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

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Homework Help Overview

The discussion revolves around a problem involving intrinsic equations, specifically showing that \( y = a \tan^3 \psi \) from the equation \( s = a \sec^3 \psi - a \). Participants express uncertainty about how to begin solving the problem and explore various mathematical approaches related to differentiation and integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using derivatives and integrals, particularly the relationship \( \frac{dy}{d\psi} = \frac{dy}{ds} \cdot \frac{ds}{d\psi} \). There are questions about how to handle constants of integration and the implications of initial conditions on the solution.

Discussion Status

There is an ongoing exploration of different methods to approach the problem, with some participants suggesting specific mathematical manipulations. While some guidance has been offered, such as integrating certain expressions, there is no clear consensus on how to proceed or resolve the issues raised.

Contextual Notes

Participants mention initial conditions \( y = 0, x = 0, \psi = 0 \) and express concerns about how these affect their ability to solve the problem. There is also a note about the use of LaTeX formatting in the discussion.

Focus
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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
 
Last edited:
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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:
 
  • #10
You can insert spaces when using latex with backslash-space "\ "

[tex]When\ y=0\ x=0\ \psi=0[/tex]
 
  • #11
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 ?
 
  • #12
[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:
 
Last edited:
  • #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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