Question

If x = a cos³t, y = a sin³t, show that `"dy"/"dx" = -(y/x)^(1/3)`.

Answer 1

x = a cos³t, y = a sin³t
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = a"d"/"dt"(cost)^3 = a.3(cost)^2"d"/"dt"(cost)`
= 3acos²t(– sint) = –3a cos²t sint
and
`"dy"/"dt" = a"d"/"dt"(sint)^3`

= `a.3(sin t)^2"d"/"dt"(sin t)`
= 3a sin²t. cos t
∴ `"dy"/"dx" = ((dy/dt))/((dx/"dt")`

= `(3a sin^2tcost)/(-3a cos^2tsint)`

= `-"sint"/"cost"`                       ...(1)
Now, x = a cos³t
∴ cos³t = `x/a`

∴ cos t = `(x/a)^(1/3)`
y = a sin³t
∴ sin³t = `y/a`

∴ cos³t = `(y/a)^(1/3)`

∴ from (1), `"dy"/"dx" = -(y^(1/3)/a^(1/3))/(x^(1/3)/a^(1/3)`

= `-(y/x)^(1/3)`

Answer 2

Alternative Method :
x = a cos³t, y = a sin³t
∴ `cos^3t = x/a, sin^3t = y/a`

∴ `cos t = (x/a)^(1/3), sin t = (y/a)^(1/3)`

∴ cos²t + sin²t = 1 gives

`(x/a)^(2/3) + (y/a)^(2/3)` = 1

∴ `x^(2/3) + y^(2/3) =a^(2/3)`
Differentiating both sides w.r.t. t, we get
`(2)/(3)x^((-1)/(3)) + (2)/(3)y^((-1)/(3)),"dy"/"dx"` = 0

∴ `(2)/(3)y^((-1)/(3))"dy"/"dx" = -(2)/(3)x^((-1)/(3)`

∴ `"dy"/"dx" = -(x/y)^(-1/3) = -(y/x)^(1/3)`