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Tangent, Normal
and Binormal Vectors

In this section we want to look at an application of derivatives for vector functions
...

In the past we’ve used the fact that the derivative of a function was the slope of the tangent line
...

Given the vector function, r ( t ) , we call r′ ( t ) the tangent vector provided it exists and
provided r′ ( t ) ≠ 0
...
Note that we really do need to require r′ ( t ) ≠ 0 in
order to have a tangent vector
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Also, provided r′ ( t ) ≠ 0 , the unit tangent vector to the curve is given by,

r′ (t )
T (t ) =
r′ (t )
While, the components of the unit tangent vector can be somewhat messy on occasion there are
times when we will need to use the unit tangent vector instead of the tangent vector
...

Solution
First, by general formula we mean that we won’t be plugging in a specific t and so we will be
finding a formula that we can use at a later date if we’d like to find the tangent at any point on the
curve
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Here is the tangent vector to the curve
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r ′ ( t ) = 4t 2 + 4 cos 2 t + 4sin 2 t
= 4t 2 + 4
The unit tangent vector is then,

T (t ) =
=

1
2

4t
2t

2t i + 2 cos t j − 2sin t k )
(
+4

4t + 4
2

i+

2 cos t
4t + 4
2

j−

2sin t
4t + 4
2

k

Example 2 Find the vector equation of the tangent line to the curve given by
r ( t ) = t 2 i + 2sin t j + 2 cos t k at t = π3
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2π
2π
⎛
π
⎞
⎛
π
⎞
⎛
π
⎞
r′ ⎜ ⎟ =
i + 2 cos ⎜ ⎟ j − 2sin ⎜ ⎟ k =
i + j − 3k
3
⎝3⎠ 3
⎝3⎠
⎝3⎠
We’ll also need the point on the line at t = π3 so,

⎛π ⎞ π
r⎜ ⎟=
i + 3 j +k
⎝3⎠ 9
2

The vector equation of the line is then,

r (t ) =

π2

2π
, 3,1 + t
,1, − 3
9
3

Before moving on let’s note a couple of things about the previous example
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However, that would have
made for a more complicated equation for the tangent line
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Do not get excited about that
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With normal functions, y is the generic letter that we used to represent
functions and r ( t ) tends to be used in the same way with vector functions
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The unit normal vector is defined to be,

N (t ) =

T ′ (t )

T ′ (t )

The unit normal is orthogonal (or normal, or perpendicular) to the unit tangent vector and hence
to the curve as well
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They will show up with some regularity in several Calculus III topics
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The binormal vector is defined to be,

B (t ) = T (t ) × N (t )

Because the binormal vector is defined to be the cross product of the unit tangent and unit normal
vector we then know that the binormal vector is orthogonal to both the tangent vector and the
normal vector
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Solution
We first need the unit tangent vector so first get the tangent vector and its magnitude
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3
3
T ′ ( t ) = 0, −
sin t , −
cos t
10
10
9
9
9
3
2
2
T ′ (t ) =
=
sin t + cos t =
10
10
10
10
The unit normal vector is then,

10
3
3
N (t ) =
0, −
sin t , −
cos t = 0, − sin t , − cos t
3
10
10
Finally, the binormal vector is,

B (t ) = T (t ) × N (t )
i
=

1
10
0

j

k

3
cos t
10
− sin t

3
−
sin t
10
− cos t

i
1
10
0

j
3
cos t
10
− sin t

3
1
1
3
2
=−
cos t i −
sin t k +
cos t j −
sin 2 t i
10
10
10
10
3
1
1
=−
i+
cos t j −
sin t k
10
10
10



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