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X = tt1+logtt2, y = tt3+2logtt - Mathematics

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Question

x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`

Sum
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Solution

Given that: x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`

Differentiating both the parametric functions w.r.t. t

`"dx"/"dt" = ("t"^2 * "d"/"dt" (1 + log "t") - (1 + log "t") * "d"/"dt" ("t"^2))/"t"^4`

= `("t"^2 * (1/"t") - (1 + log "t") * 2"t")/"t"^4`

= `("t" - (1 + log "t") * 2"t")/"t"^4`

= `("t"[1 - 2 - 2 log "t"])/"t"^4`

= `(-(1 + 2 log "t"))/"t"^3`

y = `(3 + 2 log "t")/"t"`

`"dy"/"dt" = ("t" * "d"/"dt" (3 + 2 log "t") - (3 + 2 log "t") * "d"/"dt" ("t"))/"t"^2`

= `("t"(2/"t") - (3 + 2 log "t")* 1)/"t"^2`

= `(2 - 3 - 2 log "t")/"t"^2`

= `(-(1 + 2 log "t"))/"t"^2`

∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `((-(1 + 2 log "t"))/"t"^2)/((-(1 + 2 log "t"))/"t"^3)`

= `"t"^3/"t"^2`

= t

Hence, `"dy"/"dx"` = t.

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Chapter 5: Continuity And Differentiability - Exercise [Page 110]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 48 | Page 110

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