Question

Here is the related rate problem:
At noon Ship A is 150 km west of Ship B. Ship A is sailing east at 35km/h and Ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 p.m.? I can't figure the diagram to use to help with the equation. I think it is 10^2 + 100^2 = z^2 where dx/dt is 35 and dy/dt is 25. Is this correct? I was thinking I should land up with a negative answer as the distance is being reduced at this time.

Answer 1

you are on right track and calculate dz/dt....but it would be definitely positive The distance was initially decreasing and then at a point reached it's minimum seperation and then started to increase. You cannot c all this if u think mathematically. But physics explains all this by just a imagining the figure:)

Answer 2

At (12) Ship A is 150 km west of Ship B.

A <------- 150 km ---------- B

Ship A is sailing east at 35km/h

A <------- 150 km ---------- B
--> 35km/h

Ship B is sailing north at 25 km/h.


^
| 25 km/h
A <------- 150 km ---------- B
--> 35km/h


How fast is the distance between the ships changing at 4:00 p.m.?

a formula defining placement will give us a formula to derive to find speed. Looks to be the pythag thrm for a right triangle

a^2 + b^2 = d^2 , implicit derivative gives

2a a' + 2b b' = 2d d' ; dividing off the common 2 and dividing by d solves for d'

d' = (a a' + b b')/d

Answer 3

a = 150 - 4(35)
b = 4(25)
d = sqrt(a^2+b^2)

a' is the speed of a
b' is the speed of b

everything is known to find d'. the speed at which the ships are moving with resoect to each other

Answer 4

Thanks for your affirmations. That is very helpful