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(MATH100)midterm-2001.pdf
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MATH 100 Introduction to Multivariable Calculus Mid-term Test
March 26, 2001
Total Time: 2 Hours
1.
(20 marks) Show that the intersection of the surfaces z = x 2 + y 2 and y = x 2 + z 2

is a plane curve (i.e., the curve is completely located on a plane). What kind curve is it?

2.
(20 marks) Determine whether the limit exists. If so, .nd its value. x2 . y2 x3 + y3


(a) lim (b) lim
(x,y)(0,0) x2 + y2(x,y)(0,0) x2 + y2
3. (20 marks) Let f(t) be a di.erentiable function. A parametric system, with the parameter t, is given by
. x cos t + y sin t + ln z = f(t), .x sin t + y cos t = f.(t).
This system de.nes a function z = z(x, y). Show that
..z .2 ..z .22
+= z.
.x .y
4. (20 marks) Let f(t) be a di.erentiable function, with f.(t) .= 0 for any
t. Consider the surface
z = fx2 + y2.
(a) Find the normal line equation of the surface at (x0,y0,z0), with (x0,y0) .
= (0, 0).
(b) Show that the normal line intersects to the z-axis and .nd the intersection point.
5. (20 marks) Show that the function
z = (1+ ey) cos x . yey
has no local minimum.

Solutions
1. The intersection satis.es
22
. z = x+ y, 2 + z2
y = x. The di.erence of these two equations gives (z . y)(z + y +1) = 0. We claim that for the intersection, we must have z + y +1 .
=0. In fact, if z + y + 1 = 0, the .rst equation implies .y . 1= x 2 + y 2 , or
. 1.2 3
x 2 + y += . ,
24
a contradiction. Thus, we have z = y, which is a slanted plane. In other words, the intersection is completely located on the plane z = y. Now the intersection is
. z = y, y = x2 + y2 ,
or
.z = y,
.
.
2 . 1.2 .1.2
x + y . = .
.
. 22
This shows that the intersection is made by a slanted plane and a circular cylinder. So the intersection is an ellipse.
2
2. (a) For any .xed constant k, consider the straight line y = kx that passes the origin. Since along the straight line (with x .
= 0),
x2 . y2 1 . k2
= .
x2 + y2 1+ k2
Thus, along two di.erent straight lines
y = k1x, y = k2x,
the function
2
x2 . y
x2 + y2
approaches di.erent values. Hence the limit does not exist.

(b) Take x = r cos , y = r sin , we have
x3 + y3 r3 cos3 + r3 sin3
lim = lim
22
(x,y)(0,0) x2 + yr0+ r
= lim r (cos3 + sin3 )
r0+
=0.
Here we use the fact that cos3 + sin3 is bounded:
.
cos 3 + sin3 .
.
cos .
3 + .
sin .
3 2.
3. The second equation of the parametric system indicates that t is an implicit function of x and y. Thus, if we di.erentiate the .rst equation, with respect to x and y, respectively, we have
.t .t 1 .z .t
cos t + x(. sin t) + y cos t + = f.(t) ,
.x .x z.x .x
and
x(. sin t) .t .y + sin t + y cos t .t .y + 1 z .z .y = f.(t) .t .y .

By the second equation in the parametric system, the equations above can be simpli.ed to
1 .z
cos t + =0,
z .x
and
1 .z
sin t + =0.
z .y
Thus,
.2 .2
..z ..z
+ =(