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AP Calculus BC
Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
9.4 Defining and Differentiating Vector-Valued Functions
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The vector-valued function
r
⃗
(
t
)
\vec{r}(t)
r
(
t
)
represents the position of a moving object as a function of the parameter t.
The vector-valued function
r
⃗
(
t
)
\vec{r}(t)
r
(
t
)
represents the velocity of a moving object.
False
What is a vector-valued function?
A function that assigns a vector to each input value
What is a vector-valued function?
Assigns a vector to each input
A vector-valued function can be written as
\vec{r}(t)
.
Match the component of a vector-valued function with its description:
x
(
t
)
x(t)
x
(
t
)
↔️ The x-coordinate function
y
(
t
)
y(t)
y
(
t
)
↔️ The y-coordinate function
z
(
t
)
z(t)
z
(
t
)
↔️ The z-coordinate function
What does the vector-valued function
r
⃗
(
t
)
\vec{r}(t)
r
(
t
)
represent?
Position of a moving object
A vector-valued function can be written as
r
⃗
(
t
)
=
\vec{r}(t) =
r
(
t
)
=
⟨
x
(
t
)
,
y
(
t
)
,
z
(
t
)
⟩
\langle x(t), y(t), z(t)\rangle
⟨
x
(
t
)
,
y
(
t
)
,
z
(
t
)⟩
, where
x
(
t
)
x(t)
x
(
t
)
,
y
(
t
)
y(t)
y
(
t
)
, and
z
(
t
)
z(t)
z
(
t
)
are the components
The component
x
(
t
)
x(t)
x
(
t
)
in a vector-valued function represents the x-coordinate of the function.
True
What does the vector-valued function
r
⃗
(
t
)
\vec{r}(t)
r
(
t
)
represent?
The position of a moving object
To graph a vector-valued function, you plot points defined by the parametric equations
x
(
t
)
x(t)
x
(
t
)
,
y
(
t
)
y(t)
y
(
t
)
, and
z
(
t
)
z(t)
z
(
t
)
.
Each value of
t
t
t
in a vector-valued function corresponds to a unique position vector.
True
How is the derivative of a vector-valued function calculated?
By differentiating each component function
The derivative of
r
⃗
(
t
)
=
\vec{r}(t) =
r
(
t
)
=
⟨
x
(
t
)
,
y
(
t
)
,
z
(
t
)
⟩
\langle x(t), y(t), z(t)\rangle
⟨
x
(
t
)
,
y
(
t
)
,
z
(
t
)⟩
is
r
⃗
′
(
t
)
=
\vec{r}'(t) =
r
′
(
t
)
=
⟨
x
′
(
t
)
,
y
′
(
t
)
,
z
′
(
t
)
⟩
\langle x'(t), y'(t), z'(t)\rangle
⟨
x
′
(
t
)
,
y
′
(
t
)
,
z
′
(
t
)⟩
, where
x
′
(
t
)
x'(t)
x
′
(
t
)
,
y
′
(
t
)
y'(t)
y
′
(
t
)
, and
z
′
(
t
)
z'(t)
z
′
(
t
)
are the derivatives of
x
(
t
)
x(t)
x
(
t
)
,
y
(
t
)
y(t)
y
(
t
)
, and
z
(
t
)
z(t)
z
(
t
)
.
Steps to graph the vector-valued function
r
⃗
(
t
)
=
\vec{r}(t) =
r
(
t
)
=
⟨
t
2
,
t
⟩
\langle t^{2}, t \rangle
⟨
t
2
,
t
⟩
1️⃣ Create a table of values for
t
t
t
,
x
(
t
)
x(t)
x
(
t
)
, and
y
(
t
)
y(t)
y
(
t
)
2️⃣ Plot the points on a coordinate plane
3️⃣ Connect the points to create the curve
Each value of
t
t
t
corresponds to a position vector
Steps to create the curve for a vector-valued function
1️⃣ Choose values for
t
t
t
2️⃣ Calculate
x
(
t
)
x(t)
x
(
t
)
and
y
(
t
)
y(t)
y
(
t
)
3️⃣ Plot the points
4️⃣ Connect the points to create the curve
The Sum Rule for differentiating vector-valued functions states that
d
d
t
[
u
⃗
(
t
)
+
\frac{d}{dt}[\vec{u}(t) +
d
t
d
[
u
(
t
)
+
v
⃗
(
t
)
]
=
\vec{v}(t)] =
v
(
t
)]
=
u
⃗
′
(
t
)
+
\vec{u}'(t) +
u
′
(
t
)
+
v
⃗
′
(
t
)
\vec{v}'(t)
v
′
(
t
)
True
The Product Rule for Dot Products states that
d
d
t
[
u
⃗
(
t
)
⋅
v
⃗
(
t
)
]
=
\frac{d}{dt}[\vec{u}(t) \cdot \vec{v}(t)] =
d
t
d
[
u
(
t
)
⋅
v
(
t
)]
=
u
⃗
′
(
t
)
⋅
v
⃗
(
t
)
+
\vec{u}'(t) \cdot \vec{v}(t) +
u
′
(
t
)
⋅
v
(
t
)
+
u
⃗
(
t
)
⋅
v
⃗
′
(
t
)
\vec{u}(t) \cdot \vec{v}'(t)
u
(
t
)
⋅
v
′
(
t
)
is the derivative
What is the formula for the Product Rule for Cross Products?
d
d
t
[
u
⃗
(
t
)
×
v
⃗
(
t
)
]
=
\frac{d}{dt}[\vec{u}(t) \times \vec{v}(t)] =
d
t
d
[
u
(
t
)
×
v
(
t
)]
=
u
⃗
′
(
t
)
×
v
⃗
(
t
)
+
\vec{u}'(t) \times \vec{v}(t) +
u
′
(
t
)
×
v
(
t
)
+
u
⃗
(
t
)
×
v
⃗
′
(
t
)
\vec{u}(t) \times \vec{v}'(t)
u
(
t
)
×
v
′
(
t
)
The velocity of an object is the derivative of its position vector with respect to time
True
What is the formula for arc length along a curve defined by a vector-valued function?
s
=
s =
s
=
∫
a
b
∣
∣
v
⃗
(
t
)
∣
∣
d
t
\int_{a}^{b} ||\vec{v}(t)|| dt
∫
a
b
∣∣
v
(
t
)
∣∣
d
t
If
r
⃗
(
t
)
=
\vec{r}(t) =
r
(
t
)
=
⟨
t
2
,
sin
(
t
)
,
e
t
⟩
\langle t^{2}, \sin(t), e^{t} \rangle
⟨
t
2
,
sin
(
t
)
,
e
t
⟩
, then
r
⃗
′
(
t
)
=
\vec{r}'(t) =
r
′
(
t
)
=
⟨
2
t
,
cos
(
t
)
,
e
t
⟩
\langle 2t, \cos(t), e^{t} \rangle
⟨
2
t
,
cos
(
t
)
,
e
t
⟩
is correct.
True
Match the component with its function in \vec{r}(t) = \langle t^{2}, \sin(t), e^{t} \rangle</latex>:
x
(
t
)
x(t)
x
(
t
)
↔️
t
2
t^{2}
t
2
y
(
t
)
y(t)
y
(
t
)
↔️
sin
(
t
)
\sin(t)
sin
(
t
)
z
(
t
)
z(t)
z
(
t
)
↔️
e
t
e^{t}
e
t
Graphing vector-valued functions involves plotting points defined by
parametric equations
True
What is the formula for the derivative of a vector-valued function
r
⃗
(
t
)
\vec{r}(t)
r
(
t
)
?
r
⃗
′
(
t
)
=
\vec{r}'(t) =
r
′
(
t
)
=
⟨
x
′
(
t
)
,
y
′
(
t
)
,
z
′
(
t
)
⟩
\langle x'(t), y'(t), z'(t)\rangle
⟨
x
′
(
t
)
,
y
′
(
t
)
,
z
′
(
t
)⟩
If \vec{r}(t) = \langle t^{2}, \sin(t), e^{t} \rangle</latex>, then
r
⃗
′
(
t
)
=
\vec{r}'(t) =
r
′
(
t
)
=
⟨
2
t
,
cos
(
t
)
,
e
t
⟩
\langle 2t, \cos(t), e^{t} \rangle
⟨
2
t
,
cos
(
t
)
,
e
t
⟩
is the derivative