SOLUTION: Three runner's are going around a track which is an equilateral angle triangle with side 50 meters long. If the runners are equally spread and all running counterclockwise at 20 ki

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Question 1128661: Three runner's are going around a track which is an equilateral angle triangle with side 50 meters long. If the runners are equally spread and all running counterclockwise at 20 kilometers per hour, at what rate is the distance between a pair of them changing when they are
A.leaving vertices?
B.arriving at the vertices
C.at the midpoints of the sides
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
speed, such as km per hour, or meters per second.
I expect that you are supposed to use calculus and derivatives.
The rate of change of the distance between two runners will have units of
No unit was specified, but once one set of (time, distance) units is chosen, the result can be converted as needed.

With km and hours:
t= time in hours since each runner passed a vertex
The track and positions of runners A and B,
superimposed on a pair of coordinate axes,
are shown in the sketch below,
with distances in km.

Each runner last passed a vertex t hours ago,
and is now 20t km away from that vertex.
The position of runner A is given by the coordinates
system%28x=20t%2Cy=0%29 .
The position of runner B is given by the coordinates
system%28x=0.05-cos%2860%5Eo%29%2A20t%2Cy=sin%2860%5Eo%29%2A20t%29 , or
system%28x=0.05-10t%2Cy=17.32t%29 .
The distance between runners A and B is
D%28t%29=sqrt%28%280.05-10t-20t%29%5E2%2B%2817.32t-0%29%5E2%29
D%28t%29=sqrt%28%280.05-30t%29%5E2%2B300t%5E2%29
D%28t%29=sqrt%280.0025-3t%2B900t%5E2%2B300t%5E2%29
D%28t%29=sqrt%281200t%5E2-3t%2B0.0025%29
The rate of change of that distance with time, in km per hour is
dD%2Fdt%22=%22%281%2F2sqrt%281200t%5E2-3t%2B0.0025%29%29%282400t-3%29
dD%2Fdt%22=%22%281200t-1.5%29%2Fsqrt%281200t%5E2-3t%2B0.0025%29

A. When leaving vertices t=0 ,
D%280%29=0.05km=50m , and
dD%2Fdt=-1.5%2Fsqrt%280.0025%29=-1.5%2F0.05=-30
That is highlight%28%22-30+km+%2F+hour%22%29 or
The number is negative because the distance between the runners is decreasing.
B. When arriving at the vertices,
it should be the opposite of the result above,
because it is like running the film backwards.
Since my computer will calculate for me,
I will just apply the equations/formulas found above.
20t=0.05 , so t=0.05%2F20=0.0025 is the time in hours since last passing a vertex.
That is 0.0025+hours%283600+seconds%2Fhour%29=9.00+seconds
Of course, D%280.0025hours%29=0.05km=50m .
dD%2Fdt=%281200%2A0.0025-1.5%29%2Fsqrt%281200%2A0.0025%5E2-3%2A0.0025%2B0.0025%29%22=%22%283-1.5%29%2Fsqrt%280.0075-0.0075%2B0.0025%29%22=%221.5%2Fsqrt%280.0025%29%22=%221.5%2F0.05=30
The distance between the runners is increasing at a rate of
highlight%28%2230+km+%2F+hour%22=%228.333+...+m+%2F+s%22%29

C. At the midpoints of the sides,
the positions of two runners and the vertex between them
form an equilateral triangle with sides measuring 25 meters.
the distance between the two runners is 25 meters,
and as that is the midpoint of a very symmetrical situation,
that is the minimum distance and the rate of change is zero.
Just looking art the distance formula,
you see it is the square root of a quadratic function,
and it is obvious that the quadratic function and its square root will have a minimum.
At that point, the runners have run half a side since last passing a vertex,
0.025km=25m in 0.00125hours=4.5+minutes (half of the results for part B).
The formula for rate of change tells us that
dD%2Fdt=%281200%2A0.00125-1.5%29%2Fsqrt%281200%2A0.00125%5E2-3%2A0.00125%2B0.0025%29%22=%22%281.5-1.5%29%2Fsqrt%281200%2A0.00125%5E2-3%2A0.00125%2B0.0025%29%22=%220