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I posted the question.

Set dy/dt = rate of snail = 2m / hr dx/dt = rate of tortoise = 4m / hr

|\

| \

y | \ r

| \

|____\ roughly....

x

so r = √(x2 + y2)

and dr/dt = √ (dx/dt2 + dy/dt2) right?

but how do you take into account the 2m head start of the snail?

dr/dt = √ (dx/dt2 + (dy/dt + 2)2) ???

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Q: If a snail and a tortoise start at point P the snail starts north at 1pm at 2m per hr and the tortoise starts east at 2pm at 4m per hr how fast is the distance between them changing at 3 pm?
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