Legendre transformation, geometrically conceived. Given a functional relation, y

Question

Legendre transformation, geometrically conceived. Given a functional relation, y = y(x) (1) with slope or tangent dy/dx, how can one replace the original independent variable x by the tangent t as the new independent variable while retaining all the informationcontained in equation (1) ? Conceive of the curve y = y(x) as generated by the envelope of its tangent lines, as illustrated in parts (a) and (b) of figure 10.2. If you do not have the curve in front of you, then, to construct the set of tangent lines, you need to know the y-intercept of each tangent lines as a function of the slope. That is, you need 1-1(t), where 1 denotes the y-intercept. The letter 1 is a mnemonic for Intercept. axis 0.2 Diagrams for the Legendre transformation, conceived geometrically. (a) The curve yx) (b) The curve defined by the envelope of its tangent lines. (e) One tangent line and its to x, y, t, and ).

Explanation / Answer

No , there does not

y = mx +c

from question , we understand that

y =t*x +I(t)

we cannot solve the above equation to get x and y values as we have only 1 equation to solve and 2 unknown variables

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.