Calculus related rates problem - the relation between distance and time A person B is located 350 meters east from a person A. Person A starts riding a bike north at a rate of 5 m/sec and 7 minutes later person B starts riding a bike south at a rate 3 m/sec. At what rate is the distance separating the two people changing 25 minutes after person A starts riding?
So i guess my biggest problem is that I'm having a hard time understanding this question...
 so I am given dx/dt = 3m/sec
and I have to find dy/dt when x=25?? Can someone provide a basic guideline on how to approach this question?
 A: Start by defining your axes and variables.  Let East be $+x$, North be $+y$, measured in meters with time in seconds.  If A starts at the origin, B starts at $(350,0)$  As A is riding North, his location at time $t$ is $(0,5t)$  What is B's location after he starts riding?  Now find the distance $d$ as a function of time. You are then asked for $\frac {dd}{dt}$ at 25 minutes.
A: You need to find $r(t)$ which is the distance between the two cyclists at time $t$. Then to find the rate of change of distance you take the derivative with respect to time. Let $\vec{r}_1(t)$ be the position vector of person A and $\vec{r}_2(t)$ be the position vector of person B. Let the origin be placed such that $\vec{r}_1(0)=\vec{0}$, that is, person A starts at the origin at time zero. Then using the information in the question you can determine explicit expressions for $\vec{r}_1(t)$ and $\vec{r}_2(t)$. Then you can find the distance between the two vectors for any time by finding $r(t):=|\vec{r}_1(t)-\vec{r}_2(t)|$, which is a scalar function. Then you take the derivative of that function with respect to t and evaluate it at t=25 minutes.
