Calculus Related Rates
Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT

It's All Relative
Related Rates

 
 When you find the derivative of something, you are finding the rate that one thing is changing as another changes.
 means the rate that Y is changing (dY) as X changes (dX).
 dY and dX are two separate things.
 
 We sometimes have problems where they are separated, like this:
 

 dY = (stuff)dX   or even (stuff)dY = (other stuff)dX

 
 Sometimes the variables in our problems are also functions
 of some OTHER variable
 That other variable is usually time.
 You will see that even that is no big deal.
 
 Example:
 The area of a circle is found using the equation:
 
 
 In this equation there are two variables, A and r
 Say the circle gets bigger as time passes.
 That means the area and the radius get bigger as time passes.
 The change in area can be written as dA.
 The change in time can be written as dt.
 That means the change in area as time passes can be written as:
 
 In math talk, we say:
 "the change in area with respect to time" just to sound really smart.
 
 The radius also changes as time passes.
 We can write that one as:
 
 
 If the area and the radius change as time passes,
 we say that they are "Functions of Time."
 That means neither one is really an independent variable.
 Instead, time is the independent variable.
 
 Great!, Swell!, Big Deal! So just what does that mean to us in the real world?
 
 All that means is that if we want to find the derivative of "A" or "r" 
 as time passes, 
 we want:
 
 We have always said that one way to write the derivative
 was the d(this)/d(that) notation.
 The difference now,
 is that we have a d(that) that wasn't even part of the original equation.
 
 This is a very big deal.
 It is where we really start to make calculus rock and roll.
 The only thing is when we use a variable for the d(that), 
 it should be something that changes the value of the d(this) as it changes.
 
 It is a usually a safe bet to say that d(this) changes as time passes.
 That means that d(something)/dt works just about everywhere.
 
 So let's find the derivative of the circle area equation as time passes,
 er that is, "with respect to time"
 
 The area equation is:
 So taking the derivative of each side
 with respect to time is:
 
 On the left side we have the derivative of the area with respect to time
 This is a rate. 
 It is the rate that the area changes as time passes.
 
 On the right side of the equation we have pr 2.
 p is just a number. 
 "r" stands for some function where time is the independent variable.
 We actually have r 2, which makes it one of those outside/inside things.
 That is:

 r2 = (Some unknown equation where time is the variable)2

 
 The left side of this dA/dt, is pretty much done for the moment.
 Lets work on the right side.
 We said that the r2 was an outside/inside type thing, and p was just a number.
 p will just kind of hang where it is.
 The outside part, r2 will become 2r.
 The inside part, and watch this trick, becomes:
 
 
 That makes the whole thing:
 
 
 STOP!
 Something really important just happened here.
 We took the derivative of r2 with respect to time. That part gave us:
 This is an unbelievably important point.
 
 Make sure you understand this step.
 
 Did I mention that this was important?
 Well it is.
 
 OK, the whole derivative equation is:
 
dA = 2pr dr


dt dt
 
 This means that the rate that the area of the circle is changing
 at some point in time is equal to 2 times p times the radius at that moment 
 times the rate the radius is changing at that time.
 
 There are three unknown things here:
 
 
 We need to know two of these to find the third one.
 
 A problem might go something like:
 
 A circle is increasing in radius at the rate of 2 inches per minute.
 Find the rate that the area is changing when the radius is 5 inches.
 
 So we have:
 
 Make sure that all of the units you have agree,
 like all of the lengths use the same measure,
 and all of the time uses the same measure.
 Here we have inches per minute, and inches, so we're fine.
 If the units were not the same we would have to convert values until they were.
 If the radius had been stated in feet or centimeters,
 we could convert it to inches.
 Since we only have one value with a time dimension,
 there probably won't be a problem there.
 
 To solve this problem,
 just substitute the stuff we know into the derivative equation
 
 
 and find the thing we don't know.
 So we have:
 
 OK, the problem is over but here are some points to keep in mind:
 
 Our derivative was:
 
 We COULD have written this as:
 
 While using something like A' is technically correct, 
 it doesn't give as much info as using:
 
 
 This says: "The rate that the area is changing as time changes."
 A' is not so specific. It says : "The rate that the area is changing"
 BUT it doesn't say as compared to what.
 A' isn't wrong, it's just not as accurate.
 
 When we took the derivatives of A and r, we used implicit differentiation.
 The rules of calculus say that whenever we take the derivative of something
 with respect to something OTHER than itself, we do implicit differentiation.
 That is, it becomes an outside/inside thing.
 
 So:
 BUT ...
 
 This is more of that really important stuff.
 
 Last Example:
 (for this page anyway)
 A circle is decreasing in size at the rate of 5 square inches per minute.
 At what rate is the radius decreasing when the radius is 4?
 
 Start with the equation for the area of a circle...
 
 
 Now take the derivative of this as time passes.
 OK, take the derivative with respect to time ...
 
 
 Let's write out what we know:
 
 Negative signs just mean the value is getting smaller...
 
 Now we just substitute what we know into the derivative equation
 and solve for the thing we don't know ...
 
 
 -5/8p is about -0.2
 That means the radius is getting smaller at the rate of 0.2 inches per minute
 when the radius is 4.
 The minus sign means it is getting smaller.
 

   copyright 2008 Bruce Kirkpatrick

Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT