Vector valued functions
Defining vector valued functions - suggested problems -solutions
P1: r(t) =< ln(−2 − t), 3t + 1, sin(t2) >
(a) Write the parametric equations for the function.
x(t) = ln(−2 − t)
y(t) = 3t + 1
z(t) = sin(t2)
(b) Give the domains of each of the component functions, and the domain of r(t).
Domain of x: −2 − t > 0, so t < −2. [logarithmic function] In interval notation,
t ∈ (−∞, −2).
Domain of y: t ∈ (−∞, ∞) [polynomial]
Domain of z: t ∈ (−∞, ∞) [sine function]
Domain of r(t) is the intersection of the component domains: domain of r(t):
(−∞, −2).
(c) Find three points on the function.
Answers vary; just pick some t values in the domain:
r(−3) = < ln(−2 − (−3)), 3(−3) + 1, sin((−3)2) >=< 0, −8, sin 9 >
r(−4) = < ln(−2 − (−4)), 3(−4) + 1, sin((−4)2) >=< ln 2, −11, sin 16 >
r(−10) = < ln(−2 − (−10)), 3(−10) + 1, sin((−10)2 ) >=< ln 8, −29, sin 100 >
Points (0, −8, sin 9), (ln 2, −11, sin 16), and (ln 8, −29, sin 100) are all on the function.
(d) Plot the function in MVT.
I set the t domain from −10 to −2, and left the corresponding x, y, and z as default.
3
25 − t2 j − (tan−1 t) k
P2: r(t) = t −
i
+
4
t+5
(a) Write the parametric equations for the function.
3
x(t) = t −
4
2
25
−
y(t) = t + 5t
z(t) = − tan−1 t
(b) Give the domains of each of the component functions, and the domain of r(t).
Domain of x: t − 4 6= 0, so t 6= 4 [rational]. t ∈ (−∞, 4) ∪ (4, ∞).
Domain of y: t + 5 6= 0, so t 6= −5 [rational]. t ∈ (−∞, −5) ∪ (−5, ∞).
Domain of z: t ∈ (−∞, ∞) [inverse tangent]
Domains of inverse trig functions come from ranges of corresponding trig functions
- since the range of tan x is (−∞, ∞), the domain of tan−1 x is (−∞, ∞) as well.
Another example of this is that since −1 ≤ sin x ≤ 1, the domain of sin−1 x is [−1, 1].
Domain of r(t) is the intersection of the component domains. From x 6= 4 and
x 6= −5, we have
(−∞, −5) ∪ (−5, 4) ∪ (4, ∞)
(c) Find three points on the function.
3 , 25 − 02 , − tan−1 0 >=< − 3 , 5, 0 >
r(0) = < 0 −
4 0+5
4
3 , 25 − 12 , − tan−1 1 >=< −1, 24 , π >
r(1) = < 1 −
4 1+5
7 4
25 − (−1)2
3
−1
r(−1) = < −1 − 4 , −1 + 5 , − tan −1 >=< − 53 , 6, − π4 >
π
3
π
Points (− 34 , 5, 0), (−1, 24
7 , 4 ), and (− 5 , 6, − 4 ) are all on the function.
(d) Plot the function in MVT.
The t domain is set from −7 to 6 to catch the interesting spots at t = −5 and t = 4.
That vertical line you see connecting the function shouldn’t be there (it’s the asymptote
at t = 4). There’s a hole at t = −5 that doesn’t show up in the graph - you know that
it’s there since − is not in the domain.
P3: r(t) =<
√
√
3 − t, t − 2, et >
(a) Write the parametric equations for the function.
√
x(t) = √3 − t
y(t) =
t−2
t
z(t) = e
(b) Give the domains of each of the component functions, and the domain of r(t).
Domain
Domain
Domain
Domain
of
of
of
of
x: 3 − t ≥ 0, so t ≤ 3 [even root]. (−∞, 3].
y: t − 2 ≥ 0, so t ≥ 2 [even root]. [2, ∞).
z: t ∈ (−∞, ∞) [exponential].
r(t) is the intersection of the component domains: domain of r(t): [2, 3].
√
√
2
2 − 2, e2 >=< 1, 0, e√
>√
r(2) = < √3 − 2, √
2.5
2.5
r(2.5) = < √3 − 2.5,
√ 2.5 − 32, e >=< 3 .5, .5, e >
r(3) = < 3 − 3, 3 − 2, e >=< 0, 1, e >
√ √
Points (1, 0, e2 ), ( .5, .5, e2.5), and (0, 1, e3) are all on the function.
(d) Plot the function in MVT.
Knowing the endpoints on the domain is important for simply being able to plot this one
- see what happens if you set the t values from, say t = 1 to t = 2 instead. MVT gets
very unhappy trying to plot things that don’t exist.