Question

If x = 3sint – sin 3t, y = 3cost – cos 3t, find `"dy"/"dx"` at t = `pi/3`

Answer 1

Given that: x = 3sint – sin 3t, y = 3cost – cos 3t.

Differentiating both parametric functions w.r.t. t

`"dx"/"dt" = 3 cos "t" - cos 3"t" * 3`

= 3(cos t – cos 3t)

`"dy"/"dx" = -3 sin "t" + sin 3"t" * 3`

= 3(– sin t + sin 3t)

∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `(3(- sin "t" + sin 3"t"))/(3(cos "t" - cos 3"t"))`

= `(-sin "t" + sin 3"t")/(cos "t" - cos 3"t")`

Put t = `pi/3`

`"dy"/"dx" = (- sin  pi/3 + sin 3 (pi/3))/(cos  pi/3 - cos  3 (pi/3))`

= `(- sqrt(3)/2 + sin pi)/(1/2 - cos pi)`

= `(- sqrt(3)/2 + 0)/(1/2 - (- 1))`

= `(- sqrt(3)/2)/(1/2 + 1)`

= `(- sqrt(3)/2)/(3/2)`

= `(-1)/sqrt(3)`

Hence, `"dy"/"dx" = (-1)/sqrt(3)`.