Vector-Valued Functions
Finding The Domain of a Vector-Valued Function:
When mapping real numbers onto vectors you get vector-valued functions. It is just a three-
dimensional plane curve with the following equation:
r (t ) = f (t )ˆ + g (t )ˆ + h(t )k
i j ˆ
To obtain the domain of the vector, combine the domains of f(t), g(t), and h(t).
Example:
Find the domain of the vector-valued function
1 ˆ
r (t ) = 2 cos tˆ + 2 sin tˆ +
i j k
t +1
1. Find the domain of each component.
Domain of f(t) = 2 cos t: (-∞,∞)
Domain of g(t) = 2 sin t: (-∞,∞)
Domain of h(t)= 1/(t+1): t ≠ -1
2. Combine the domains of the components in Step 1 together.
Domain of r(t): (-∞,-1) U (-1,∞)
Sketching A Vector-Valued Function:
To sketch the graph of the vector, just pick values for t and plug into the i, j, and, k components
of the vector to get x, y, and z values.
Example:
Sketch the graph of the vector-valued function
3 ˆ
r (t ) = tˆ + t 2 ˆ + tk
i j
2
1. Make a table of values to get components of the vector.
t f(t) g(t) h(t)
-1 -1 1 -3/2
0 0 0 0
1 1 1 3/2
2 2 4 3
2. Sketch the graph using the values from Step 1 above.
z
y
x
Differentiation and Integration of Vector-Valued Functions:
To find the derivative of a vector-valued function, just take the derivative of each component of
the vector.
Example:
Find the derivative of the vector-valued function
r (t ) = 4 t ˆ + 2 cos tˆ + 2 sin tk
i j ˆ
1. Finding the derivative with respect to t of the vector yields:
1
r ' (t ) = 4 t −1 / 2 ˆ + 2(− sin t )ˆ + 2(cos t )k
2
i j ˆ
2. Simplify.
( )
r '(t ) = 2t −1 / 2 ˆ + (−2 sin t )ˆ + (2 cos t )k
i j ˆ
2
= 1 / 2 ˆ − 2 sin tˆ + 2 cos tk
i j ˆ
t
To integrate a vector-valued function, just integrate each component of the vector.
Example:
Integrate the vector-valued function
r (t ) = 4 t ˆ + 2 cos tˆ + 2 sin tk
i j ˆ
1. Integrating with respect to t of the vector yields:
t 3 / 2 ˆ
3 / 2 i + 2(sin t ) j + 2(− cos t )k
r (t ) = 4
ˆ ˆ
2. Simplify.
2t 3 / 2 ˆ
3 i + (2 sin t ) j + (−2 cos t )k
r (t ) = 4
ˆ ˆ
3/ 2
8t ˆ
= i + 2 sin tˆ − 2 cos tk
j ˆ
3
Applications of Vector-Valued Functions: Position, Velocity, Speed, and Acceleration
Assume that the position vector is given by
r (t ) = x(t )ˆ + y (t )ˆ + z (t )k .
i j ˆ
1. We can calculate velocity by taking the first derivative of the position vector with respect to
time.
v (t ) = r '(t ) = x' (t )ˆ + y ' (t )ˆ + z ' (t )k
i j ˆ
Since velocity is speed in a given direction, then it must be a vector quantity.
2. We can calculate speed by taking the magnitude of the velocity vector.
v (t ) = r '(t ) = [x' (t )]2 + [ y ' (t )]2 + [z ' (t )]2
Since speed is a magnitude of a vector, then it must be a scalar quantity (number).
3. We can calculate acceleration by taking the first derivative of the velocity vector with respect
to time (also is the second derivative of the position vector).
a(t ) = v '(t ) = r '' (t ) = x' ' (t )ˆ + y ' ' (t )ˆ + z ' ' (t )k
i j ˆ
Acceleration is also a vector quantity.
Example:
Given the following position vector
3
r (t ) = t 3 ˆ − ˆ ,
i j
t
calculate the velocity, speed, and acceleration at t = 1.
1. Take the first derivative of the position vector r.
v (t ) = r ' (t ) = 3t 2 ˆ − 3(−t −2 )ˆ
i j
−1
v (t ) = 3t 2 ˆ − 3 2 ˆ
t
i j
3
= 3t 2 ˆ + 2 ˆ
i j
t
2. Plug in the value of t to get the velocity vector at that point.
3 ˆ
v (1) = 3(1) 2 ˆ +
i j
(1) 2
= 3ˆ + 3ˆ
i j
3. Take the velocity vector v from Step 1 and take the magnitude of it.
3
+ 2
2
v (t ) = (3t )
2 2
t
9
= 9t 4 +
t4
4. Plug in the value of t to get the speed at that point.
9
v (1) = 9(1) 4 + = 9 + 9 = 18 = 3 2
(1) 4
5. Take the first derivative of the velocity vector to get the acceleration vector a.
( )
a(t ) = v ' (t ) = r ' ' (t ) = 6tˆ + 3 − 2t −3 ˆ
i j
−2
a(t ) = 6tˆ + 3 3 ˆ
t
i j
6
a(t ) = 6tˆ − 3 ˆ
i j
t
2. Plug in the value of t to get the acceleration vector at that point.
Example:
Find the position vector with
a(t ) = ˆ + ˆ, v (0) = 5ˆ, and r (0) = 0 .
i j j
v (t ) = ∫ a(t )dt = ∫ ˆ + ˆ dt
1. Find the velocity vector v by integrating the acceleration vector a with respect to time.
( )
i j
v (t ) = (t + C1 )ˆ + (t + C 2 )ˆ + C 3k
i j ˆ
2. Using the given information, find the particular solution of the velocity vector.
--Using the information above for the velocity vector, plug in the value of t (in this
example, t = 0).
v (0) = (0 + C1 )ˆ + (0 + C 2 )ˆ + C 3 k
i j ˆ
v (0) = C1ˆ + C 2 ˆ + C 3 k
i j ˆ
--Substitute the value of left side of the equation with the vector it is equal to. In this
example, substitute the vector 5j in for v(0).
5ˆ = C1ˆ + C 2 ˆ + C 3k
j i j ˆ
--Solve for the constants C1, C2, and C3 by equating like components together.
5ˆ = C1ˆ + C 2 ˆ + C 3k ⇒ 0ˆ + 5ˆ + 0k = C1ˆ + C 2 ˆ + C 3 k
j i j ˆ i j ˆ i j ˆ
0 = C1
5 = C2
0 = C3
Thus, the velocity vector is
v (t ) = tˆ + (t + 5)ˆ .
i j
r (t ) = ∫ v (t )dt = ∫ tˆ + (t + 5)ˆ dt
3. Find the position vector r by integrating the velocity vector v with respect to time.
( i j )
t2 t2
r (t ) = + C1 ˆ + + 5t + C 2 ˆ + C 3k
2 i 2 j
ˆ
4. Using the given information, find the particular solution of the position vector.
--Using the information above for the position vector, plug in the value of t (in this
example, t = 0).
r (0) = (0 + C1 )ˆ + (0 + 0 + C 2 )ˆ + C 3k
i j ˆ
r (0) = C1ˆ + C 2 ˆ + C 3 k
i j ˆ
--Substitute the value of left side of the equation with the vector it is equal to. In this
example, substitute the vector 0 = 0i + 0j + 0k in for r(0).
0 = C1ˆ + C 2 ˆ + C 3 k
i j ˆ
--Solve for the constants C1, C2, and C3 by equating like components together.
0 = C1ˆ + C 2 ˆ + C 3 k ⇒ 0ˆ + 0ˆ + 0k = C1ˆ + C 2 ˆ + C 3 k
i j ˆ i j ˆ i j ˆ
0 = C1
0 = C2
0 = C3
t2 ˆ t2
Thus, the position vector is r (t ) = i + + 5t ˆ .
2 j
2