=========================preview======================
(MATH100)quiz01sol-1a.pdf
Back to MATH100 Login to download
======================================================
MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Quiz 01 Solutions: Vectors, Vector-valued Functions Duration: 40 minutes
Name: ID No.: Tutorial Section: T1A
Answer ALL of the following 4 short answer questions. Show all your work for full credit.
1. (2 Marks) What is the magnitude of the projection of the vector .1, 2, 3. onto the vector .4, 3, 2. ?
Solution: The vector projection of b onto a, denoted projab, is de.ned as
.a b .

projab = a.
a a
Let a = .4, 3, 2. and b = .1, 2, 3., so that the projection of b onto a is

.4, 3, 2..1, 2, 3. 4+6+6 16
.4, 3, 2. = .4, 3, 2. = .4, 3, 2..
.4, 3, 2..4, 3, 2. 16+9+4 29 Hence, the magnitude of the above vector is

.
16 .
16 1616
.4, 3, 2.= 16+9+4= 29= .
29 29 2929
|a b|
In fact the magnitude of the vector projection is simply .projab. =.
.a.
R3
2. (3 Marks) The equations of planes below de.ne a line of intersection L in , x + y =1,x + z =1.
Find an equation which de.nes a plane that is perpendicular to L and passes through the point (1, 1, 1).
R3
Solution: Rewrite the equations that de.ne the line L in in parametric form, as follows: x = t, y = .t, z =1 . t. More re.ned, write them in vector form as .x, y, z. = .t, .t, 1 . t. = .0, 0, 1. + t .1, .1, .1.. Now, this line will be a normal to the required plane. Moreover, the required plane ax + by + cz = d has a normal .a, b, c.. The equation of plane becomes x . y . z = d. The plane also passes through the point (1, 1, 1) so the constant d can be determined by
(1) . (1) . (1) = d =. d = .1.
Hence the equation of plane is
x . y . z = .1.

R3
3. (2 Marks) The position vector to a curve in is given by
.cos t, cos t, 2 sin t..

What is the unit tangent vector at t =?
3

Solution: The tangent vector at t = can be found by evaluating the value of the derivative of each
3

component of the position vector at t =:
3

.. sin , . sin , 2 cos . = .. , . , ..
3 32
33 3222
The magnitude of the above vector is


.
3 32 .
s.3 .2 .3 .2 . 2 .2 r331
.. , . , .= . + . + = ++=2.
222 2 22442
Normalize the above vector by dividing each component by its own magnitude, so the unit tangent vector is

6 61
.. , . , ..
4 42
R3
4. (3 Marks) The coordinates of an object moving in at time t are
2
23
x = t, y = t, z = t.
3
What is the distance traveled by the object between t = 0 and t =3?

Solution: The distance traveled by the object between t = 0 and t = 3 is given by the length of the arc created by the object in from t =0 to t = 3. The length, L, of an arc in from
R3 R3
t = a to t = b is given by
Z b s.dx .2 .dy .2 .dz .2
L = ++ dt.
a dt dtdt
Since
dxdy dy 2
=1, =2t, =2t,
dtdt dt
the distance traveled by the object is given by
Z 3 Z 3 L = p(1)2 + (2t)2 + (2t2) dt = p1+4t2 +4t4 dt 00
Z 3 Z 32
= p(2t2 + 1)2 dt = (2t + 1) dt 00
.2 3 C3 2
= t + t = (27)