[jammysmt]

Hi, I've been stuck on this question for the past 30 mins:

A curve is given by the parametric equations:

x=4Sin^3(t)
y=cos(2t)

Show that dx/dy = -3sin(t)

I've tried (dx/dt)/(dy/dt) method and changing x into 3sint -sin3t to be able to differentiate. I've also tried substitution. These methods don't seem to work and I need a bit of help. Any help offered appreciated. Thanks!

[f1mad]

dy/dt= -2sin2t *what is sin2t equal to?*

dx/dt = 12sin^2tcost

dx/dy is indeed equal to (dx/dt)/(dy/dt).

[jammysmt]

Yea, I already have dy/dt= -4sintcost using sin2x rule.

I'm struggling how you got dx/dt....

[f1mad]

(Original post by jammysmt)
Yea, I already have dy/dt= -4sintcost using sin2x rule.

I'm struggling how you got dx/dt....

Chain rule:

x=4Sin^3(t)

dx/dt= 4* d/dt(sin^3(t))

To differentiate sin^3t think of it as (sint)^3 then apply the chain rule.

[justinawe]

(Original post by jammysmt)
Yea, I already have dy/dt= -4sintcost using sin2x rule.

I'm struggling how you got dx/dt....

d/dx(uⁿ) = nu^n-1(du/dx)

Using that,

d/dx(sinⁿ x) = n(sin^n-1 x)(cos x)

[jammysmt]

Ah, finally got it. Thanks very much for your help

[RevisionSuccess]

x = 4sin^3(t)
y = cos(2t)
dx/dt = 12sin^2(t)cos(t)
dy/dt = -2sin(2t)
dx/dt / dy/dt:
12sin^2(t)cos(t)
-2sin(2t)
= 12sin^2(t)cos(t)
-4sintcost sin(2t) = 2sin(t)cost(t)
the cos(t)'s cancel out. 12/-4 = -3 and sin^2(t)/sin(t) = sin(t)
therefore equal to -3sin(t)

hope i helped

[jsmith6131]