



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: 





r^{2}
= (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 r^{2}
was an outside/inside type thing, and p
was just a number. 


p
will just kind of hang where it is. 


The
outside part, r^{2}
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 r^{2} 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: 











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 
