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

Sorta Kinda
Approximation with Derivatives

 

 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 r2

 
 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 r2

 as the radius changes
dA = 2pr

dr
 
 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:
V = 4 p r3

3
 
 The derivative of the volume as the radius changes is:
 
dV = 4p r2

dr
 
 We can multiply both sides of the equation by dr
 that gives us:

 dV = 4p r2 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:
 
dY

 = (Some Stuff)


dX
 
 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

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