Reason: Set and
, that is
Solving this linear system, we get a unique solution x=2, y=-1, which gives us the critical point (2,-1).
Reason: Let (a,b, c) be the point on the paraboloid closest
to (3,0,0). The square of the distance
from a point (x,y,z) on the paraboloid to the point (3,0,0)
is given by .
So (a,b) is a point at which the function
is minimized. We compute ,
and set them equal to zero to obtain
From the second equation we have . But
is
always positive. So y=0. Substituting y=0 in the first equation, we get
. The only real root of this equation is x=1. Substituting
x=1 and y=0 into
, we have z=1. Thus (1,0,1) is the
point on the paraboloid closest to (3,0,0). The distance between these
points is
.
Let be the polar coordinates. Then
is (e) 0, and
is (a) 1.
Reason: Since and
, we have
,
,
and
. In terms of polar coordinates, we have
and
.
So, by the chain rule, we have
Reason: ,
which is (6, -2, -8) at (1,1,2). Also, a unit vector in the direction of
is given by u where
So,
( <-- Actually, the answer is 6 over the square root of 14).
(c) is a normal vector to S at P.
A comment on (e): we may correct
this statement by saying that (not
) is in
the direction of the steepest descent.
Reason: The normal direction of the plane is given by the
vector at P, that is (4,-1, -4). Thus the
equation of the tangent plane is
Proof. Taking partial derivatives on both sides of F(f(x,y,z), g(x, y,z))=0 and applying the chain rule, we have
We can rewrite the above three identities in the vector form as
. This shows that
and
are linearly dependent, since
tells us that at least one of
,
is nonzero at each point.