Question

 यदि x = asin2t (1 + cos2t)  और y = b cos2t (1–cos2t) तो दर्शाइए कि, x = `pi/4` पर;`("dy"/"dx") = "b"/"a"`

Answer 1

दिया गया है: x = asin2t (1 + cos 2t) और y = bcos2t (1 – cos 2t).

दोनों प्राचलिक फलनों को अलग करना 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")]`

t = `pi/4` रखिये

∴ x = `pi/4` पर;`("dy"/"dx") = "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"`

इसलिए, x = `pi/4` पर;`("dy"/"dx") = "b"/"a"`