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Maths A level help!!!

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tc32
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#1
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#1
How do you solve this question and explanation please!! my exam is in a few days time
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RDKGames
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(Original post by tc32)
How do you solve this question and explanation please!! my exam is in a few days time
For the first part use some type of interval bisection method.

For the second one, rearrange for x^3, then divide by x before square rooting both sides in order to arrive at the needed form, then explain how it converges onto the root.
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tc32
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(Original post by RDKGames)
For the first part use some type of interval bisection method.

For the second one, rearrange for x^3, then divide by x before square rooting both sides in order to arrive at the needed form, then explain how it converges onto the root.
I'm more interested in the first part. How do you get 2 and 3?
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RDKGames
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(Original post by tc32)
I'm more interested in the first part. How do you get 2 and 3?
Pick an arbitrary interval, test the two ends, and cut it down using your intuition until you get the equation to change sign between two integers.
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tc32
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(Original post by RDKGames)
Pick an arbitrary interval, test the two ends, and cut it down using your intuition until you get the equation to change sign between two integers.
thanks very much. I've got a couple of other questions to ask if you don't mind it's just that i'm writing pure maths on monday and i'm revisiting all the problem questions i've encountered from past papers and you seem to be the only one giving valuable responses.
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RDKGames
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(Original post by tc32)
thanks very much. I've got a couple of other questions to ask if you don't mind it's just that i'm writing pure maths on monday and i'm revisiting all the problem questions i've encountered from past papers and you seem to be the only one giving valuable responses.
Go ahead and ask
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tc32
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(Original post by RDKGames)
Go ahead and ask
this one? i'm more interested in the second part
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RDKGames
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(Original post by tc32)
this one?
\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx} and \frac{dt}{dx}=(\frac{dx}{dt})^{-1}

and I'm sure you can do the differentials, they're simple chain rules.

For the last one it's a simple case of substituting for y-b=m(x-a) where a,b are respective points with parameter t, and m is the gradient with the same parameter.
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tc32
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[QUOTE=RDKGames;67987740]\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx} and \frac{dt}{dx}=(\frac{dx}{dt})^{-1}

and I'm sure you can do the differentials, they're simple chain rules.

For the last one it's a simple case of substituting for y-b=m(x-a) where a,b are respective points with parameter t, and m is the gradient with the same parameter.[/QUO

the problem is finding a and b
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RDKGames
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(Original post by tc32)
the problem is finding a and b
Not really much of a problem. What do you think a and b are when it says "with parameter t" ?
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tc32
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(Original post by RDKGames)
Not really much of a problem. What do you think a and b are when it says "with parameter t" ?
1/cos^3t and tan^3t ?
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RDKGames
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(Original post by tc32)
1/cos^3t and tan^3t ?
Yep.
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tc32
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(Original post by RDKGames)
Yep.
i'm failing to rearrange them to get the final equation
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RDKGames
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(Original post by tc32)
i'm failing to rearrange them to get the final equation
Show your attempt. Could be the part where you need to factor out a negative tan(t) and you do not switch both signs inside the bracket.
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tc32
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#15
(Original post by RDKGames)
Show your attempt. Could be the part where you need to factor out a negative tan(t) and you do not switch both signs inside the bracket.
don't worry i've got it. i had to factor out tant and use some trigonometry identities so that the cos^2t cancels out. thanks a lot for the help.
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RDKGames
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(Original post by tc32)
don't worry i've got it. i had to factor out tant and use some trigonometry identities so that the cos^2t cancels out. thanks a lot for the help.
Ah good, no problem.
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tc32
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#17
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#17
(Original post by RDKGames)
Ah good, no problem.
i've got another question. I seem to be getting 2 + 2x instead of 4 + 4x.
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