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Maths A level help!!!

How do you solve this question and explanation please!! my exam is in a few days time
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 x3x^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.
Reply 2
Original post by RDKGames
For the first part use some type of interval bisection method.

For the second one, rearrange for x3x^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?
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.
Reply 4
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.
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 :smile:
Reply 6
Original post by RDKGames
Go ahead and ask :smile:


this one? i'm more interested in the second part
(edited 7 years ago)
Original post by tc32
this one?


dydx=dydtdtdx\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx} and dtdx=(dxdt)1\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 yb=m(xa)y-b=m(x-a) where a,b are respective points with parameter t, and m is the gradient with the same parameter.
Reply 8
Original post by RDKGames
dydx=dydtdtdx\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx} and dtdx=(dxdt)1\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 yb=m(xa)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
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" ?
Reply 10
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 ?
Original post by tc32
1/cos^3t and tan^3t ?


Yep.
Reply 12
Original post by RDKGames
Yep.


i'm failing to rearrange them to get the final equation
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.
Reply 14
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.
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. :smile:
Reply 16
Original post by RDKGames
Ah good, no problem. :smile:


i've got another question. I seem to be getting 2 + 2x instead of 4 + 4x.

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