By Barry Cipra
This strange, yet very priceless, ''how-not-to'' booklet comes in a moment version which removes so much involuntary mistakes.
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Sample text
Now this might seem mildly contradictory (as if any contradiction is ever mild). For example, the integral 7 13 sin lOx cos 3x dx = - cos 7x — — cos 13x 4- C makes perfectly good sense (in spite of being incorrect) in spite of the fact t h a t you're taking cosines of x. All this means is t h a t χ cannot be assigned a dimension in this problem. At least not as it stands. But how did we get this integral? From a formula, of course: / τ ι a—b sin ax cos bx dx = —-— cos(a — b)x a+b — cos(a + b)x + G.
2 [a] V = ^h Rr ^ 2 + R - r + r) 2 + Rr + r ) [d] V = ^nh{R [e] V = -Kh(R ό 2 2 We do so by noting t h a t all except [e] violate the accepted dimensions 3 of volumes: length χ length χ length = (length) . T h e variables R, r, and h are all lengths. But then answer [a] has 2 4 dimensions (length) χ length χ length = (length) ; t h a t can't be. Like35 2 wise for answers [b] and [c], which have dimensions length /(length) = -1 2 2 (length) and length χ (length) /length = (length) = area, respectively.
I would probably still get close to full credit for having written down the equation (***). One more comment about this problem. Replacing the d a t a 3 and 9π with symbols k and Vb allows us to see beyond the immediate problem. For instance, from (***) we can see t h a t , if the rate k is increased, the rate dr/dt is increased. This of course makes a lot of sense, and helps convince us t h a t (***) is the correct formula. Is there anything interesting we can do with Vo? Yes: We can set it equal to zero!