- #1
Benny
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Hi, I'm stuck on the following implicit function question.
Q. Find the values of t_1, t_2 such that every solution theta is determined as a C^1 function of t_1 and t_2.
The equation is [itex]\theta ^3 + t_1 \theta + t_2 = 0[/itex].
Ok from what I gather we want [itex]\theta = g\left( {t_1 ,t_2 } \right)[/itex]. This one is a little different to the others I've done before and I think one of the conditions I need to check, for this particular case is:
[tex]
\frac{{\partial F}}{{\partial \theta }} = 3\theta ^2 + t_1 \ne 0,F\left( {t_1 ,t_2 ,\theta } \right) = \theta ^3 + t_1 \theta + t_2
[/tex] where the not equal to zero condition is satisified at a specific point (..,..,..).
Of course if I want to apply the implicit function theorem I need a specific point (t_1,t_2), maybe call it (x,y) to avoid ambiguity. Usually I'm given
one but in this case I'm not so I'm at a loss as to what I need to do. I was told that this question requires a bit of thought but I'm too stupid to work this out so can someone help me out? Any help would be good thanks.
Q. Find the values of t_1, t_2 such that every solution theta is determined as a C^1 function of t_1 and t_2.
The equation is [itex]\theta ^3 + t_1 \theta + t_2 = 0[/itex].
Ok from what I gather we want [itex]\theta = g\left( {t_1 ,t_2 } \right)[/itex]. This one is a little different to the others I've done before and I think one of the conditions I need to check, for this particular case is:
[tex]
\frac{{\partial F}}{{\partial \theta }} = 3\theta ^2 + t_1 \ne 0,F\left( {t_1 ,t_2 ,\theta } \right) = \theta ^3 + t_1 \theta + t_2
[/tex] where the not equal to zero condition is satisified at a specific point (..,..,..).
Of course if I want to apply the implicit function theorem I need a specific point (t_1,t_2), maybe call it (x,y) to avoid ambiguity. Usually I'm given
one but in this case I'm not so I'm at a loss as to what I need to do. I was told that this question requires a bit of thought but I'm too stupid to work this out so can someone help me out? Any help would be good thanks.
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