[Tisabiki]

A curve has parametric equations: x = 2cos2t y = 6sint

a) find the gradient of the curve at the point where t = pi/3 (done this)

b) find a cartesian equation of the curve in the form:
y = f(x), -k < x < k

No idea what to do for b =S

[foolsihboy]

Is there any way you can eliminate t from the equations, maybe using a trigonometric identity?

[Tisabiki]

Tried using cos2t as 1 - 2sin^2t, but it didn't seem to work out.

[foolsihboy]

As far as I'm aware that should work out...

What did you get for 2sin^2(t) then?

[Tisabiki]

Erm... y = 6 - root(18x)
Can't find a mark scheme to check, but no idea where the K comes into it x_X

[nitinkf]

it should be y=3root(2-x)
x=2(1-2sin^2t) y=6sint
y^2=36sin^2t
(y^2)/36=sin^2t

x=2-4sin^2t
sub y
x=2-((Y^2)/9)

multiply through by 9

9x=18-(Y^2)
Y=root(18-9x)

thats what i get atleast

so -2<x<2

[Toneh]

(Original post by nitinkf)
it should be y=3root(2-x)
x=2(1-2sin^2t) y=6sint
y^2=36sin^2t
(y^2)/36=sin^2t

x=2-4sin^2t
sub y
x=2-((Y^2)/9)

multiply through by 9

9x=18-(Y^2)
Y=root(18-9x)

thats what i get atleast

so -2<x<2

I concur.

[willowaller]

Put x=2cos2t into the form x=2(2(cost)^2-1) or x=4(cost)^2-2

then y=6sint into Y^2=36(sint)^2

Then into identity (Sint)^2 + (cost)^2 = 1 in order to eliminate paramiter

The constant bit (K) is to show that you undersatnd the maximum and minimum x values of the cartesian equation.

[Tisabiki]

Got it now, thanks all

[OL1V3R]

For this one I got


EDIT: Just realised I forgot the 6! Dammit...

[revilowaldow]

I've got the same question but can anyone help me with part a? I used the chain rule on it and and the whole dx/dt=1/(dt/dx) but I want to check my answer? :3

[ghostwalker]

(Original post by revilowaldow)
I've got the same question but can anyone help me with part a? I used the chain rule on it and and the whole dx/dt=1/(dt/dx) but I want to check my answer? :3

I make it

If you get anything else and want someone to check it, then post your working.