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Zorodius

Units of a derivative?

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Can someone point out the error in my reasoning here? I feel like it should be obvious to me, but it''s not. I want to find how much a circular object''s surface area increases when its radius increases. Since A = πr², then dA / dr = 2πr, by the power rule. So, say for instance the radius was 4 cm, then the change in area with respect to radius would be cm. This is where I''m confused. Increasing or decreasing the area by anything other than square units seems like a nonsensical operation. Were I to take the derivative with respect to something else, say time, then dA / dt = 2πr * dr/dt, and the units would then be in square centimeters or square feet, etc. Can someone help explain how to interpret this correctly? I just don''t understand how the change in area with respect to anything can be in anything but square units.

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Because you''re forgetting that you''re multiplying by unit in order to use the derivative.

If you take dA/dt the units are square centimeters per second.

For dA/dr the units are square centimeters per centimeters, otherwise known as plain old centimeters.

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Guest Anonymous Poster
quote:
Original post by Zorodius
Can someone point out the error in my reasoning here? I feel like
Can someone help explain how to interpret this correctly? I just don''t understand how the change in area with respect to anything can be in anything but square units.


The derivative doesn''t give you the change in area, but the RATE of change in area. To get the actual change you''ll have to integrate the rate, which will get you back the the correct unit.

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one could also argue that because it's an instantaneous rate of change, it shouldn't be in square units anymore. that's because an area is really just a bunch of one dimensional 'lines' or in this case radii, and the smallest possible change in area couldn't ever be an area if it's made up of smaller units.

if you want your brain to fry, then think about something else. say radius is changing at .5 units per second. then that means the da/dt =s 2PI r * (1/2)

but think about it, if the instantaneous rate of change of area is the smallest amount that the area can change, then it is impossible to get a unit that is 1/2 the instantaneous rate. that means one instant it changes by 1, and then the 'next instant' it can't change at all



[edited by - shadow12345 on April 4, 2004 4:41:44 PM]

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quote:
Original post by shadow12345
one could also argue that because it''s an instantaneous rate of change, it shouldn''t be in square units anymore. that''s because an area is really just a bunch of one dimensional ''lines'' or in this case radii, and the smallest possible change in area couldn''t ever be an area if it''s made up of smaller units.

A line, too, is a set of points with an infinite number of elements.

Thinking of an area as a bunch of one dimensional lines seems like a logical trap. For instance, a square with an area of one square unit is infinitely larger than a line with a length of one unit. It''s still exactly the way I think about it, though

quote:
Original post by shadow12345
but think about it, if the instantaneous rate of change of area is the smallest amount that the area can change, then it is impossible to get a unit that is 1/2 the instantaneous rate. that means one instant it changes by 1, and then the ''next instant'' it can''t change at all

Correct me if I''m mistaken, but an instantaneous rate of change gives you information like, "the area increases by four square centimeters for every one second". It''s not the smallest amount of change possible, it''s the unit change. You could multiply by 1/10th of a second to find out how much the area changed during that time. The change that occurs is continuous, it doesn''t advance a whole unit in one instant and then stand totally still the next.

quote:
Original post by Anonymous Poster
The derivative doesn''t give you the change in area, but the RATE of change in area. To get the actual change you''ll have to integrate the rate, which will get you back the the correct unit.

Eh? Integrating would give me my original function, +C. If I want to find the actual change, I multiply by the change in the independent variable, correct?

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Guest Anonymous Poster
quote:
Original post by Zorodius
Eh? Integrating would give me my original function, +C. If I want to find the actual change, I multiply by the change in the independent variable, correct?

No, you take the definite integral of the rate of change. Of course that will include your original function, since the change in area is simply the difference A2 - A1 = π(r2)² - π(r1)².

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Guest Anonymous Poster
Maybe this looks better...

A2 - A1 = pi(r2)² - pi(r1)²

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quote:
Original post by shadow12345
one could also argue that because it''s an instantaneous rate of change, it shouldn''t be in square units anymore. that''s because an area is really just a bunch of one dimensional ''lines'' or in this case radii, and the smallest possible change in area couldn''t ever be an area if it''s made up of smaller units.


Actually, since a true infinitesimally thin ''line'' never ever has any area, it is impossible to build up nonzero area from a bunch, even an infinite number, of ''lines''. An area is a continuous thing, and it is made of an infinite number of "differential" areas. These differential areas are kind of like lines, e.g., they are long and very thin. But, they are not infinitely thin. And the fact that they are not infinite thin means: a) they have nonzero area; and b) therefore, have square units.

The concept of a differential element is fundamental to integral Calculus.

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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well grhodes, I think the problem with calculus is that it takes into account a mode of thinking that actually means that both of us are more or less correct. You're right, it's not actually infinitely thin, it's the limit as the thickness approaches infinitely thin, or as I stated above, the smallest amount the quantity can *possibly* changed.

If you've got some function started at x = 0, and you grab one side of it and start sweeping out the function at one unit per second, the change in area is the height of the function, f(x), multiplied by some change in x, or dx. This itself isn't a rate, the rate of change of area is:

(f(x) * dx) / dx

if you take the limit as dx approaches zero, the dx cancels, leaving you f(x), which is not an area, but it *is* the smallest amount the area can change, which implies that the actual area under the curve must be made up of a whole bunch of these infinitely small f(x)s (I say area because its easy, just assume f(x) is always above the x axis and never goes negative).

so, yeah, takes some getting used to.

EDIT:
and, coming back to what i said earlier, if you sweep an area out at 1/2 unit per second, then that means the rate change of area is:

1/2f(x)

but, as I showed above, f(x) is the smallest amount the area could change (or the limit to (f(x)*dx)/dx, which is the rate of change of area), subsequently one instant there must be no change of area, and the next instant there must be a change of area of f(x), because you can't have a change of area of less than the limit of (f(x) * dx) / dx as dx approaches 0 (or more conveniently f(x)).

(The only reason I kept saying 'the limit of(f(x) * dx) / dx as dx approaches zero' is that as you said, f(x) has no area)

[edited by - shadow12345 on April 9, 2004 8:22:42 PM]

[edited by - shadow12345 on April 9, 2004 8:26:42 PM]

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shadow*,

Good analysis. Actually, in the limit the differential areas are infinitely thin as you point out. It is still possible to have square units. Just happens that there are zero of them. The thickness dimension is zero units times the length dimension that is nonzero = area of zero units squared.

Graham Rhodes
Principal Scientist
Applied Research Associates, Inc.

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cool thanks. I hope to get to some of the more difficult calculus soon. We''re just starting 3D volumes, which is the coolest stuff I''ve ever done in school.

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