From barnett@math1.cims.nyu.edu Mon Jan 27 11:33:06 2003
Date: Mon, 27 Jan 2003 11:29:37 -0500 (EST)
From: Alexander Barnett <barnett@math1.cims.nyu.edu>
Reply-To: barnett@cims.nyu.edu
Subject: Calc III: corrections to today


Hi - 2 little corrections to today's lecture:

i) I compared change of variable dh/dx = (1/x') dh/dt to like going from
km/hr to miles/hr. A more correct analogy is converting from km/minute
to km/hr. (The denominator changes units). Of course the conversion
factor, x', can vary along the curve, unlike hours/minute which is always
60.

ii) In the final surface of revolution example, the catenary
	x = sinh^{-1}(t)
	y = sqrt(1 + t^2)
has y-value y_0 = cosh(1) at the "grabbing points" x_0 = +-1.
I had mistakenly said sqrt(2).
Remember that then you are shifting the curve down by y_0, so that you can
revolve it about the usual x-axis. This gives y = sqrt(1 + t^2) - y_0.
To make positive you then take the absolute value, giving
	y = y_0 - sqrt(1 + t^2),
which you then use in the regular formula.
You should get the final answer for area A = 2.pi.cosh(1).sinh(1).

Do let me know any future slips if you can spot them.
Thanks,
		Alex

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