Question

Section B Bay Swim (10 points total) A person swims in a bay of parabolic shape f(x) = and would like to swim back to shore on the shortest path possible. The person is at position A = (0,3). The aim is to find the landing point P = (x, f(x)) with x > 0 that minimises the distance to swim. f(x)= A=(0,3) e P= (x, f(x)) Problem 4 (2 points) ? What distance D(x ) wil the person need to swim to reach an arbitrary landing point P = (x, f(x)) on the shore when f (x) = ? D(x) =

Answer 1

Given figure is Y fialna A(0,3) P(x,x)) X А person Swims in bay of parabolic shape fine The person is at position A = (0,3) let P(x, to be any anbitary point on parabola. DG)-AP & We-o) ²+ (2x) - 3 ) (fix) - 372) e D'a) Ap.o. (J x 4 (+3)=) DEG) = AY" - 5 x + (x * - 3)² Ap² = x²+ -3)? wish to find minimum distance. gex) x?+ (x2 -3)? suppose gle) 2 We will maxima – minima method find it using

g'(x) 2x 2(2-3) (2) gla 2% + 2(4-3) for extremizing , 2x + 24 (2-3) = 0 2x (1 + x2 -3) 2* ( *--) - * ( *2-4) 2x 2 08 x = -2 in ou so x = 2, g") = 2 + 2(2).+ 262 - :) 2 + 2x² 4x² - 6 3x² - 4 g"6) co) = -4 Lo for minimization, guco) g" (2) = 364) -4 = 870 = 3(4) -4 = 8 >0 9"60)

X = 2 and x=-2 point of minima. are and (+2, 2) (2, 2) landing points (+2, 2) reach that distance is minimene D(x) = ✓ **+ (223)