Differentiating parametric Equations with respect to x?

Question

ok now here is a tough one (i think)

x = (cos (t))^3
y = (sin^3 (t))^3

The questions asks for dy/dx IN TERMS of x, which i am unable to do. I understand the normal ways of differentiating parametric equations, but cannot apply them to this question where I can't isolate the x/y equations into t = ..... So i thought i would have to differentiate implicitly, but to do that i would have to fine a common relationship between the two functions in order to get rid of the 'nasty and horrid trigonometry functions'.

Please help me!

OMG you guys rock my socks! Thank you ever so much!

Answer 1

I am not 100% if this is right.
I tried this:

(dy/dt) = (dy/dx) * (dx/dt) =>

(dy/dx) = (dy/dt) / (dx/dt)

dy/dt = 9sin^8(t)*cost ( y = sin^9(t))
dx/dt = 3cos^2(t)*(-sint)

=> dy/dx = - 3sin^7(t)/cost
x = cos^3(t) => cost = x^1/3

cost = x^1/3 => cos^2(t) = x^2/3 => 1 - sin^2(t) = x^2/3
=> sint = (1 - x^2/3)^1/2

=> dy/dx = -3(1 - x^2/3)^7/2 / x^1/3

Answer 2

What you wrote seems to be
x=(cos t)^3
y= (sin t)^9
x^1/3 =cos t
y^1/9= sin t
squaring and sum
x^2/3 +y^2/9 =1

Derivating 2/3 x^-1/3 +2/9 y^-7/9 *y´=0

y´=- (2/3 x^-1/3 )/(2/9 y^-7/9) = 3 y^7/9 *x^-1/3

Answer 3

x = (cos t)³
dx/dt = 3(cos t).(- sin t)
dx/dt = - 3.cos t.sin t
y = (sin³ t)³ as stated in question.
dy/dt = 3 (sin³ t)².3 sin²t.cos t
dy/dt = 9.sin^(6)t.sin² t.cos t
dy / dt = 9 sin^(8)t.cos t
dy/dx = 9 sin^(8) t.cos t / 3 cos² t(- sin t)
dy / dx = - 3 sin^(7)t / cos t
dy/dx = - 3 tan t.sin^(6) t

Answer 4

x = (cos(t))^3; x' = -3(cos(t))^2 sin(t)
y = (sin(t))^9; y' = 9 (sin(t))^8 cos(t)
dy/dx = y'/x' = -3 (sin(t))^7 / cos(t) = -3 y^(8/9) / x^(1/3)

Answer 5

"a" is to relate these to equations to each other, so that you can substitute the value of x in the eqn of y below so you can get an equation of y. which is by following the steps: a= x/(θ²-3cos2θ) so y=[ (sin²3θ-4θ)/(θ²-3cos2θ) ]x

Answer 6

i'm shure if you want to go down this road but you coud say that
t=arccos (cube root (x))
so y=sin^3(arccos(cube root(x))^3

dy/dx= (3 sin^2(arccos(cube root(x))^3)*

cos((arccos(cube root(x))^3)*

3 arccos ((arccos(cube root(x))^2)*

(-1/sqrt(1-(cube root(x)^3)^2)*

1/3(x)^(-2/3)