draw a sketch of the situation,
at a time of t hours after 9:00 am
let A1 be the position of Aida
let I1 be the position of Ian
Join A1 and I1, to get a right-angled triangle
let D be the distance between them
D^2 = (6t)^2 + (20-8t)^2
= 36t^2 + 400 - 320t + 64t^2
= 100t^2 - 320t + 400
2D dD/dt = 200t - 320
dD/dt = (200t-320)/(2D)
= 0 for a minimum of D
200t = 320
t = 1.6
So they are closest after 1.6 hours or 1 hour and 36 minutes past 9:00 am
making it 10:36 am
This is a standard question for this topic of Calculus.
In my opinion, you should really know how to do this question
Ian's house is located 20km north of Aida's house. At 9am, one saturday, Ian leaves his house and jogs south at 8km/hour. At the same time Aida leaves her house and jogs east at 6km/h. When are Ian and Aida closest together given that they both run for 2.5 hours? The answer is t=1.6 hours @ 10:36 am
PLEASE SHOW sTEP bY STEP!! i don't understand how to do this
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