College Calculus 1
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!!!!!

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). 
I have no fricken clue. That's why I fricken hatee MATH. LIKE HOLY FRICK! WHY THE FRICK DO WE HAVE TO LEARN?

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