Question

Given  x = k1e^a1t : y = k2e^a2t it is observed that the curvature function obtained is zero. What is the relation between a1 and a2?

(a) a1 ≠ a2

(b) a1 = a2

(c) a1 = (a2)^2

(d) a2 = (a1)^2

This question was addressed to me by my school principal while I was bunking the class.

I want to ask this question from Curvature topic in division Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (b) a1 = a2

To explain: Using formula for Curvature in parametric form

k=\(\left |\frac{y”x’-y’x”}{((x’)^2+(y’)^2)^{\frac{3}{2}}} \right |\)

Equating y”x’-y’x” to zero (given curvature function is zero) we get

\(\frac{y”}{y’}=\frac{x”}{x’}\)

Put x=k1e^a1t:y=k2e^a2t

\(\frac{k_1(a_1)^2.e^{a_1t}}{k_1.a_1.e^{a_1t}}=\frac{k_2(a_2)^2.e^{a_2t}}{k_2.a_2.e^{a_2t}}\)

a1=a2