Respuesta :
Answer:
a) Particle has a constant speed of 4, b) Velocity and acceleration vector are orthogonal to each other, c) Clockwise, d) False, the particle begin at the point (0,1).
Step-by-step explanation:
a) Let is find first the velocity vector by differentiation:
[tex]\vec v = \frac{dr_{x}}{dt} i + \frac {dr_{y}}{dt} j[/tex]
[tex]\vec v = 4\cdot \cos 4t\, i - 4 \cdot \sin 4t \,j[/tex]
[tex]\vec v = 4 \cdot (\cos 4t \, i - \sin 4t\,j)[/tex]
Where the resultant vector is the product of a unit vector and magnitude of the velocity vector (speed). Velocity vector has a constant speed only if magnitude of unit vector is constant in time. That is:
[tex]\|\vec u \| = 1[/tex]
Then,
[tex]\| \vec u \| = \sqrt{\cos^{2} 4t + \sin^{2}4t }[/tex]
[tex]\| \vec u \| = \sqrt{1}[/tex]
[tex]\|\vec u \| = 1[/tex]
Hence, the particle has a constant speed of 4.
b) The acceleration vector is obtained by deriving the velocity vector.
[tex]\vec a = \frac{dv_{x}}{dt} i + \frac {dv_{y}}{dt} j[/tex]
[tex]\vec a = 16\cdot (-\sin 4t \,i -\cos 4t \,j)[/tex]
Velocity and acceleration are orthogonal to each other only if [tex]\vec v \bullet \vec a = 0[/tex]. Then,
[tex]\vec v \bullet \vec a = 64 \cdot (\cos 4t)\cdot (-\sin 4t) + 64 \cdot (-\sin 4t) \cdot (-\cos 4t)[/tex]
[tex]\vec v \bullet \vec a = -64\cdot \sin 4t\cdot \cos 4t + 64 \cdot \sin 4t \cdot \cos 4t[/tex]
[tex]\vec v \bullet \vec a = 0[/tex]
Which demonstrates the orthogonality between velocity and acceleration vectors.
c) The particle is rotating clockwise as right-hand rule is applied to model vectors in 2 and 3 dimensions, which are associated with positive angles for position vector. That is: [tex]t \geq 0[/tex]
And cosine decrease and sine increase inasmuch as t becomes bigger.
d) Let evaluate the vector in [tex]t = 0[/tex].
[tex]r(0) = \sin (4\cdot 0) \,i + \cos (4\cdot 0)\,j[/tex]
[tex]r(0) = 0\,i + 1 \,j[/tex]
False, the particle begin at the point (0,1).