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प्रश्न
If x = 3sint – sin 3t, y = 3cost – cos 3t, find `"dy"/"dx"` at t = `pi/3`
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उत्तर
Given that: x = 3sint – sin 3t, y = 3cost – cos 3t.
Differentiating both parametric functions w.r.t. t
`"dx"/"dt" = 3 cos "t" - cos 3"t" * 3`
= 3(cos t – cos 3t)
`"dy"/"dx" = -3 sin "t" + sin 3"t" * 3`
= 3(– sin t + sin 3t)
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
= `(3(- sin "t" + sin 3"t"))/(3(cos "t" - cos 3"t"))`
= `(-sin "t" + sin 3"t")/(cos "t" - cos 3"t")`
Put t = `pi/3`
`"dy"/"dx" = (- sin pi/3 + sin 3 (pi/3))/(cos pi/3 - cos 3 (pi/3))`
= `(- sqrt(3)/2 + sin pi)/(1/2 - cos pi)`
= `(- sqrt(3)/2 + 0)/(1/2 - (- 1))`
= `(- sqrt(3)/2)/(1/2 + 1)`
= `(- sqrt(3)/2)/(3/2)`
= `(-1)/sqrt(3)`
Hence, `"dy"/"dx" = (-1)/sqrt(3)`.
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