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Question
If x = 2cos4(t + 3), y = 3sin4(t + 3), show that `"dy"/"dx" = -sqrt((3y)/(2x)`.
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Solution
x = 2cos4(t + 3), y = 3sin4(t + 3)
∴ `cos^4(t + 3) = x/(2), sin^4(t + 3) = y/(3)`
∴ `cos^2(t + 3) = sqrt((x)/(2)), sin^2(t + 3) = sqrt((y)/(3)`
∵ cos2(t + 3) + sin2(t + 3) = 1
∴ `sqrt((x)/(2)) + sqrt((y)/(3)` = 1
Differentiating x and y w.r.t. t, we get
`(1)/sqrt(2)"d"/"dx"(sqrt(x)) + (1)/sqrt(3)"d"/"dx"(sqrt(y))` = 0
∴ `(1)/sqrt(2) xx (1)/(2sqrt(x)) + (1)/sqrt(3) xx (1)/(2sqrt(y))."dy"/"dx"` = 0
∴ `(1)/(2sqrt(3).sqrt(y))."dy"/"dx" = -(1)/(2sqrt(2).sqrt(x)`
∴ `"dy"/"dx" = -(sqrt(3).sqrt(y))/(sqrt(2).sqrt(x)`
= `-sqrt((3y)/(2x)`.
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