Find dydxdydx if : x = 2 cot t + cos 2t, y = 2 sin t – sin 2t at t = π4 - Mathematics and Statistics

प्रश्न

Find `dy/dx` if : x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = `pi/(4)`

बेरीज

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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Derivatives of Implicit Functions

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 2.4 Q 2.3 Q 2.5

पाठ 1: Differentiation - Exercise 1.4 [पृष्ठ ४८]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

पाठ 1 Differentiation
Exercise 1.4 | Q 2.4 | पृष्ठ ४८

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