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प्रश्न
Find `dy/dx` if : x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = `pi/(4)`
उत्तर
2 cos t + cos 2t, y = 2 sin t – sin 2t
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = "d"/"dt"(2cost + cos2t)`
= `2"d"/"dt"(cost) + "d"/"dt"(cos2t)`
= `2(-sin t) + (- sin 2t)."d"/"dt"(2t)`
= – 2 sin t – sin 2t x 2 x 1
= – 2 sin t – 2 sin 2t and,
`"dy"/"dt" = "d"/"dt"(2sint - sin2t)`
= `2"d"/"dt"(sint) - "d"/"dt"(sin2t)`
= `2cost - cos2t."d"/"dt"(2t)`
= 2 cos t – cos 2t x 2 x 1
= 2 cos t – 2 cos 2t
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(2cost - 2cos2t)/(-2sint - 2sin2t)`
= `(cost - cos2t)/(-sint - sin2t)`
∴ `(dy/dx)_("at" t = pi/4)`
= `(cos pi/4 - cos pi/2)/(-sin pi/4 - sin pi/2)`
= `(1/sqrt(2) - 0)/(-1/sqrt(2) - 1`
= `(-1)/(1 + sqrt(2)`
= `(-1)/(1 + sqrt(2)) xx (1 - sqrt(2))/(1 - sqrt(2)`
= `(-(1 - sqrt(2)))/(1 - 2)`
= `1 - sqrt(2)`.
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