Posted by **Stuck** on Thursday, June 17, 2010 at 9:31pm.

"At 9am ship A is 50 km east of ship B. Ship A is sailing north at 40km/h and ship B is sailing south at 30km/h. How fast is the distance between them changing at noon?"

I can never get these type of questions where two ships are heading in different directions and one is always ahead...please help!

- Calculus -
**Reiny**, Thursday, June 17, 2010 at 10:14pm
For most of these, the key is to have a good diagram.

In this problem I would place B at the origin and A on the 50 axis somewhere.

label AB as 50

draw a vertical up from A to C to show the path of A

Draw a vertical down from B to D to show the path of B

Join DC, this is our distance between them.

Now we need a righ-angled triangle, so extend CA to E so that AE = BD. Join DE

Let t hours be any time after 9:00 am

then BD = 30t and AC = 40t, making CE = 70t

We know DE = 50

DC^2 = CE^2 + DE^2

= 4900t^2 + 2500

differentiate with respect to t

2 DC (dDC/dt) = 9800t

dDC/dt = 9800t/2DC

when t- 3 (noon)

DC^2 = 4900(9) + 2500

DC = 215.87

so dDC/dt = 9800(3)/(2(215.87)) = 68.1 km/h

- Calculus -
**Quidditch**, Thursday, June 17, 2010 at 10:16pm
This same question was answered Wednesday for your 10:58pm post.

Check back a few pages to see it.

- Calculus -
**Stuck**, Friday, June 18, 2010 at 5:10pm
Thank you both! However, the problem is I can't seem to visualize the diagram correctly. I can't see where the right triangle is coming from. But no worries, I will look at more example problems and see if I can figure these questions out.

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