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# Related rate given find the time it occurs

Thu 24 Oct, 2013 12:55 pm
Ships A and B leave the same port. Ship A leaves at 10:00 am going north at 12 mpr
Ship B leaves at 10:30 am going west at 16 mpr
At what TIME is the rate of change of the distance between the ships equal to 18.86 mph.
Here is what I have so far,
Distance ship A goes is A, distance ship B goes is B, distance between the ships is S. dA/dt = 12 mph. dB/dt = 16 mph. dS/dt = 18.86
A = 12 t. B= 16t- 8
I am unable to calculate the time at which dS/dt = 18.86 mph
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fresco

1
Thu 24 Oct, 2013 01:32 pm
@Gamma,
Are you sure of your wording ?
I think the relative velocity is 12 mph for first 30 min and 20mph (by Pythagoras) thereafter. At no time does it appear to be 18.86 mph.
Maybe the question is "at what time is the distance between the ships 18.86 miles ?"
Gamma

1
Thu 24 Oct, 2013 03:58 pm
@fresco,
Hi, Thanks for replying... The problem does ask for the time that the rate of separation is equal to 18.86 mph. The problem is from a calculus text from 1972. The answer from the back of the text, is 11:00 am , but I have no idea how this is calculated. The portion of the distance ship A has as a head start is 6 miles and this becomes proportionally smaller as time passes. If you work backwards and put the relative values in using the answer of time equallying 1 hour dS/dt is indeed 18.86,
S dS/dt= A dA/dt + B dB/dt
S*18.86= 12(12) + 8(16). And S equals 14.422
I am sure the problem is do-able, just tricky, probably really simple too
but finding the relevant equational relationship to calculate this time given only the rates has proven to be exceedingly frustrating. I am sure I am overlooking something. I hoped someone would see a way to solve it.
The way it is asked, this problem is relegating time to a dependant variable status, in most other problems, time is an independant variable, so solving it might involve differentiating (implicitly) with respect to something other than time. I have attempted solutions with this in mind but have had no luck.

fresco

1
Fri 25 Oct, 2013 05:39 am
@Gamma,
Suggestion:
Draw a pair of axes at right angles representing position from the port (at the origin). Then at time t, A will be at 12t and the vertical axis and B will be at 16 (t-0.5) on the horizontal axis. So the distance S, between them, will given by
S= sqrt{ (12t)^2 + ((16(t-0.5)^2)} by Pythagoras.
Differentiate that and equate it to 18.86 to solve for t.
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fresco

1
Fri 25 Oct, 2013 06:22 am
@Gamma,
...and working backwards for t=1, it works !
(I didn't fancy actually solving the equation )
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engineer

1
Fri 25 Oct, 2013 06:25 am
@Gamma,
The relative velocity between these two ships is always a constant and doesn't change - 20mph. I think you are reading something wrong here.

You got this correct: S dS/dt= A dA/dt + B dB/dt, but B=4A/3 and S=5A/3 so when you substitute in, all the A's drop out so once again, no solution.
fresco

1
Fri 25 Oct, 2013 06:31 am
@engineer,
Thats was my first impression too but the question is not about the relative velocity between the ships, its about the rate of change of the difference in positions from the port. Follow my solution above which concurs with the given answer of 11.oo.
Lordyaswas

1
Fri 25 Oct, 2013 06:33 am
@fresco,
Is that GMT?

If not, I think that's where I'm going wrong.
fresco

1
Fri 25 Oct, 2013 06:38 am
@Lordyaswas,
Its certainly based on a flat earth !
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