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Sage Rat
01-16-2006, 12:59 AM
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.3 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 = 2πr2 out of his butt. Then for the equation where h and r will both vary by 1 inch to create the 400 in.3 difference he uses dV/dr = 3πr2. Neither of these does he show where he is coming to these equations.

V = πr2h 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 = πr2*r, which becomes πr3. Then derived it would be dV/dr = 2πr2--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.

ultrafilter
01-16-2006, 01:53 AM
If V = pr2h and r = h, then V = pr3 and dV/dr = 3pr2. The 2pr2 is probably a misprint.

Sage Rat
01-16-2006, 02:39 AM
which becomes πr3. Then derived it would be dV/dr = 2πr2
3πr2, doh. Alright that was my error there (derivation is of course n * x n-1 not (n - 1) * x n - 1)

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 2πr2 in the book, but plugging in h as a constant gives us:

V = πr2h (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
= 2πr2

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! :cool:

Sage Rat
01-16-2006, 02:45 AM
which becomes πr3. Then derived it would be dV/dr = 2πr2
3πr2, doh. Alright that was my error there (derivation is of course n * x n-1 not (n - 1) * x n - 1)

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 2πr2 in the book, but plugging in h as a constant gives us:

V = πr2h (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
= 2πr2

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! :cool:

Dr_Paprika
01-16-2006, 03:07 AM
I'm not convinced it is a misprint. The author is discussing two separate cases.

V=pi*r^2*h

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 = 2*pi*r^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=pi*r^2*h=pi*r^3

dV/dr= 3*pi*r^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.)