~RELATED RATES~

~A rate (or rate of change) in calculus is instantaneous unless it is specified as average.  For example, ds/dp will give the rate of change of quantity s wrt p. So, derivatives are calculated one way or another.

~Time derivatives also give rates since they measure how a certain quantity changes with time. Velocity is the derivative of a distance wrt to time, or ds/dt. A positive ds/dt tells us that s is increasing and a negative ds/dt will tell us s is decreasing.

~Other quantities, such as volumes, areas, lengths of objects, & so on, also could be increasing or decreasing and, consequently, can have a measured derivative value.

~In dynamic systems (quantities changing) involving several quantities, the rates involved (derivatives of these quantities wrt time), are all tied together. This is where the terminology of "related rates" comes into play.

~Since derivatives give instantaneous rates, getting an equation tying these rates together, in effect, allows us to "freeze" time when a value of time is used. (this will describe the dynamic system at that instant). This is a very powerful tool of calculus, since this allows us to analyze the problem at any instant we desire.

~How does one get this related rate equation?

~Use the following procedure carefully and, hopefully, you'll be successful.

1)  Draw a sketch & label it with the constants & variables involved in    
    your problem.  Label all changing quantities variables.

2) State, in mathematical terms, what is known (given) and what you   
     are after in your problem.

3) Find an equation between the variables of your problem that hold for    
    all times. This could be any valid mathematical formula from the
    theorem of Pythagorus to a similar triangle relationship to a trig
    relationship, or some other one.  Many times, this relationship is a     
    basic one, but not always.

4) Take the derivative implicitly of both sides of this equation with   
     respect to time. This is your related rate equation. Then, substitute
     into this equation all information given and solve for the desired    
    derivative (rate). There should be enough info to find that desired
    rate. Sometimes you'll have to go to the general relationship
    (in step 3) to get values for other quantities.

~Time for a very simple example: (more involved examples will be covered in class.

 A point is moving on the parabola,  y = x². The abscissa (x value) is increasing at a rate of 2 inches/sec. How fast is the ordinate value (y value) changing  at the point where x = 5 inches?

1) A sketch is nice, but not necessary.

2) dx/dt = 2.  Find dy/dt when x = 5

3) The relationship between x and y is given by  its equation y = x²

4) dy/dt = 2x dx/dt   (implicitly differentiating both sides wrt time)
    Since the instant we want is x=5 and we know dx/dt =2, just sub
    into this related rate equation. We get,  dy/dt = 20 inches/sec. for
    our answer.
   ~Note: we can see that the rate y changes not only depends on dx/dt but on the value of x also.

~Note:  Not all related rate problems are this simple. Some are quite challenging. However, follow the procedure and your chances of getting it right are very good.