Sven starts walking due south at 6 feet per second from a po

Sven starts walking due south at 6 feet per second from a point 140 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 190 feet west of the intersection. Write an expression for the distance d between Sven and Rudyard t seconds after they start walking. d = When are Sven and Rudyard closest? (Round your answer to two decimal places.) t = What is the minimum distance between them? (Round your answer to two decimal places.) d =

Solution

a)

d =[(140 -6t)² +(190-4t)²]

b) closest when d is minimum

d is minimum when f(t)=(140 -6t)² +(190-4t)² is minimum

f \'(t)=0 for minimum

f \'(t)=(2(140 -6t)(-6)) + (2(190-4t)(-4)) =0

f \'(t)=104t -3200=0

t=400/13

f \"(t)=104 >0

so f(t) is minimum when t= 400/13 =30.77s

minimum distance =[(140 -6(400/13))² +(190-4(400/13))²]

minimum distance =80.43