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Thursday, August 23, 2007

Differentials for Dummies

This is (I think) too good to be hidden as comment #3 under “An Uncomfortable Teacher”.

Here’s something I handed my calculus teacher in high school:

Solve: A d2y/dx2 + B dy/dx + C y = 0

Multiplying through by dx2:
A d2y + B dy dx + C y dx2 = 0

This is a quadratic in dx:
dx = (-B dy +/- sqrt(B2 dy2 – 4AC y d2y)) / 2C y

Therefore
dx/dy = (-B +/- sqrt(B2 – 4AC y d2y/dy2)) / 2C y

But d2y/dy2 = d/dy (dy/dy) = 0

dx/dy = (-B +/- sqrt(B2)) / 2C y = (-B +/- B) / 2C y

dx/dy = 0 OR (-B/C) (1/y)

Therefore
EITHER x is constant (highly doubtful)
OR x = (-B/C) log y + K
IN WHICH CASE y = exp((-C/B) (x – K)) = K’ exp((-C/B) x)

Done!

Needless to say, I did not receive a convincing explanation of why this derivation was invalid.

3 comments:

  1. Wabulon,

    I used to think I was a geek.

    ReplyDelete
  2. Anonymous9:00 PM

    My favorite bit is: "EITHER x is constant (highly doubtful)"

    ReplyDelete
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