Download Misteaks. And How to Find Them Before the Teacher Does by Barry Cipra PDF

By Barry Cipra

This strange, yet very helpful, ''how-not-to'' e-book comes in a moment variation which gets rid of so much involuntary mistakes.

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No one would ever make such mistakes, right? ) But how about the following integral: A simple change of variables (which unfortunately we wrote in a miserable, unreadable scrawl) changes this integral into 3 u u J u e ^ = Jue du = (u- l)e u + C (integrating by parts for the last equality). Now what was t h a t stupid _ 1 x change of variables? Well, since we started with e / , we should wind x x u p with e~ l '. So t h a t ' s what u must be: u = — 1/x. W h a t did you expect? Thus we get the "answer" (Unfortunately, this answer is also wrong: If you stick in definite limits, like 1 and oo, you get f h'-'""* - (τ - 0 «-"ΙΓ - <-*-° - H * " ' - ; - which is negative, whereas the original function is positive.

The mistakes you make will be as unique as your fingerprints or your handwriting (unless you're copying from someone else's paper; I once had the dubious pleasure of nailing a cheater who had copied verbatim from another student's test paper, including a mistake so distinctive t h a t the odds against two people making it independently were surpassed only by the odds against the cheater passing the course on his own). Nevertheless, there are errors t h a t are common enough t h a t , perhaps by pointing t h e m out here, we can take steps toward their control.

F(x) = e^ c o sx = • f'(x) = -^ei/cosx z COS X (Hint: Look at small values of x. ) 3. Point out the errors in the a t t e m p t s to solve the following max-min problem: A cylinder is to be built to hold 1000 cubic inches of highly compressed pedagogical hot air. W h a t is the smallest possible surface area (top, bottom, and shaft) for such a cylinder? h 2 2 2 2 a. V = nr h, A — nr + 2nrh = nr + 2/r, dA/dr = 2πτ-2ν = 0, 3 2 γ ( τ γ _ r ) — ο , r = π, A = π + 2 / π . ) 3 2 2 2 b. rr = 0, 4nr(l + 1500r) = 0, r = - 1 / 1 5 0 0 , 2 h = ΙΟΟΟπτ· = ΙΟΟΟπ/2250000 = π/2250, A = 2π/225000 2 2 π / ( 1 5 0 0 χ 2250).

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