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Question
Find `"dy"/"dx"` if : x = t2 + t + 1, y = `sin((pit)/2) + cos((pit)/2) "at" t = 1`
Solution
x = t2 + t + 1, y = `sin((pit)/2) + cos((pit)/2)`
Differentiating x and y w.r.t. t, we get
`"dx"/"dt" = "d"/"dt"(t^2 + t + 1)`
= 2t + 1 + 0 = 2t + 1
and
`"dy"/"dt" = "d"/"dt"[sin(pit/2)] + "d"/"dt"[cos(pi/t)]`
= `cos((pit)/2)."d"/"dt"((pit)/2) + [-sin((pit)/2)]."d"/"dt"((pit)/2)`
= `cos((pit)/2) xx pi/(2) xx 1 - sin ((pit)/2) xx pi/(2) xx 1`
= `pi/(2)[cos((pit)/2) - sin((pit)/2)]`
∴ `"dy"/"dx" = (("dy"/"dt"))/(("dx"/"dt")`
= `(pi/(2)[cos((pit)/2) - sin((pit)/2)])/(2t + 1)`
∴ `(dx/dy)_("at" t = 1)`
= `(pi/(2)[cos pi/2 - sin pi/2])/(2(1) + 1)`
= `(pi/(2)(0 - 1))/(3)`
= `- pi/(6)`.
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