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