



Suppose
we have a circle with a radius of ...



maybe
40 inches. 


How
would we do it? 





We
would use the equation for the area of a circle, of course! 





A
= p
r^{2}






So
we would have: 


A
= p
(40)^{2}



A
= 1600p



A
= 5026.5






But
what if we wanted to find the area of a ring one inch wide 


with
an inside radius of 40 inches? 





We
could find the area of a circle with a radius of 41 inches 


and
subtract from that the area of a circle with radius of 40 inches ... 






A
= p
(40 + 1)^{2}  p
(40)^{2} 



A
= 1681p
 1600p 



A
= 81p 





That
works fine, and we get the right answer. 


But
when the equations get messy, this method might get tough. 





There
is a way to use calculus to approximate the area of a ring, or a
shell, 


or a tube
or
anything for which we have the equation for the basic shape. 





Think
of the area of the ring as the change (or difference) in the area of
a circle 


from
one with a radius of 40 inches to one with a radius of 41 inches. 


That
is, DA. 


And
the difference in the radius (41  40) as Dr. 





If
Dr
is very small compared to r, 


derivatives
give us a good approximation of the area of the ring. 





Take
the derivative of the area equation 


A
= p
r^{2}



as
the radius changes 








Did
you see that we didn't need to use implicit differentiation 


on the
right side of the equation since
the "with respect to" variable was r. 





We
can multiply both sides of this equation by dr, to get: 





dA
= 2p
r dr






That
last little move didn't look like much, 


but
it will come in VERY HANDY in later topics. 





This
equation translates in English to: 


"The
change in area is equal to two times pi times the original radius 


times the change in the radius" 





Plugging
the numbers we have into this equation, we have: 





dA
= 2p(40)(1)



dA
= 80p






That
was pretty close to the actual answer A
= 81p
but took less effort. 


This
method will work with any shape, 


but
the thickness of the ring or shell needs to be small 


compared
to the radius itself. 





Example: 


Use
the approximation method to find the volume of material 


that
goes into making a ball 0.1 inches thick that has an inside radius
of 5 inches. 





The
equation for the volume of a sphere is: 








The
derivative of the volume as the radius changes is: 











We
can multiply both sides of the equation by dr 


that
gives us: 


dV
= 4p
r^{2} dr






Substituting
in the values: 


r =
5 and dr = 0.1






That
gives us: 


dV
= 4p(5)^{2}
(0.1)



dV
= 10p
cubic inches






This
is the approximate volume of the material the ball is made of. 


It
is not the volume inside the ball. 


The
volume inside the ball is much greater. 





Things
like dA, or dr or dY are called differentials, 


and
any equation that contains one or more of these 


can be called a
differential equation. 


(Impressive
name, eh?) 





If
we have any equation that looks like: 











We
can multiply both sides by dX and get: 





dY = (Some
Stuff)dX 





If
X is an independent variable, then dX = 1 so this is no big deal 


BUT,
if dY and dX both stand for some complex stuff, 


the process is very
important. 





"Differential
Equations" is supposed to be some big nasty scary topic. 


You've
already solved some and didn't even know it! 


It
can't be all that bad then, can it? 





copyright 2005 Bruce Kirkpatrick 
