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Give at least one point where the following vector function r(t) = -sin t, 2e^…

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Wize University Calculus 2 Textbook > Vector Functions

Vector Functions

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Give at least one point where the following vector function
r(t)=⟨−sin⁡t,2et,−t⟩r(t) = \langle -\sin t, 2e^t, -t\rangle
has tangent vectors normal to the plane whose equation is
x−2y+z=0x - 2y + z = 0

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
3 The position of an object is given by
r⃗(t)=ln⁡(sin⁡t)i⃗−tk⃗\vec r(t) = \ln(\sin t)\vec i -t\vec k
for 0<t<π. 0 < t < \pi. Determine the minimum speed of the object.
2 The position of an object is given by
r⃗(t)=ln⁡(sin⁡t)i⃗−tk⃗\vec r(t) = \ln(\sin t)\vec i -t\vec k
Determine the velocity, acceleration, and speed at time t.
1 The position of an object is given by
r⃗(t)=etsin⁡ti⃗+etcos⁡tj⃗−etk⃗\vec r(t) = e^t\sin t\vec i + e^t\cos t\vec j-e^t\vec 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:
r⃗(t)=ln⁡(cos⁡t)i⃗+sin⁡(2t)j⃗−cos⁡2tk⃗ 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
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) or t∈(−∞,π]∪(13,∞)t \in (-\infty, \pi] \cup ( 13, \infty) .
v(x)=⟨x9x2−x4,−log⁡(−x3)⟩v(x) = \left \langle \frac{x}{\sqrt{9x^2 - x^4}}, - \log \left ( - \frac{x}{3} \right )\right \rangle
The following curves intersect when t=0t = 0 . What is the angle between these two curves:
r1(t)=⟨t,1+t2,cos⁡(t)+t⟩r2(t)=⟨sin⁡(t),cos⁡(t),et⟩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
Let's say that a vector function r(t)r(t) that describes a curve CC has the property that for all tt , 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.