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(MATH100)quiz01sol-1c.pdf
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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: T1C
Answer ALL of the following 4 short answer questions. Show all your work for full credit.
1. (3 Marks) Find a vector n perpendicular to the plane P determined by the points A(1, .1, 0),
B(2, 1, .1), and C(.1, 1, 2). Solution: The vector n = AB AC is perpendicular to the plane because it is perpendicular to both vectors
AB = OB . OA = .2, 1, .1...1, .1, 0. = .1, 2, .1.
and
AC = OC . OA = ..1, 1, 2...1, .1, 0. = ..2, 2, 2..
Hence,

.
i jk .
n = AB AC =1 2 .1 =(4+2) i . (2 . 2) j + (2 + 4) k,
.22 2
or

n =6i +6k.
2. (2 Marks) Find the parametric equations of the line tangent to the graph of
r(t)= ti + t 2j
at the point (1, 1) on the curve.

Solution: If r(t)= ti + t2j, then
r .(t)= i +2tj.
At (1, 1), we have t = 1 and so
r .(1) = i +2j.
Therefore a vector equation of the tangent line is given by
r(t)= r0 + tv
= r(1) + t r .(1)
=(i + j)+ t (i +2j)
= (1+ t) i + (1 + 2t) j,
or in parametric form

x =1+ t, y =1+2t.
R3
3. (2 Marks) The coordinates of an object moving through are 1
x = at sin t, y = at cos t, z = bt2 ,
2 for time t> 0, where a, b, and c are constants. What is the speed of the object at any time t ?
Solution: Let r(t) be the position vector that describes the moving object and is de.ned as
1

r(t)= .at sin t, at cos t, bt2..
2
Di.erentiating the above position vector gives the velocity vector
r .(t)= .a sin t + at cos t, a cos t . at sin t, bt..

The speed of the object at any time t is the magnitude of the tangent vector r .(t), which is given by
.r .(t).
= ..a sin t + at cos t, a cos t . at sin t, bt..
= p(a sin t + at cos t)2 +(a cos t . at sin t)2 +(bt)2

2t22
= p(a2 sin2 t +2a2t sin t cos t + acos2 t)+(acos2 t . 2a2t cos t sin t + a2t2 sin2 t)+ b2t2 , or
.r (t). = pa2 + a2t2 + b2t2 .
R3
4. (3 Marks) Let C be the curve in de.ned by the parametric equations
x(t)=2t, y(t) = ln t, z(t)= t 2
fortin [1, 2]. What is the length of C ?

R3
Solution: The length of a curve C in is the integral of the square root of the sum of the squares of the derivatives of each parametric curve over the interval [a, b]. Thus, the derivatives of the above parametric equations are
x .(t) = 2, y .(t) = 1 , z .(t) = 2t.
t
So that the length of curve C, denoted L, is given by
Z 2

L = p[x.(t)]2 +[y.(t)]2 +[z.(t)]2 dt
1
Z 2 r 1
= [2]2 +[ ]2 + [2t]2 dt
t
1
= Z 2 1 r 4 + 1 t2 + 4t2 dt
= Z 2 s.2t + 1 .2 dt = Z 2 .2t + 1 . dt
1 t 1 t

or
2
= .t + ln t.2 = (4+ln2) . (1 + ln1)
1
L = 3+ln2.