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(MATH102)[2006](f)midterm~ma_lyf^_10422.pdf
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HKUST MATH 102
Midterm Examination

Multivariable and Vector Calculus
6 Nov 2006
Answer ALL 5 questions
Time allowed C 120 minutes
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 /20
2 /20
3 /20
4 /20
5 /20
Total /100


Problem 1 (20 points) Your Score:
(b) (i)

(a) Assume a, b and c are three dimensional vectors and if
(a b) c = .a + b + .c.

Use su.x notation to .nd ., and . in terms of the vectors a, b and c. Can you say something about the direction of the vector (a b) c.
(b) (i) Find the distance (in terms of n, r0 and r1 only) from the point r1 to the plane (r . r0 ) n = 0.
(ii) Use (i) or otherwise, .nd the distance d between the two parallel planes determined by the equations Ax + By + Cz = D1 and Ax + By + Cz = D2 .
(iii) Use (ii) or otherwise, .nd equations for the planes that are parallel to x +3y . 5z =2 and lie three units from it.
Solution:
(a)
[(a b) c]i = ijk(a b)j ck
= ijkjpq ap bq ck
= jkijpq ap bq ck
=(kp iq . kq ip)ap bq ck

= ak bick . aibk ck

i.e. (a b) c =(a c)b . (b c)a, i.e. = a c, . = .b c and . = 0.
The resulting vector of (a b) c is a linear combination of the vectors a and b, hence it lies on the plane containing the vectors a and b.
d = r1 . r0 |cos |
= r1 . r0 n.|cos |
= |(r1 . r0 ) n.|

(ii) First locate a point on P1 , i.e. we can let y = z = 0, then x = D1 /A, i.e. r1 =(D1 /A, 0, 0) D1 . D2
is a point on P1 . Similarly, r2 =(D2 /A, 0, 0) is a point on P2 . Then r1 . r2 = i
A (A, B, C)
and n.= . Therefore, from (i),
(A2 + B2 + C2)1/2
|A (D1 . D2 )/A|
d =
(A2 + B2 + C2)1/2 |D1 . D2 |
= .
A2 + B2 + C2

(iii) Use (ii), here A = 1, B = 3, C = .5 and D1 = 2. The above equation becomes:
|2 . D2 ||2 . D2 |
3= = .
.12 +32 +(.5)2 35
So

3 35= |2 . D2 | or 2 . D2 = 3 35.

So D2 =2 3 35 and the equation of the two planes are:
x +3y . 5z =2 3 35.

Problem 3 (20 points) Your Score:
2

.. x2 . y. (i) Let z = f (x, y)= .. |x|.|y| .. .|x|.|y|, use the fundamental theorem of partial di.erentiation
. xy if (x, y) = (0., 0)

Let f (x, y)= x2 + y2 to .nd fx(0, 0) and fy (0, 0).
. 0 if(x, y) = (0, 0).

(ii) Is the function f (x, y) in (i) di.erentiable at (0, 0). Why?
(i)
Find fx(x, y) and fy (x, y) for (x, y) = (0., 0).

(ii)
Find the partial derivatives fx(0,y) and fy (x, 0). Hint: You may use the following theorem to do part (ii), the function f (x, y) is di.erentiable at the point