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Section 10.2: Derivatives and Integrals of Vector Functions
Practice HW from Stewart Textbook (not to hand in)
p. 707 # 3-21 odd, 29-35
Differentiation of Vector Functions
Differentiation of vector valued functions are done component wise in the natural way.
Thus, for
1. 2D Case: If r(t ) = f (t) i + g (t) j, then r ′(t ) = f ′ (t) i + g ′(t) j
2. 3D Case: If r(t ) = f (t) i + g (t) j + h (t) k , then r ′(t ) = f ′ (t) i + g ′(t) j + h ′(t ) k
Example 1: Find the derivative of the vector function
r(t ) = <
1
, 16t ,
t
t >.
Solution:
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Example 2: Find the derivative of the vector function
r(t ) = e 2t i + sin 2 t j + ln(t 2 + 1) k
Solution:
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Note: Look at properties involving the derivative of vector value functions on p. 705
Theorem 3 of Stewart text.
Tangent Vector to a Vector Valued Function
Recall that the derivative provides the tool for finding the tangent line to a curve. This
same idea can be used to find a vector tangent to a curve at a point. We illustrate this idea
in the following example.
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Example 3: For the vector function r(t ) = e t i + e − t j,
a. Sketch the plane curve with the given vector equation.
b. Find r ′(t )
c. Sketch the positive vector r(t ) and the tangent vector r ′(t ) at t = 0.
Solution:
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Example 4: Find the parametric equations for the tangent line to the curve with the
parametric equations x = t 2 − 1, y = t 2 + 1, z = t + 1 at the point (-1, 1, 1).
Solution:
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Note: Sometimes, it is convenient the normalize a vector tangent to a vector valued
function. This gives the unit tangent vector.
Unit Tangent Vector
Given a vector function r on an open interval I, the unit tangent vector T(t) is given by
T ( t) =
1
r ′(t)
=
r ′(t) where r ′( t ) ≠ 0
| r ′(t) | | r ′(t) |
Example 5: Find the unit tangent vector T(t) for r(t ) = 2 cos t i + sin 6t j + 2t k at
t=
π
6
.
Solution: The unit tangent vector is given by the formula
T ( t) =
1
r ′(t)
=
r ′(t) .
| r ′(t) | | r ′(t) |
Using r(t ) = 2 cos t i + sin 6t j + 2t k, we see that
r ′( t ) = −2 sin t i + 6 cos 6t j + 2k = < −2 sin t , 6 cos 6t , 2 >
and
| r ′( t ) |= (−2 sin t ) 2 + (6 cos 6t) 2 + (2) 2 = 4 sin 2 t + 36 cos 2 6 t + 4
Thus,
T ( t) =
At t =
1
r ′(t) =
| r ′(t) |
π
6
, we have
1
2
2
4 sin t + 36 cos 6t + 4
< −2 sin t , 6 cos 6t , 2 >
(continued on next page)
6
π
T( ) =
6
=
=
1
2
2
4 sin (π / 6) + 36 cos (6π / 6) + 4
< − s 2in(π / 6) , 6 cos(6π / 6), 2 >
1
1
< −2( ) , 6(−1), 2 > (Note that sin(π / 6) = 1 / 2 , cos(6π/6) = cosπ = -1)
2
(1 / 2) 2 + 36(−1) 2 + 4
1
< −1, − 6, 2 >
1
4( ) + 36(1) + 4
4
1
=
< −1, − 6, 2 >
1 + 36 + 4
1
=
< −1 , − 6, 2 >
41
1
6
2
= <−
>
,−
,
41
41 41
The following graph plots in 3D space the vector function r(t) and the corresponding unit
tangent vector T(t) evaluated at t =
π
6
.
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Integrals of Vector Functions
Integrals of Vector Valued Functions are computed component wise.
∫ r(t )dt = ∫ f (t )dt i + ∫ g (t )dt j
+ h (t) k , then ∫ r (t )dt = ∫ f (t )dt i + ∫ g (t )dt
1. 2D Case: If r(t ) = f (t) i + g (t) j, then
2. 3D Case: If r(t ) = f (t) i + g (t) j
Example 6: Evaluate the integral
∫ (3t
2
j + ∫ h(t )dt k
i + 4t j − 8t 3 k )dt
Solution:
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π
Example 7: Evaluate the integral
3
2 t
∫ (t e i + sin t j − 2t k )dt
0
Solution:
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Example 8: Find r(t) if r ′(t ) = 2t j + t k and r(0) = i + j.
Solution: Writing r ′(t ) = 2t j + t k = 2t j +
1
(t ) 2
k , we see that
r (t ) = ∫ r ′(t )dt
= ∫ ( 2t
=2
2
1
i+t2
k )dt
3
t2
t
i+
k+C
3
2
2
3
2
= t i+ t2 k +C
3
2
3
2
Thus, r (t ) = t i + t 2 k + C and we need to use the initial condition r(0) = i + j to find the
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constant vector C. We see that
2
3
2
i + j = r ( 0) = ( 0) i + ( 0) 2 k + C
3
2
which gives C = i + j. Thus, substituting for C gives
3
2
r (t ) = t i + t 2 k + (i + j) ,
3
2
which, when combining like terms, gives the result.
3
2
r (t ) = (t + 1)i + j + t 2 k
3
2
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