MATH 3251, Midterm 3, Fall 1996, Solutions of problems
November 21, 1996.
1. One can compute it directly.
Hence 
.
There also exists a shorter way.
The function 
satisfies
which implies
2. This is a particular case of problem 52 in Sec. 12.5. By implicit differentiation, we get
3. From the reminder the tangent planes in question are
and
A vector parallel to the line of their intersection we find as
. Thus the answer is
4. By differentiating the equation we get
which at points where 
, yields 2x-y=0.
Thus at any point on the curve at which , we
have y=2x and 
. Therefore, at these points
The corresponding values of y are 0 and 
.
Thus we got two points
(0,0) and (1/8,1/4). Since we are not interested in (0,),
the answer is (1/8,1/4).
By the way, one may notice that for 
,
we have 
at the origin,
so that the implicit function theorem
is not applicable, and we cannot speak about 
for
(0,0) without additional investigations.
An appropriate investigation shows that when x is close to zero,
there are three continuous
functions y=y(x) such that y(0)=0 and
. Two of them are not differentiable
at zero, so that y'(0) is not defined let alone the equality
y'(0)=0. For the third one, the derivative at zero is one, again
not zero.