Question

Find `"dy"/"dx"` if : x = t² + t + 1, y = `sin((pit)/2) + cos((pit)/2) "at"  t = 1`

Answer 1

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)`.