calculus related rates

Question

A police car traveling south toward Sioux Falls, Iowa, at 160km/h pursues a truck traveling east away from Sioux Falls at 140km/h. At time 𝑡=0, the police car is 60km north and the truck is 30km east of Sioux Falls.

Calculate the rate at which the distance between the vehicles is changing at 𝑡=10 minutes.

(Use decimal notation. Give your answer to three decimal places.)

Answer 1

h^2=x^2 +y^2

hh' = xx' +yy'

h' = (xx'+yy')/h= rate of change of distance between police and the truck

but recalculate x, y and h at t=10 minutes

at t =0 y = 60, y'=160, x=30, x'=140, h=sqr(x^2 +y^2)

at t=10 minutes = 1/6 hour

y= 60-160/6= 60-26 2/3 = 33 1/3 = new y value = 100/3

new x = 30 +140/6 = 30+23 1/3 = 53 1/3 = 160/3

x' and y' remain the same

new h = sqr(100/3^2 +160/3^2)=(10/3)sqr356

plug the new values in to solve for h'

= (100/3(160)+160/3(140))/(10sqr356)/3

= (3840)sqr356

= about -203.600 kmh

(no guarantees there're no errors above. it's tedious calculations, possible arithmetic mistakes throughout. 200 kmh looks too large, maybe a decimal point error somewhere. for large enough t, h' will become 20 kmh, when both are going due east. Until then max h' would seem to be in the range of 160 kmh)

basic method is find x,y,h, at t=1/6, plug in those values plus x' and y'

into h' = (xx' +yy')/h

all derivatives are with respect to time