Examples from Section 13.2
#18: Find the unit tangent vector T(t) at the point with the given value of the
parameter t.
r(t) = ⟨ 4√(t), t2, t ⟩ , t = 1
Recall, the unit tangent vector is T(t) =
r’(t)
|r’(t)|
r(t) = ⟨ 4√(t), t2, t ⟩
r’(t) = ⟨
2
,
t
2t, 1 ⟩
r’(1) = ⟨ 2, 2, 1 ⟩
|r’(1)| =
4+4+1 =3
T(1) = ⟨ 23, 23, 13 ⟩
Example similar to #23 - #29
Find the parametric equations for the tangent line to the curve with the given
parametric equations at the specified point.
x = t2 – 1,
y = t2 + 1, z = t + 1,
(-1,1,1)
At (-1, 1, 1) it is obvious that t = 0.
r(t) = ⟨ t2 – 1, t2 + 1, t + 1 ⟩
r’(t) = ⟨ 2t, 2t, 1 ⟩
r’(0) = ⟨ 0, 0, 1 ⟩ = v the direction vector of the tangent line.
Eqn of tangent line: ⟨ x, y, z ⟩ = ⟨ -1, 1, 1 ⟩ + t⟨ 0, 0, 1 ⟩
Parametric equations of the line: x = -1, y = 1, z = 1 + t.
#32: At what point do the curves
r1(t) = ⟨ t, 1 – t, 3 + t2 ⟩ and r2(s) = ⟨ 3 – s, s – 2, s2 ⟩ intersect?
Find their angle of intersection correct to the nearest degree.
We need x1 = x2, y1 = y2, and z1 = z2, thus
t = 3 – s, 1 – t = s – 2, and 3 + t2 = s2.
Simple algebra gives us: t = 1, s = 2.
r1(1) = ⟨ 1, 0, 4 ⟩ = r2(2).
The curves intersect at the point (1, 0, 4).
The angle of intersection of the two curves can be found by finding the angle of
intersection of the two tangent lines.
For r1(t) the direction vector of the tangent line is r’1(1) = ⟨ 1, -1, 2 ⟩
For r2(s) the direction vector of the tangent line is r’2(2) = ⟨ -1, 1, 4 ⟩
⟨ 1, -1, 2 ⟩  ⟨ -1, 1, 4 ⟩
-1 – 1 + 8
6
1
3
θ
=
=
|⟨ 1, -1, 2 ⟩| |⟨ -1, 1, 4 ⟩| cosθ
(1 + 1 + 4)
=
6 3 cosθ
=
cosθ
≈
54.7°
1/2
1/2
(1 + 1 + 16)
cosθ
1
4
2t


#34: Evaluate this integral:   0 i + 1 + t2 j + 1 + t2 k dt


0
Recall
4


2 dt = 4arctant + C
1+t
 2t dt
2

2 = ln(1 + t ) + C
1
+
t

So, the value of the original integral is:
1
⟨ 0, 4(arctant), ln(1 + t2) ⟩ |0
=
⟨ 0, 4(arctan1 – arctan0), ln2 – ln1 ⟩
=
⟨ 0, π, ln2 ⟩
d
#48: Find an expression for dt [u(t)  (v(t) x w(t))]
I am going to leave the argument (t) out of the functions. The ’ indicates a
derivative with respect to t.
[u  (v x w)]’
=
u’  (v x w) + u  [v x w]’
=
u’  (v x w) + u  (v’ x w) + u  (v x w’)
=
u’  (v x w) - u  (w x v’) + (u x v)  w’
=
u’  (v x w) – (u x w)  v’ + (u x v)  w’
=
u’  (v x w) + v’  (w x u) + w’  (u x v)
The primary advantage that the fifth line has over the 2nd line is that it is more
aesthetically pleasing.