Solve Intrinsic Equations: y=atan^3ψ | Step-by-Step Guide

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

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

\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}
 
Fermat said:
try using,
\frac{dy}{d\psi}= \frac{dy}{ds}\cdot \frac{ds}{d\psi}
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.

\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.
 
Fermat said:
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.
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...
When y=0 x=0 \psi=0
But I still can't solve it...:cry:
 
  • #10
You can insert spaces when using latex with backslash-space "\ "

When\ y=0\ x=0\ \psi=0
 
  • #11
Focus said:
Oh I forgot to write this bit...
When y=0 x=0 \psi=0
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 ?

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

If it's that one, do you have any working to show ?
 
  • #12
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 :confused:
 
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  • #13
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.
 
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