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प्रश्न
If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that `("dy"/"dx")_("at t" = pi/4) = "b"/"a"`
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उत्तर
Given that: x = asin2t (1 + cos 2t) and y = bcos2t (1 – cos 2t).
Differentiating both the parametric functions w.r.t. t
`"dx"/"dt" = "a"[sin2"t" * "d"/"dt" (1 + cos 2"t") + (1 + cos 2"t") * "d"/"dt" sin 2"t"]`
= a[sin 2t .(– sin 2t) + (1 + cos 2t)(cos 2t).2]
= a[2(cos22t – sin22t + 2 cos 2t]
= a[2 cos22t – sin22t) + 2 cos 2t]
= a[2 cos 4t + 2 cos 2t] ....[∵ cos 2x = cos2x – sin2x]
= 2a[cos 4t + cos 2t]
y = b cos 2t (1 – cos 2t)
`"dy"/"dx" = "b"[cos 2"t" * "d"/"dt" (1 - cos 2"t") + (1 - cos 2"t") * "d"/"dt" (cos 2"t")]`
= b[cos 2t . sin 2t.2 + (1 – cos 2t).(– son 2t).2
= b[sin 4t – 2 sin 2t - 2 sin 2t + 2 sin 2t cos 2t]
= b[2 sin 4t – 2 sin 2t]
= 2b (sin 4t – sin 2t)
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
= `(2"b"[sin 4"t" - sin2"t"])/(2"a"[cos 4"t" + cos 2"t"])`
= `"b"/"a" [(sin 4"t" - sin 2"t")/(cos 4"t" + cos 2"t")]`
Put t = `pi/4`
∴ `("dy"/"dx")_("at t" = pi/4) = "b"/"a" [(sin 4(pi/4) - sin 2* (pi/4))/(cos 4(pi/4) + cos 2*(pi/4))]`
= `"b"/"a" [(sin pi - sin pi/2)/(cos pi + cos pi/2)]`
= `"b"/"a" [(0 - 1)/(-1 + 0)]`
= `"b"/"a"((-1)/(-1))`
= `"b"/'a"`
Hence, `("dy"/"dx")_("at t" = pi/4) = "b"/"a"`.
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