Mat 272 Calculus III
Updated on 2/5/08
Dr. Firoz
Solution: The parametric form of the curve is
cos ,
sin ,
x
t y
t z
t
=
=
=
, which
represents a curve that spirals around a circular cylinder with level curves
2
2
1
x
y
+
=
.
The curve is known as circular helix. See the figure at page # 851 in you text.
5.
Find the vector function that represents the curve of intersection of the cylinder
2
2
1
x
y
+
=
and the plane
2
y
z
+
=
Solution: Consider
cos ,
sin
x
t y
t
=
=
, then
2
sin
z
t
=

, 0
2
t
π
≤
≤
.
Now ( )
cos ,sin ,2
sin
cos
sin
(2
sin )
r t
t
t
t
i
t
j
t
k
t
=<

>=
+
+

6.
Find the vector equation of a line thru the points P(1, 0, 1) and Q(2, 3, 1)
Solution: Use the formula thru two given points as
0
1
( )
(1
)
, 0
1
r t
t r
tr
t
=

+
≤
≤
We have ( )
(1
)
1,0,1
2,3,1
, 0
1
r t
t
t
t
=

<
> + <
>
≤
≤
7.
Identify the curve ( )
cos10 ,
sin10 ,
t
t
t
r t
e
t e
t e



=<
>
The curve is a spiral around a cone whose level curves are circles
2
2
t
x
y
e

+
=
.
Observe that
2
2
2
x
y
z
+
=
Section 13.2 Derivatives and Integrals of Vector Functions
For given vector function
( )
( ),
( ), ( )
( )
( )
( )
r t
f t
g t
h t
f t i
g t j
h t k
=<
>=
+
+
The derivative is defined as
( )
( ),
( ),
( )
( )
( )
( )
r t
f
t
g t
h t
f
t i
g t j
h t k
′
′
′
′
′
′
′
=<
>=
+
+
and the
integral is defined as
( )
( )
,
( )
,
( )
r t dt
f t dt
g t dt
h t dt
=<
>
∫
∫
∫
∫
Smooth curve
: A curve given by a vector function
( )
r t
on an interval I is called smooth
if
( )
r t
′
is continuous and
( )
0
r t
′
≠
, where 0 is a zero vector.
Examples:
1. Given
3
1
( )
1
,
,sin
1
r t
t
t
t
=
+

, find
( )
r t
′
and
( )
r t dt
∫
Solution:
2/3
2
1
1
( )
,
,cos
3
(
1)
r t
t
t
t

′
=


and
4/3
1
2
3
3
( )
,ln(
1)
,
cos
4
r t dt
t
t
c
t
c
t
c
=
+
+

+

+
∫
2.
Find a unit tangent vector at (1, 0, 0) to the vector
3
( )
1
,
,sin 2
t
r t
t
te
t

=<
+
>