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Question
Find `"dy"/"dx"`, if : `x = cos^-1(4t^3 - 3t), y = tan^-1(sqrt(1 - t^2)/t)`.
Solution
`x = cos^-1(4t^3 - 3t), y = tan^-1(sqrt(1 - t^2)/t)`
Put t = cosθ.
Thenθ = cos–1t.
∴ x = cos–1(4cos3θ – 3cosθ ),
y = `tan^-1(sqrt(1 - cos^2θ)/cosθ)`
∴ x = `cos^-1(cos3θ),y = tan^-1((sinθ)/(cosθ)) = tan^-1(tanθ)`
∴ x = 3θ and y = θ
∴ x = 3cos–1t and y = cos–1t
Differentiating x and y w.r.t. x, we get
`"dx"/"dt" = 3"d"/"dt"(cos^-1)`
= `3 xx (-1)/sqrt(1 - t^2)`
= `(-3)/sqrt(1 - t^2)`
and
`"dy"/"dt" = "d"/"dt"(cos^-1t)`
= `(-1)/sqrt(1 - t^2)`
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(((-1)/(sqrt(1 - t^2))))/(((-3)/(sqrt(1 - t^2)))`
= `(1)/(3)`.
Alternative Method :
x = cos–1 (4t3 – 3t), t = `tan^-1(sqrt(1 - t^2)/t)`
Put t = cosθ.
Then x = cos–1(4cos3θ – 3cosθ),
y = `tan^-1(sqrt(1- cos^2θ)/cosθ)`
∴ x = cos–1 (cos3θ), y = `tan^-1((sinθ)/(cosθ)) = tan^-1(tanθ)`
∴ x = 3θ, y = θ
∴ x = 3y
∴ y = `(1)/(3)x`
∴ `"dy"/"dx" = (1)/(3)"d"/"dx"(x)`
= `(1)/(3) xx 1`
= `(1)/(3)`.
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