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3 The position of an object is given by r(t) = ln(sin t) i -t k for 0 < t <…
Related Topics
Wize University Calculus 2 Textbook > Vector Functions
Vector Functions
3 Activities
3 The position of an object is given by
r
⃗
(
t
)
=
ln
(
sin
t
)
i
⃗
−
t
k
⃗
\vec r(t) = \ln(\sin t)\vec i -t\vec k
r
(
t
)
=
ln
(
sin
t
)
i
−
t
k
for
0
<
t
<
π
.
0 < t < \pi.
0
<
t
<
π
.
Determine the minimum speed of the object.
Answer
I don't know
Check Submission
More Vector Functions Questions:
Given the following vector equation, sketch out the curve that it generates.
r
(
t
)
=
⟨
cos
(
t
)
,
t
⟩
r(t) = \langle \cos (t), t \rangle
r
(
t
)
=
⟨
cos
(
t
)
,
t
⟩
2 The position of an object is given by
r
⃗
(
t
)
=
ln
(
sin
t
)
i
⃗
−
t
k
⃗
\vec r(t) = \ln(\sin t)\vec i -t\vec k
r
(
t
)
=
ln
(
sin
t
)
i
−
t
k
Determine the velocity, acceleration, and speed at time
t
.
1 The position of an object is given by
r
⃗
(
t
)
=
e
t
sin
t
i
⃗
+
e
t
cos
t
j
⃗
−
e
t
k
⃗
\vec r(t) = e^t\sin t\vec i + e^t\cos t\vec j-e^t\vec k
r
(
t
)
=
e
t
sin
t
i
+
e
t
cos
t
j
−
e
t
k
Determine the velocity, acceleration, and speed at time
t
.
Find the Cartesian equation of the tangent line to the following curve at the indicated point:
r
⃗
(
t
)
=
ln
(
cos
t
)
i
⃗
+
sin
(
2
t
)
j
⃗
−
cos
2
t
k
⃗
at
t
=
π
/
6
\vec r(t) = \ln(\cos t) \vec i + \sin(2t) \vec j- \cos^2 t \vec k\quad \text{ at } t = \pi/6
r
(
t
)
=
ln
(
cos
t
)
i
+
sin
(
2
t
)
j
−
cos
2
t
k
at
t
=
π
/6
Practice: Domain of a vector-valued function 2
Find the domain of the following function. Express the domain as an interval, or a union of intervals. i.e.
t
∈
[
−
3
,
3
)
t \in [-3, 3)
t
∈
[
−
3
,
3
)
or
t
∈
(
−
∞
,
π
]
∪
(
13
,
∞
)
t \in (-\infty, \pi] \cup ( 13, \infty)
t
∈
(
−
∞
,
π
]
∪
(
13
,
∞
)
.
v
(
x
)
=
⟨
x
9
x
2
−
x
4
,
−
log
(
−
x
3
)
⟩
v(x) = \left \langle \frac{x}{\sqrt{9x^2 - x^4}}, - \log \left ( - \frac{x}{3} \right )\right \rangle
v
(
x
)
=
⟨
9
x
2
−
x
4
x
,
−
lo
g
(
−
3
x
)
⟩
The following curves intersect when
t
=
0
t = 0
t
=
0
. What is the angle between these two curves:
r
1
(
t
)
=
⟨
t
,
1
+
t
2
,
cos
(
t
)
+
t
⟩
r
2
(
t
)
=
⟨
sin
(
t
)
,
cos
(
t
)
,
e
t
⟩
r_1(t) = \langle t, 1 + t^2, \cos(t) + t \rangle \quad\quad\quad r_2(t) = \langle \sin(t), \cos(t), e^t \rangle
r
1
(
t
)
=
⟨
t
,
1
+
t
2
,
cos
(
t
)
+
t
⟩
r
2
(
t
)
=
⟨
sin
(
t
)
,
cos
(
t
)
,
e
t
⟩
Let's say that a vector function
r
(
t
)
r(t)
r
(
t
)
that describes a curve
C
C
C
has the property that for all
t
t
t
,
r
(
t
)
r(t)
r
(
t
)
is perpendicular to its tangent vector. Demonstrate that the curve must be completely contained on the surface of a sphere centered at the origin.
Give at least one point where the following vector function
r
(
t
)
=
⟨
−
sin
t
,
2
e
t
,
−
t
⟩
r(t) = \langle -\sin t, 2e^t, -t\rangle
r
(
t
)
=
⟨
−
sin
t
,
2
e
t
,
−
t
⟩
has tangent vectors normal to the plane whose equation is