Question

Two ships leave port at noon. One ship sails east at 6mi/hr and the other sails north at 8mi/hr. At what rate are the two ships separating 2 hours later?

Answer 1

Just to be sure, this is from a calculus class, correct?

Answer 2

Yes! =]

Answer 3

related rates lol.

Answer 4

Awesome, so how far are you able to get on this on your own?

Answer 5

im not sure if this is a right first step but is it first to add 6 and 8 then multiply 2 hours each ? im really bad with application problems so im not so sure.

Answer 6

Not quite, but that's a good step in the right direction since distance is rate times time. This would work if they were both heading in opposite directions, but they're heading at a right angle to each other, so we have to do something a little different, I'll show you.

Answer 7

Ohhh oops yeah i thought it was opposite.

Answer 8

i think the first step should be to draw a right triangle ?

Answer 9

Yeah exactly, good job. =) Draw one out and label everything you can, and I'll help you if you get stuck.

Answer 10

thanks I will, im so dumb i thought they were going opposite from each other.

Answer 11

brb trying on my own.

Answer 12

Haha no problem, you seem to know how to figure things out pretty well.

Answer 13

Okay so would db/dt equal 6 and dh/dt = 8 ?

Answer 14

after drawing the right triangle i came up with those labels

Answer 15

am i going in the right track ?

Answer 16

That sounds perfect to me. =)

Answer 17

oh okay good!

Answer 18

Im really slow at trying to figure it out so i may take a while to get back to you with my results so just stay here until im ready lol.

Answer 19

Yeah, take your time. I'm just working on my own problems here, trying to figure stuff out too haha.

Answer 20

would I have to find the perimeter of the triangle to get my answer ?

Answer 21

or is that not part of the steps ?

Answer 22

Nope, it's not necessary to find the perimeter. It might be useful to use the pythagorean theorem though.

Answer 23

oh so how do i find the rate of change then ?

Answer 24

i thought i had to find the perimeter of the triangle to find the rate of change using differenitation

Answer 25

I used dp/dt

Answer 26

Draw your picture on here and label it and I'll show you. You're not looking for the change in perimeter, you're looking for the change in distance between the two boats which is basically the change of the hypotenuse.

Answer 27

Oh okay so i just wasted my time then lol.

Answer 28

okay one sec.

Answer 29

OHHH wait so instead of just adding i multiply ? o-o

Answer 30

cause the equation i did was

Answer 31

dp/dt = dx/dt +dy/dt + 1/2(x^2+y^2)^-1/2(2x(dx/dt)+2y(dy/dt)

Answer 32

sorry it looks messed up without the parentheses

Answer 33

Yeah, I'm not really sure what's going on there, but it doesn't seem to be quite right. Here, I'll set you straight!

Basically we just need a relationship between the boats and we also need to keep an eye on what we want. We want the change in distance between the two boats. So here:

So if we look at the problem, we're basically given da/dt and db/dt and we want to find dc/dt. So using a^2+b^2=c^2 we can take the derivative of that and solve for dc/dt and plug in all the other values we know.