Posted by Please Help! on .
A runner sprints around a circular track of radius 150 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 300 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 300 m? (Round to 2 decimal places.
My teacher tried helping me with this problem and only confused me more. I drew a diagram. I used the formula c^2=a^2+b^22abcosC. Right now I have 2L(dL/dt)=2(150)(300)sin(angle)(dangle/dt) I think (dangle/dt) is 7/150. This would give you (solving for dL/dt:
(dL/dt)= (45000sin(angle)(7/100))/300
HELP!!!!!

College Calculus 1 
MathMate,
I suggest the following approach.
Let the friend (outside of the tracks) be at the origin.
Express the x and y coordinates of the runner in terms of t, for example,
X(t)=rcos(st/r)+d
Y(t)=rsin(st/r)
s=speed in m/s
r=radius
d=given distance
Express the distance between the two persons in terms of t:
L(t)=√(X(t)²+Y(t)²)
Differentiate L(t) with respect to t, which gives L'(t), the rate of change of distance as a function of t.
Finally, find the times at which the two persons are 300m apart, namely, solve for t at which
L(t)=300.
The answers are t1=39 and t2=96 seconds approx.
L'(t1) and L'(t2) are the required answers.
I get about ±6.8 m/s, which is reasonable considering that the runner runs at 7m/s and the points in question are not far from the points of tangency (where L'(t) should give ±7m/s). 
College Calculus 1 
Shayne,
I have no fricken clue. That's why I fricken hatee MATH. LIKE HOLY FRICK! WHY THE FRICK DO WE HAVE TO LEARN?

College Calculus 1 
Anonymous,
JEEZZZ,
dont understand."
it will be easier to learn by talking...
im confused with the signs hahaha!!!