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(MATH100)worksheet-ch2-sol.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Class Worksheet Chapter 2 Vector-Valued Functions
1. (Vectors) Find vectors that satisfy the stated conditions : (a) Same direction as v = .2i +3j and three times the length of v. (b) Length 2 and oppositely directed to v = .3i +4j + k.
Solution
(a)
A vector, in the same direction of v and three times the length of v, is simply
3v = .6i +9j.
(b)
A unit vector in the direction of v is
v .3i +4j + k 341
= = . i + j + k.
.v. 9+16+1 26 26 26
Hence the required vector is given by
.v . 682
2 . = i . j . k.
.v. 26 26 26
2. (Vectors) Let u =2i . j and v =4i +2j. Find the scalars c1 and c2 such that c1u + c2v = .4j.
Solution In order to .nd the scalars c1 and c2 such that c1u + c2v = .4j, we write
c1u + c2v = (2c1 +4c2) i +(.c1 +2c2) j
= .4j.
Thus, we obtain a simple system of two linear equations in two unknowns c1 and c2,
( 2c1 +4c2 =0 , .c1 +2c2 = .4 .
After solving, we have
c1 = 2 and c2 = .1.
3. (Vectors) Find two unit vectors in 2-space parallel to the line y =3x + 2.
Solution By direct observation the points P1 = (0, 2) and P2 =(. 23 , 0) lie on the line y =3x + 2. The vectors ended with these two points are
.. 2 .. 2
P1P2 = .. , .2. and P2P1 = . , 2..
33
The corresponding lengths are the same:
.. .. r4 r40 2
.P1P2. = .P2P1. = +4= = 10.
9 93
After normalizations, we get two unit vectors parallel to the line y =3x + 2. They are
.. ..
P1P2 13 P2P1 13
= .. , . . and = . , ..
.. ..
||P1P2|| 10 10 ||P2P1|| 10 10
4. (Vector function) Find the intersection point between the line r(t)= t i + (1 + 2t) j . 3t k and the plane 3x . y . z = 2.
Solution The line in space which has the vector form: r(t)= x(t) i + y(t) j + z(t) k, where
x(t)= t, y(t)=1+2t, z(t)= .3t
intersects the plane 3x . y . z = 2 when
3
3(t) . (1 + 2t) . (.3t)=2 or t = .
4
The intersection point is therefore
35 9
( ,, . ).
42 4
5. (Vector function) Find r(t) given that r .(t)= t2 i +2t j and r(0) = i + j. Solution Integrating r .(t) to obtain r(t) gives
3
Z .Z 2 t2
r(t)= r (t) dt =(t i +2t j) dt = i + t j + c0.
3
To .nd the constant vector c0 we substitute t = 0 in the above and use the given value of r(0). We obtain i + j = r(0) = 0+0+ c0 so that c0 = i + j. Thus 3
t
r(t)=( +1) i +(t 2 + 1) j.
3
6. (Graph of vector function) Sketch the graph of
r(t)= t i + t j + sin t k
and show the direction of increasing t.
Solution We can write the vector function as r(t)= x(t) i + y(t) j + z(t) k, where x(t)= t, y(t)= t, and z(t) = sin t. After eliminating the variable t, we know that the graph is indeed the intersection curve between the surfaces z = sin x and z = sin y in 3-space.
z
z
z = sin x
z = sin y
y
xy
The intersection curve of the above surfaces is sketched as follows, where the arrows indicate the direction of increasing t.
z
y