Petter cycles straight north towards a crossroads. He starts 11 meters away, and the distance to the junction after t seconds is equal to 11-t^2 meters. At the same time Anne runs straight east from the intersection at a constant speed of 2m/s?

At what time is the distance between Petter and Anne at least?

The answer is 3 seconds, but how??

1 Answer
Harish Chandra Rajpoot
Jul 26, 2018

#3\ sec#

Explanation:

The distance traveled by Anne in time #t# from intersection point

#=2t#

The distance of Peter from intersection point at any time #t#

#=11-t^2#

At any instant of time #t#, the distance say #D# between Peter & Anne will be equal to the hypotenuse of a right triangle with legs #2t# & #11-t^2# which is given as

#D=\sqrt{(2t)^2+(11-t^2)^2}#

#D^2=4t^2+121+t^4-22t^2#

#D^2=t^4-18t^2+121#

Let #D^2=S# a new variable then

#S=t^4-18t^2+121#

#{dS}/dt=4t^3-36t#

#{d^2S}/dt^2=12t^2-36#

For minimum or maximum, #{dS}/dt=0#

#4t^3-36t=0#

#4t(t^2-9)=0#

#t(t-3)(t+3)=0#

#t=3, 0, -3 #

But, #t>0# hence we get #t=3\ sec#

Setting #t=3# in second differential, we get

#{d^2S}/dt^2=12(3)^2-36=72>0#

Since, #{d^2S}/dt^2>0# hence the value of #S# is minimum at #t=3# i.e. the distance #D# between Peter & Anne is minimum at the time #t=3\ sec#

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