From zeilberg@math.rutgers.edu  Mon Nov 27 18:54:41 2006
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Date: Mon, 27 Nov 2006 18:54:19 -0500 (EST)
From: Doron Zeilberger <zeilberg@math.rutgers.edu>
Message-Id: <200611272354.kARNsJS09915@math.rutgers.edu>
To: zeilberg@math.rutgers.edu
Cc: lpudwell@math.rutgers.edu, scottsch@math.rutgers.edu
Subject: MathIsFun; Today's quiz and other stuff
Status: RO
Content-Length: 2570


Dear Students,

   1. A corrected version of the handout can be downloaded from

 http://www.math.rutgers.edu/~zeilberg/calc3/ho166.pdf

where all the typos are corrected (I hope!) plus I follow
my own advise to "PLUG-IT-IN AS SOON AS POSSIBLE".
(In the previous version I first computed r_u x r_v for
general u and v which you need to do for questions about
Surface Area (and coming up shortly, Surface-Integrals)
but definitely NOT for questions about TANGENT PLANEs,
where the focus is on ONE (specific) POINT.

 2. For those who scored less than 70 on Exam 2, the
"deal" has been amended, see

 http://www.math.rutgers.edu/~zeilberg/calc3/MakeUp1.html
 
 3. I have already posted the COMPLETE solutions for Exam 2 for
   sections 1-3
 http://www.math.rutgers.edu/~zeilberg/calc3/mt2aSol.pdf

In a few days, I hope to post those for sections 4-6

  http://www.math.rutgers.edu/~zeilberg/calc3/mt2bSol.pdf

4. Regarding today's "quiz" almost everyone got the first
one right, and almost everyone got a right, but impolite
answer for the second: 

  -4x=0

The right and polite answer is x=0 (see below).


--------Answers to the "quiz" of Nov. 27, 2006----------

1. Set-up, but do not evaluate, an integral for 
the surface area of the surface z=xy above
the region R in the xy-plane , where

R={ (x,y) | 0 <= x <= 1  , 0 <= y <= x^3 }


Solution to 1. f(x,y)=xy. So f_x=y, f_y=x , and 

Surface Area= 
IntegralSign IntegralSign_R sqrt(1+(f_x)^2+(f_y)^2) dA
=
IntegralSign IntegralSign_R sqrt(1+y^2+x^2) dA

Converting the double-integral to an ITERATED integral, we get

Surface Area= 

IntegralSign_0^1 IntegralSign_0^{x^3} sqrt(1+y^2+x^2)  dy dx

This is the answer.

2. Find an equation for the tangent plane to the given
parametric surface  at the specified point.

x=u^2 , y=2uv z=3u-2v \;  u=0 , v=1;

Solution to 2.


r=<u^2,2uv,3u-2v>

The POINT is

r(0,1)=<0^2,2(0)(1),3(0)-2>=<0,0,-2>=
So the POINT is (0,0,-2).

r_u=<2u,2v,3> 
r_v=<0,2u,-2>

Plugging-in the SPECIFIC u=0 v=0 we get, that at OUR POINT

r_u(0,1)=<0,2,3>

r_v(0,1)=<0,0,-2>

              | i   j   k |
r_u x r_v =   | 0   2   3 |  = i(-4)-j(0)+k(0) = -4i
              | 0   0  -2 |

So N=<a.b,c>=<-4,0,0>

Using a(x-x0)+b(y-y0)+c(z-z0)=0 we get

(-4)(x-0)+ 0(y-0)+ 0(z--2)=0

which simplifies to -4x=0

WHICH SIMPLIFIES TO (BE POLITE!)

x=0 .

Ans.: the tangent plane is x=0 (alias the yz-plane).


Note: Most people got a RIGHT (but IMPOLITE) answer: -4x=0

It is good manners to perform TRIVIAL SIMPLIFICATIONS 
(even w/o a calculator). I hope that you all know
that 0/(-4)= 0.



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