Question

Exercise 2: Show that the parabola y = ax2 + bx + c, a + 0, b and c are constants, has its largest curvature at its vertex and that it has no minimum curvature.

Answer 1

Solution :- we have to show that the parabola y = ax²t badcato, b and a are constants, has its largest curuature as its verter and that it has no minimum curvature . y = anat batch = y' = 2an+b & y" - 2a for verten y'=2an+b=0 an=- we know that cuneature kca)= Y [4+ (!) 2 1/2 kui zan. (1+ 2an+b)273 = 20 (1 + (2onth)212 k' (u) = 2ax (+3) [1+(2431+012.2 (20x+6) 201 = -12a (2an+b) [1 +(20x +b)275/2 90 H!! Cau) = -12 0 [ 2+ (zaan tus]?. (201) –(202492 [ 24 Conjunto - 2 (2an+b). 20} [2+ (24x+b)?]$ | = - 12 / 24 (1+ (a + b))_ loa (243) - (2x +LÝ CI+ C 227+5)?js for critical points of kro= 0 - 3 a (zandb [1t (2an+b) 2 2

200+b= 0 = M- at *" ( r= ) = -bha -12a82a Caton-loato [1+0]"h? - - 12a (2a - o) = – 24 a2 co so, kcals ich matinum at a= - b So there is no minimum value of kca).