The temperature (in ) of a reaction is an invertible differentiable function of time…

MATHEMATICS · Single correct · CBSE · Mathematics · Continuity and Differentiability · Continuity and Differentiability

Problem

The temperature TT (in ∘ ⁣C^{\circ}\!\text{C}) of a reaction is an invertible differentiable function of time tt (in seconds). At t=10t=10 s, the data are T=150T=150 and dTdt=3  ∘ ⁣C ⁣/ ⁣s\dfrac{dT}{dt}=3\;^{\circ}\!\text{C}\!\big/\!\text{s}. Determine dtdT\dfrac{dt}{dT} at T=150 ∘ ⁣CT=150\,^{\circ}\!\text{C}. (a) 3  s/∘ ⁣C3\;\text{s}\big/^{\circ}\!\text{C} (b) −13  s/∘ ⁣C-\dfrac13\;\text{s}\big/^{\circ}\!\text{C} (c) 13  s/∘ ⁣C\dfrac13\;\text{s}\big/^{\circ}\!\text{C} (d) 115  s/∘ ⁣C\dfrac1{15}\;\text{s}\big/^{\circ}\!\text{C}

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