9.1 Vector valued functions or vector functions
Parametric Curve
Space Curve
For interesting animations of space curves go to:
http://math.bu.edu/people/paul/225/fall05/class6.html
9.1
Graph the curve traced by the given vector function.
r ( t ) = 4, 2 cos ( t ) , 2 sin ( t )
> spacecurve([4, 2*cos(t), 3*sin(t)], t=0..10, thickness=3, color=black);
z
x
y
Graph the curve traced by the given vector function.
r ( t ) = t cos ( t ) , t sin ( t ) , t 2
> spacecurve([t*cos(t), t*sin(t), t^2], t=0..10, thickness=3, color=black);
9.1
Find parametric equations of the tangent line to the given curve
at the indicated value of t .
r (t ) = t 3 − t,
6t
2
, ( 2t + 1)
t +1
r ′ ( t ) = 3t 2 − 1,
r ′ ( t ) = 3t 2 − 1,
r ′ (1) = 3 − 1,
6 ( t + 1) − 6t
( t + 1)
6
( t + 1)
2
2
t =1
, 2 ( 2t + 1) ⋅ 2
, 4 ( 2t + 1)
6
3
, 4 ( 3) = 2, ,12
2
( 2)
2
This is a direction vector for the tangent line, we need a pt. on the line.
r (1) will give the point of tangency.
r (1) = 0,3, 9 so the point is ( 0,3,9 )
pt.
dir
x
=
0
+
y
=
3
+
z
=
9
+
2t
3
t
2
12t
9.1
Evaluate the given integral.
Find the length of the curve traced
4
by the given vector function on the
indicated interval.
∫(
)
2t + 1 i − t j + sin (π t ) k dt
0
4
∫
4
2t + 1 dt i − ∫ t dt j +
0
0
4
4
4
∫ sin (π t ) dt k
r ( t ) = 2t , ln t , t 2
0
Arc Length → s = ∫ r ′ ( t ) dt
=
1 32
2 3
1
9 − 1 , − 4 2 − 0 , − ( cos ( 4π ) − cos 0 )
π
3
3
(
)
26 16
,− ,0
3
3
(
b
a
4
3
1
2 32
1
2
= ( 2t + 1) , − t , − cos (π t )
3
3 0 π
0
0
=
1≤ t ≤ 4
)
1
r ′ ( t ) = 2, , 2t
t
1
+ 4t 2
2
t
r′ (t ) = 4 +
r′ (t ) =
r′ (t ) =
4t 2 + 1 + 4t 4
4t 4 + 4t 2 + 1
=
t2
t2
( 2t
2
+ 1)
2
t2
2t 2 + 1
1
r′ (t ) =
⇒ r ′ ( t ) = 2t +
t
t
4
4
1

Arc Length → s = ∫ r ′ ( t ) dt = ∫  2t +  dt
t
1
1
4
s = ( t 2 + ln t ) = 16 + ln 4 − 1
1
s = 15 + ln 4
9.1