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# 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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Makes sense, thanks!

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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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I would have also said that if you take dA / dr = 2πr and rearrange it to dA = 2πrdr, you also get the right units.

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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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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 = &#960;(r2)² - &#960;(r1)².

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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.