{dy}{ds} = sin\psi[/tex]

So,

[tex]y = \int sin\psi\cdot \frac{ds}{d\psi}\ d\psi[/tex]

You should be able to do similar to find an expression for x.
 
Fermat said:
I suppose that, in effect, that was what I was doing.
[tex]\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}[/tex]
[tex]\mbox{We know that\ }\frac{dy}{ds} = sin\psi[/tex]
So,
[tex]y = \int sin\psi\cdot \frac{ds}{d\psi}\ d\psi[/tex]
You should be able to do similar to find an expression for x.
Ah ok thanks a lot, I can do the x myself, I just didn't know how to start :rolleyes:
 
Let me know how you get rid of the constant of integration. Ta.
 
Oh I forgot to write this bit...
[tex]When y=0 x=0 \psi=0[/tex]
But I still can't solve it...:cry:
 
You can insert spaces when using latex with backslash-space "\ "

[tex]When\ y=0\ x=0\ \psi=0[/tex]
 
Focus said:
Oh I forgot to write this bit...
[tex]When y=0 x=0 \psi=0[/tex]
But I still can't solve it...:cry:
When you say you can't solve it, do you mean you can't do the integral for the x-function ?

[tex]x = \int cos\psi\cdot \frac{ds}{d\psi}\ d\psi[/tex]

If it's that one, do you have any working to show ?
 
[tex]y= \int 3sin^{2}\psi sec^{4}\psi d\psi[/tex]
[tex]u=sec \psi[/tex]
[tex]du/(tan\psi sec\psi) = d\psi)[/tex]
some canceling and stuff
[tex]y= \int 3sin\psi u^{2} du[/tex]
I can't get rid of the sin psi

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