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Suppose we have a circle with a radius of ... |
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| maybe 40 inches. | ||||||||
| How would we do it? | ||||||||
| We would use the equation for the area of a circle, of course! | ||||||||
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A = p r2 |
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| So we would have: | ||||||||
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A = p (40)2 |
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A = 1600p |
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A = 5026.5 |
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| But what if we wanted to find the area of a ring one inch wide | ||||||||
| with an inside radius of 40 inches? | ||||||||
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| 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 ... | ||||||||
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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 | ||||||||
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A = p r2 |
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| as the radius changes | ||||||||
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| 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: | ||||||||
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dA = 2p r dr |
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| 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: | ||||||||
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dA = 2p(40)(1) |
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dA = 80p |
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| 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: | ||||||||
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| The derivative of the volume as the radius changes is: | ||||||||
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| We can multiply both sides of the equation by dr | ||||||||
| that gives us: | ||||||||
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dV = 4p r2 dr |
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| Substituting in the values: | ||||||||
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r = 5 and dr = 0.1 |
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| That gives us: | ||||||||
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dV = 4p(5)2 (0.1) |
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dV = 10p cubic inches |
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| 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: | ||||||||
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| We can multiply both sides by dX and get: | ||||||||
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dY = (Some Stuff)dX |
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| 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 |
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