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(MATH100)math100_09S_mid.pdf
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Math100 Introduction to Multivariable Calculus
Spring 2009
C Midterm Examination C

Name:

Student ID:


Lecture Section:

.
There are FIVE questions in this midterm examination.

.
Answer all questions.

.
You may write on both sides of the paper if necessary.

.
You may use a HKEA approved calculator. Calculators with symbolic calculus functions are not allowed.

.
The full mark is 100.


1. The top of a pyramid is cut o. by a plane. The vertices of the truncated pyramid are:

A:(6,6,0), B:(. 6,6,0), C:(. 6,. 6,0), D:(6,. 6,0),
E:(3,3,6), F:(. 3,3,6), G:(. 2,. 2,8), H:(2,. 2,8).


(a) Find the area of the triangle EFG. [6 points]
Solution

.
.

The area of . EFG is half of the area of the parallelogram generated by EF and EG.So, 1.
.
. 1.
area(. EFG)= EF EG = [(. 3,3,6) . (3,3,6)] [(. 2,. 2,8) . (3,3,6)]
22 1. 1 .
= (. 6,0,0) (. 5,. 5,2) = (0,12,30)
22
= . (0,6,15). = 261=3 29

(b) Find the equation of the plane passing through the vertices E, F, G, H. [6 points]
Solution
.
.


A normal vector of the plane is EF EG =(0,12,30). A linear equation representing the plane is: (0,12,30) (x . 3,y . 3,z . 6) = 0 .. 2y +5z =36
(c) Write a vector parametric equation of the line passing through A and E, and also a vector parametric equation of the line passing through B and F. Using your vector parametric equations to .nd the missing top vertex of the original pyramid. [8 points]
Solution

Vector parametric equation for the line passing through A, E: [2 pts]
.r
=(6,6,0) + t[(3,3,6) . (6,6,0)] = (6,6,0) + t(. 3,. 3,6) Vector parametric equation for the line passing through B, F: [2 pts]

.r
=(. 6,6,0) + t[(. 3,3,6) . (. 6,6,0)] =(. 6,6,0) + t(3,. 3,6) To .nd their intersection point, consider (6,6,0) + t(. 3,. 3,6) = (. 6,6,0) + s(3,. 3,6) .. (6 . 3t,6 . 3t,6t)=(. 6+3s,6 . 3s,6s)


. . 3t +3s =12
.
3t . 3s =0 .. s = t =2
.
.

6t =6s i.e., the missing vertex is (0,0,12). [4 pts]
2. The velocity vector of a moving particle at time t 0isgiven by d.r =(3t2 , 10t, 4t2). At t =3,
dt

the particle is at the point (2, . 15, 18).
d2.r

(a) Find the acceleration vector of the moving particle. [4 points]
dt2

Solution
d2.r

=(6t, 10, 8t)
dt2

(b) Find the position vector .r(t) of the particle. [8 points]
Solution . .
.r(t)= d.r dt =3t2dt, 10tdt, 4t2dt

dt
4
32 3
= t, 5t, tC

3
Putting t =3, we have

(2, . 15, 18) = (27, 45, 36) + .C =(. 25, . 60, . 18)
C .. .
4
.r(t)= t3 . 25, 5t2 . 60,t3 . 18
3

(c) Find the arc length of the curve traced out by the particle during the time interval 0 t 6. [8 points]
Solution

66 . 6 . arc length = ..r (t). dt =9t4 + 100t2 +16t4dt = 25t4 + 100t2dt 00 0
..
6 .. 6
5 5 400 10 . 40
=5tt2 +4dt =(t2 +4)3/2 = (80 10 . 8) =
333
00


3. The equation x2 +2xy3 . 112 = 0 de.nes y as a function of x implicitly. Evaluate dy