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(math102)[2005](s)final~936^_10421.pdf
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HKUST MATH 102
Final Examination

Multivariable Calculus
Answer ALL 8 questions
Time allowed C 3 hours
Directions C This is a closed book examination. No talking or whispering are allowed. Work must be shown to receive points. An answer alone is not enough. Please write neatly. Answers which are illegible for the grader cannot be given credit.
Note that you can work on both sides of the paper and do not detach pages from this exam packet or unstaple the packet.
Student Name:
Student Number:
Tutorial Session:
Question No.
Marks 1
/12 2
/12 3
/12 4
/14 5
/12 6
/12 7
/14 8
/12
Total
/100
Problem 1 (12 points) Your Score:
3
(a)
If f (x, y)=(x+ y2) 31 , .nd fx(0, 0) and fy(0, 0).

(b)
Let z = f (x, y), where x = g(t) and y = h(t).


(i) Show that
d ..z . .2zdx .2zdy d ..z . .2zdx .2z dy
=+ and =+ .
dt.x .x2 dt .y.x dt dt .y .x.y dt .y2 dt
d2z
(ii) Use the formulas in part (i) to help .nd a formula for .
dt2
Solution:
y
0

(a)
-0.4

-0.2

0.4

1 0.2
3
0.2

f (0 + h, 0) . f (0, 0) (h3 + 0) . 0 h 0.4

x

0

fx(0, 0)= lim =lim =lim =1.
h.0 h h.0 h h.0 h 0.5
-0.2
-0.4

3
f (0, 0+ h) . f (0, 0) (0+ h2) 1 . 0 z0
fy(0, 0)= lim =lim =lim h.1/3 ...
h.0 h h.0 h h.0 -0.5
Hence, fy(0, 0) does not exist.
(b) (i) Since z = f (x, y) zx or zy are also functions of x and y. Therefore, by Chain rule: d ..z . . dx . dy.2zdx .2z dy
=(zx)+(zx)= + ,
dt.x .x dt.y dt .x2 dt .y.x dt z, z x, zy
and
y
d ..z . . dx. dy .2zdx .2z dy x
=(zy )+(zy)= + .
dt .y .x dt .y dt .x.y dt .y2 dt
tt
(ii)
dz .z dx .z dy
=+
dt .x dt .y dt d2z .z d2x dxd ..z . .z d2y dyd ..z .
=+ ++
dt2 .x dt2 dtdt .x .y dt2 dtdt .y .z d2x.2z .dx .2 .2z dxdy .z d2y.2z .dy .2
=+ +2 ++ .
.x dt2 .x2 dt .y.x dtdt .y dt2 .y2 dt
C1C
(b)
(a) Find the equation of the tangent plane at the point (.1, 1, 0) to the surface y
x 2 . 2y 2 + z 3 = .e .z .
C 4 y=-2x/3+10/3
(b) Find the absolute extrema of the function 3
z = f (x, y) = xy . 5 3 x . 3y on the closed and bounded set R, where R is the triangular region with vertices (.1, 0), (.1, 4), x A B x=-1 -1 1 3 542 2 1 Rfx = y . 5 3 fy = x . 3
and (5, 0). y=0fxy = 1 = fyx
Solution: fxx = 0 = fyy
D = fxxfyy . (fxy )2 = 0 . 1 = .1.
(a) Let F (x, y, z) = x2 . 2y2 + z3 + e.z = 0. This is a level surface and F = (2x, .4y, 3z 2 . e .z ). For critical points, f = (fx, fy) = (0, 0) x = 3, y = 5/3. This point is outside the domain R (because when x = 3, y = . 2 3 x + 10 3 = 4 3 < 5 3 .)
At (.1, 1, 0), F = (.2, .4, .1). This vector is normal to the level surface at the point (.1, 1, 0). On the boundary of domain R
Hence, the required tangent plane is Along AB, y = 0 for .1 . x . 5, and u1(x) = f (x, 0) = .5x/3; (no critical points).
(x + 1, y . 1, z . 0) (.2, .4, .1) = 0 .2(x + 1) . 4(y . 1) . z = 0 Along AC, x = .1 for 0 . y . 4, and u2(y) = f (.1, y) = 5/3 . 4y; (no