Please explain why a(t)=4-2t and b(t)=4t in this question. Or answer this question in detailed with steps in another method

MAT135H5 December 2019 Final Examination v1 5. (8 marks) Peter leaves his house at noon and bikes straight south at 4 km/h. Peter's friend also ste at noon at another house, and bikes straight east towards Peter's house at 2 km/h and arrives at Pet house at 2 pm. At what time are the two friends closest together? ther Friend Peters At t hours after noon , let alt ): = Friend's distance act ) 6 (+ ) from Peter'shouse ( KM ) and 6 (+ ) = Peter's distance . Peter from Peter's house ( KM) . We want to find the minimum of the distance d(t) between Peter & Friend in the interval [0, 2]. . First, let's check the value of d(t) at the endpoints t = 0, t= 2: - The distance d(0 ) is just the distance between the houses. Since it took Friend 2 hours to travel this distance at 2 ur , it follows that dlo ) = 2.2 = 4 . - The distance d( 2) is just Peter's distance from his house of 2 pm. Since he's traveling at 4 KM it follows that d ( 2) = 4.2= 8. . Now , let's check for critical points in ( 0 , 2 ) . Note that so act ) = 4 - 2t and b (+ ) = 4t, while d (+ )? = alt ) 2 + 6 (+)?, So that differentiation yields "d ( + ) d ' ( + ) = act )a' ( +) + 6 ( + 16 ( + = -2(4-2+)+4.4t In particular , there is a critical = 20t-8. point at t= 2/5, where d ( 2/5 ) = 1 14 - 4 , 2 + ( 4/ ) =8/