I’m teaching myself the calculus from Calculus Made Easy (1998 ed.) I’ve only just begun, and have been thrown off by a bit of simple math on page 63. Thompson shows us two examples of determining for what length of h and r where h=r, a one inch difference in size will cause a 400 in.[sup]3[/sup] difference in volume of a cylinder.
The first example is where h is equal to r initially but remains a constant (i.e. the one inch variation will only be used on r.) To show this he pulls the equation dV/dr = 2πr[sup]2[/sup] out of his butt. Then for the equation where h and r will both vary by 1 inch to create the 400 in.[sup]3[/sup] difference he uses dV/dr = 3πr[sup]2[/sup]. Neither of these does he show where he is coming to these equations.
V = πr[sup]2[/sup]h is the original equation which he would have started from to create the derivation. If h = r and varies with it, I would think this would be V = πr[sup]2[/sup]*r, which becomes πr[sup]3[/sup]. Then derived it would be dV/dr = 2πr[sup]2[/sup]–but that’s the case where h is supposed to remain as a constant, while as I had expected it to be the case where h varies.
If V = [symbol]p[/symbol]r[sup]2[/sup]h and r = h, then V = [symbol]p[/symbol]r[sup]3[/sup] and dV/dr = 3[symbol]p[/symbol]r[sup]2[/sup]. The 2[symbol]p[/symbol]r[sup]2[/sup] is probably a misprint.
3πr[sup]2[/sup], doh. Alright that was my error there (derivation is of course n * x [sup]n-1[/sup] not (n - 1) * x [sup]n - 1[/sup])
So then I understand it if h=r and h varies with r, but how does he come to the first equation where h = r at it’s initial value but doesn’t vary with r over an inch? This one is 2πr[sup]2[/sup] in the book, but plugging in h as a constant gives us:
V = πr[sup]2[/sup]h (where h is constant)
dV/dr = 2hπr
…ah. Then we go that h = r at this point (i.e. after the derivation) and multiply through.
= 2rπr
= 2πr[sup]2[/sup]
Cooool. Alright well I guess this is the problem of trying to teach yourself on the train with nothing to write on. But thanks much for the response! 
3πr[sup]2[/sup], doh. Alright that was my error there (derivation is of course n * x [sup]n-1[/sup] not (n - 1) * x [sup]n - 1[/sup])
So then I understand it if h=r and h varies with r, but how does he come to the first equation where h = r at it’s initial value but doesn’t vary with r over an inch? This one is 2πr[sup]2[/sup] in the book, but plugging in h as a constant gives us:
V = πr[sup]2[/sup]h (where h is constant)
dV/dr = 2hπr
…ah. Then we go that h = r at this point (i.e. after the derivation) and multiply through.
= 2rπr
= 2πr[sup]2[/sup]
Cooool. Alright well I guess this is the problem of trying to teach yourself on the train with nothing to write on. But thanks much for the response!
I’m not convinced it is a misprint. The author is discussing two separate cases.
V=pir^2h
In case one, h is a constant. In this case,
V=(pi*h)*r^2
dV/dr = (pi*h) 2r
If h=r, then dV/dr = 2pir^2
In the other case, h and r both vary. This is where you could use the h=r in the original equation to get
V=pir^2h=pi*r^3
dV/dr= 3pir^2.
(It probably isn’t worth discussing what happens if h is a function of r, partial derivatives and the chain rule in for an introductory example.)