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November 26, 2014

November 26, 2014

Posted by **Please Help!** on Friday, October 1, 2010 at 3:33pm.

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^2-2abcosC. 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**, Friday, October 1, 2010 at 10:28pmI 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**, Monday, October 4, 2010 at 7:56pmI have no fricken clue. Thats why I fricken hatee MATH. LIKE HOLY FRICK! WHY THE FRICK DO WE HAVE TO LEARN?

- College Calculus 1 -
**Anonymous**, Thursday, October 21, 2010 at 8:55pmJEEZZZ-,-

dont understand-.-"

it will be easier to learn by talking...

im confused with the signs hahaha!!!

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