Please help Calculus problem!?
A runner sprints around a circular track of radius 130 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 260 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 260 m? (Round your answer to two decimal places.)
Comments
Set up a rectangular coordinates system at the center of the circular path.
The runner is called A & his friend is called B who is located on the +ve x-
axis at a distance 260 m from the center O. A is somewhere on the circle
at a distance s m from B & is running at a speed 7m/s in the counterclockwise
direction along the path.
Let <AOB=C, by the cosine rule, in triangle AOB
s^2=130^2+260^2-2(130)(260)cosC-----(1)
=>
2ss'=2(130(260)(sinC)C'
=>
s'=7(130)(260)sinC/(130s)
=>
s'=7(260)sinC/s---------(2)
When s=260, from(1), we have
cosC=130^2/[2(130)(260)]
=>
cosC=1/4
=>
C=1.318 or 4.965 rad.
Hence, there are 2 solutions for s' from (1) & (2):
C=1.318=>
s'=7(260)sin(1.318)/sqr[130^2+260^2-
2(130)(260)/4]
=>
s'=7sin(1.318)=6.78 m/s
C=4.965=>
s'=7(260)sin(4.965)/sqr[130^2+260^2-
2(130)(260)/4]
=>
s'=7sin(4.965)= -6.78 m/s
Ans. When s=260m, the speed of the distance AB are
at 6.78 m/s away from B or 6.78 m/s close to B.