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Question
Find `"dy"/"dx"` if : x = t + 2sin (πt), y = 3t – cos (πt) at t = `(1)/(2)`
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Solution
x = t + 2sin (πt), y = 3t – cos (πt)
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = "d"/"dt"[t + 2sin(pit)]`
= `"d"/"dt"(t) + 2."d"/"dt"[sin(pit)]`
= `1 + 2 xx cos(pit)."d"/"dx"(pit)`
= 1 + 2cos(πt) x π x 1
= 1 + 2π cos (πt)
and
`"dy"/"dt" = "d"/"dt"[3t - cos(pit)]`
= `3 xx 1 - [- sin(pit)]."d"/"dt"(pit)`
= 3 + sin (πt) x π x 1
= 3 + π sin (πt)
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(3 + pi sin(pit))/(1 + 2pi cos(pit)`
∴ `(dy/dx)_("at" t = 1/2)`
= `(3 + sin(pi/2))/(1 + 2picos(pi/2)`
= `(3 + pi xx 1)/(1 + 2pi(0)`
= 3 + π.
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