# Calculus

posted by Stuck

"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?"

1. Reiny

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

2. Quidditch

Check back a few pages to see it.

3. Stuck

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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