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 Suppose car A is 110 miles north of an intersection
and traveling south at 50 mph. Suppose car B is 100
miles east of the same intersection and traveling west at
20 mph. 1a.) At what rate are the cars approaching each
other after 1 hr? 1b.) After 3 hrs


1.)  What do you need to find:  ${dr \over dt}$

2.)  What formula can you use which involves the variable of interest (in 
this case $r$)? 

\centerline{$x^2 + y^2 = r^2$}

3.) What do you know about the variables in the above equation?  In 
particular are any of the constant?

Note whether or not a variable represents a constant is very important 
when taking the derivative. 

In this case, $x, y, r$ are NOT constants.  We do know that $x(0) = 100$, 
$y(0) = 110$, $x'(t) = -20$, and $y'(t) = 
-50$.  Note the sign of the derivative indicates direction.  Make sure you 
use the correct sign.  In this case we 
decided that the positive $x$ direction is east and the positive $y$ 
direction represents north.

4.)  Take derivative of both sides w.r.t. variable $t$ of  $x^2 + y^2 = 
r^2$
\w
\centerline{$2x{dx \over dt} + 2y{dy \over dt} = 2r{dr \over dt}$}
\w
\centerline{$x(t){dx \over dt} + y(t){dy \over dt} = r(t){dr \over dt}$}

In this case ${dx \over dt}$ and ${dy \over dt}$ are constant w.r.t time, 
but  ${dr \over dt}$ is a function of $t$ 
which is not constant.

${dx \over dt} = -20$, and ${dy \over dt} = -50$
\w\w
\centerline{Hence $-20x(t) - 50y(t) = r(t){dr \over dt}$}
\eject
a.) At what rate are cars approaching each other after 1 hour? 
\u
Note $x(1) = 100 - 20 = 80$ and $y(1) = 110 - 50 = 60$.

Use  $x^2 + y^2 = r^2$ to find $r(1)$:  $r(1) = \sqrt{80^2 + 60^2}$ 

\rightline{$= \sqrt{10^2(8^2 + 6^2)}$ 
$= 10\sqrt{64 + 36} 
= 10\sqrt{100} 
= 100$.}

Hence when $t = 1$:  $100 {dr \over dt} = -20(80) - 50(60)$

Thus $ {dr \over dt} = -2(8) - 5(6) = -16 - 30 = -46$

Note answers to word problems must use words (including units):

\centerline{After 1 hour, the cars approach each other at the rate of 46 
mph}

Note the symbol for the negative sign is not included in the answer.  
Instead we use it to determine that the cars 
are getting closer instead of moving apart.


b.) After 3 hours
\u
$x(3) = 100 - 3(20) = 40$ and $y(3) = 110 - 3(50) = -40$.
\u
The negative sign indicates that car A has crossed the intersection and 
hence is 40 miles south of the intersection 
when $t = 3$.  


Use  $x^2 + y^2 = r^2$ to find $r(3)$:  

\centerline{$r(3) = \sqrt{40^2 + (-40)^2} 
= \sqrt{40^2(2)} 
= 40\sqrt{2}$.}

Hence when $t = 3$:  $40\sqrt{2} {dr \over dt} = -20(40) - 50(-40)$

Thus $ {dr \over dt} = {-20 + 50 \over \sqrt{2}} = {30 \over \sqrt{2}}$

Note answers to word problems must use words (including units):

\centerline{After 3 hour, the cars moving apart at the rate of ${30 \over 
\sqrt{2}}$ mph}

Since $ {dr \over dt}$ is positive, we know cars are  moving apart instead 
of getting closer.


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