 Solution

At $700^{\circ }\mathrm{C}$ the undercooling below the eutectoid temperature is small, so the free-energy driving force for forming pearlite is weak and nucleation is difficult. At $300^{\circ }\mathrm{C}$ the driving force is larger, but carbon diffusion is slow, limiting the coupled growth of ferrite and cementite. Their competition produces the nose of the isothermal transformation diagram near $550^{\circ }\mathrm{C}$ and $1\mathrm{s}$. A roughly linear quench from $740$–$800^{\circ }\mathrm{C}$ must pass left of this nose, requiring a rate of order $2\times 10^2$–$3\times 10^2\mathrm{K}{\mathrm{s}}^{-1}$, then continue below the martensite-start region near $240^{\circ }\mathrm{C}$. The tangent to the full start curve, rather than just the nose, refines this estimate; a true continuous-cooling diagram would be preferable. Martensite forms by a coordinated lattice-shape change without long-range carbon partitioning, so its formation is a displacive transformation. The tetragonal-to-monoclinic transformation of zirconia is a nonmetallic example.