# calculus

posted by on .

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

• calculus - ,

let the time passed since noon be t hours.
So their paths form a right-angled triangle, with sides
(110 + 25t) and 15t
let d be the distance between them, then
d^2 = (110+25t)^2 + (15t)^2
2d dd/dt = 2(110+25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)
d^2 = 210^2 + 60^2 = 47700
d = √47700

dd/dt = ( 50(210) + 30(60) )/(2√47700) = 28.16 mph

• calculus - ,

That's not right...

• calculus - ,

You are right, I see my error.
I have ship A going west instead of east.

let the time passed since noon be t hours.
So their paths form a right-angled triangle, with sides
(110 - 25t) and 15t
let d be the distance between them, then
d^2 = (110-25t)^2 + (15t)^2
2d dd/dt = 2(110-25t)(25) + 2(15t)(15)

when t = 4 , (4:00 pm)
d^2 = 10^2 + 60^2 = 3700
d = √3700

dd/dt = ( 50(10) + 30(60) )/(2√3700) = 18.9 mph

• calculus - ,

Yeah, I originally got 18.9 km/h, but the website is not accepting my answer. Is there something else?

• calculus - ,

2d dd/dt = 2(110-25t)(-25) + 2(15t)(15)

• calculus - ,

you guys are retards

• calculus - ,

some kid is a friendless autist