Answer :
### Given Data:
- Parametric equations for the particle:
[tex]\[ x = 4 \alpha \cos^3(\mu), \quad y = 4 \alpha \sin^3(\mu), \quad z = 3 \phi \cos(2 \mu) \][/tex]
### Part (a): Finding the Velocity and Acceleration
1. Velocity Components
- The velocity vector [tex]\(\mathbf{v}\)[/tex] is obtained by differentiating the position vector with respect to time [tex]\(t\)[/tex]:
[tex]\[ v_x = \frac{dx}{dt} = \frac{d}{dt}[4 \alpha \cos^3(\mu)] \][/tex]
[tex]\[ v_x = 4 \alpha \cdot 3 \cos^2(\mu) \cdot (-\sin(\mu)) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_y = \frac{dy}{dt} = \frac{d}{dt}[4 \alpha \sin^3(\mu)] \][/tex]
[tex]\[ v_y = 4 \alpha \cdot 3 \sin^2(\mu) \cdot \cos(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_y = 12 \alpha \sin^2(\mu) \cos(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_z = \frac{dz}{dt} = \frac{d}{dt}[3 \phi \cos(2 \mu)] \][/tex]
[tex]\[ v_z = 3 \phi \cdot (-2 \sin(2 \mu)) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_z = -6 \phi \sin(2 \mu) \cdot \frac{d\mu}{dt} \][/tex]
Simplifying the expressions, the velocity components become:
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu) \][/tex]
[tex]\[ v_y = 12 \alpha \sin^2(\mu) \cos(\mu) \][/tex]
[tex]\[ v_z = -6 \phi \sin(2 \mu) \][/tex]
2. Acceleration Components
- The acceleration vector [tex]\(\mathbf{a}\)[/tex] is the time derivative of the velocity vector:
[tex]\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}[-12 \alpha \cos^2(\mu) \sin(\mu)] \][/tex]
Expanding using the product rule:
[tex]\[ a_x = -12 \alpha \left[\cos^2(\mu) \cdot \frac{d}{dt}[\sin(\mu)] + \sin(\mu) \cdot \frac{d}{dt}[\cos^2(\mu)]\right] \][/tex]
Simplifying:
[tex]\[ a_x = -12 \alpha \left[ \cos^2(\mu) (\cos(\mu) \cdot \frac{d\mu}{dt}) + \sin(\mu) \cdot 2 \cos(\mu) \cdot (-\sin(\mu)) \cdot \frac{d\mu}{dt} \right] \][/tex]
[tex]\[ a_x = -12 \alpha \cos(\mu) \cdot \frac{d\mu}{dt} \left[ \cos^2(\mu) - 2 \sin^2(\mu) \right] \][/tex]
Removing the dependency on [tex]\( \frac{d\mu}{dt} \)[/tex]:
[tex]\[ a_x = -12 \alpha [\cos^2(\mu) - 2 \sin^2(\mu)] \cos(\mu) \][/tex]
[tex]\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}[12 \alpha \sin^2(\mu) \cos(\mu)] \][/tex]
Expanding using the product rule:
[tex]\[ a_y = 12 \alpha \left[\sin^2(\mu) \cdot \frac{d}{dt}[\cos(\mu)] + \cos(\mu) \cdot \frac{d}{dt}[\sin^2(\mu)]\right] \][/tex]
Simplifying:
[tex]\[ a_y = 12 \alpha \left[ \sin^2(\mu) (-\sin(\mu) \cdot \frac{d\mu}{dt}) + \cos(\mu) \cdot 2 \sin(\mu) \cos(\mu) \cdot \frac{d\mu}{dt} \right] \][/tex]
[tex]\[ a_y = 12 \alpha \sin(\mu) \cdot \frac{d\mu}{dt} \left[ 2 \cos^2(\mu) - \sin^2(\mu) \right] \][/tex]
[tex]\[ a_y = 12 \alpha [2 \cos^2(\mu) - \sin^2(\mu)] \sin(\mu) \][/tex]
[tex]\[ a_z = \frac{dv_z}{dt} = \frac{d}{dt}[-6 \phi \sin(2 \mu)] \][/tex]
[tex]\[ a_z = -6 \phi \cdot 2 \cos(2 \mu) \cdot \frac{d\mu}{dt} \][/tex]
Simplifying:
[tex]\[ a_z = -12 \phi \cos(2 \mu) \][/tex]
### Part (b): Evaluating the Dot Product [tex]\( \mathbf{r} \cdot \mathbf{v} \)[/tex]
The dot product is given by:
[tex]\[ \mathbf{r} \cdot \mathbf{v} = x v_x + y v_y + z v_z \][/tex]
Substitute the given expressions:
[tex]\[ x = 4 \alpha \cos^3(\mu), \quad y = 4 \alpha \sin^3(\mu), \quad z = 3 \phi \cos(2 \mu) \][/tex]
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu), \quad v_y = 12 \alpha \sin^2(\mu) \cos(\mu), \quad v_z = -6 \phi \sin(2 \mu) \][/tex]
Calculate the individual products:
[tex]\[ x v_x = (4 \alpha \cos^3(\mu)) (-12 \alpha \cos^2(\mu) \sin(\mu)) \][/tex]
[tex]\[ x v_x = -48 \alpha^2 \cos^5(\mu) \sin(\mu) \][/tex]
[tex]\[ y v_y = (4 \alpha \sin^3(\mu)) (12 \alpha \sin^2(\mu) \cos(\mu)) \][/tex]
[tex]\[ y v_y = 48 \alpha^2 \sin^5(\mu) \cos(\mu) \][/tex]
[tex]\[ z v_z = (3 \phi \cos(2 \mu)) (-6 \phi \sin(2 \mu)) \][/tex]
[tex]\[ z v_z = -18 \phi^2 \cos(2 \mu) \sin(2 \mu) \][/tex]
Using trigonometric identities:
[tex]\[ \cos(2 \mu) \sin(2 \mu) = \frac{1}{2} \sin(4 \mu) \][/tex]
[tex]\[ z v_z = -18 \phi^2 \cdot \frac{1}{2} \sin(4 \mu) \][/tex]
[tex]\[ z v_z = -9 \phi^2 \sin(4 \mu) \][/tex]
Thus, the dot product [tex]\( \mathbf{r} \cdot \mathbf{v} \)[/tex] becomes:enzen
[tex]\[ \mathbf{r} \cdot \mathbf{v} = -48 \alpha^2 \cos^5(\mu) \sin(\mu) + 48 \alpha^2 \sin^5(\mu) \cos(\mu) - 9 \phi^2 \sin(4 \mu) \][/tex]
Therefore, the velocity, acceleration, and the dot product vector for the particle are as derived above.
- Parametric equations for the particle:
[tex]\[ x = 4 \alpha \cos^3(\mu), \quad y = 4 \alpha \sin^3(\mu), \quad z = 3 \phi \cos(2 \mu) \][/tex]
### Part (a): Finding the Velocity and Acceleration
1. Velocity Components
- The velocity vector [tex]\(\mathbf{v}\)[/tex] is obtained by differentiating the position vector with respect to time [tex]\(t\)[/tex]:
[tex]\[ v_x = \frac{dx}{dt} = \frac{d}{dt}[4 \alpha \cos^3(\mu)] \][/tex]
[tex]\[ v_x = 4 \alpha \cdot 3 \cos^2(\mu) \cdot (-\sin(\mu)) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_y = \frac{dy}{dt} = \frac{d}{dt}[4 \alpha \sin^3(\mu)] \][/tex]
[tex]\[ v_y = 4 \alpha \cdot 3 \sin^2(\mu) \cdot \cos(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_y = 12 \alpha \sin^2(\mu) \cos(\mu) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_z = \frac{dz}{dt} = \frac{d}{dt}[3 \phi \cos(2 \mu)] \][/tex]
[tex]\[ v_z = 3 \phi \cdot (-2 \sin(2 \mu)) \cdot \frac{d\mu}{dt} \][/tex]
[tex]\[ v_z = -6 \phi \sin(2 \mu) \cdot \frac{d\mu}{dt} \][/tex]
Simplifying the expressions, the velocity components become:
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu) \][/tex]
[tex]\[ v_y = 12 \alpha \sin^2(\mu) \cos(\mu) \][/tex]
[tex]\[ v_z = -6 \phi \sin(2 \mu) \][/tex]
2. Acceleration Components
- The acceleration vector [tex]\(\mathbf{a}\)[/tex] is the time derivative of the velocity vector:
[tex]\[ a_x = \frac{dv_x}{dt} = \frac{d}{dt}[-12 \alpha \cos^2(\mu) \sin(\mu)] \][/tex]
Expanding using the product rule:
[tex]\[ a_x = -12 \alpha \left[\cos^2(\mu) \cdot \frac{d}{dt}[\sin(\mu)] + \sin(\mu) \cdot \frac{d}{dt}[\cos^2(\mu)]\right] \][/tex]
Simplifying:
[tex]\[ a_x = -12 \alpha \left[ \cos^2(\mu) (\cos(\mu) \cdot \frac{d\mu}{dt}) + \sin(\mu) \cdot 2 \cos(\mu) \cdot (-\sin(\mu)) \cdot \frac{d\mu}{dt} \right] \][/tex]
[tex]\[ a_x = -12 \alpha \cos(\mu) \cdot \frac{d\mu}{dt} \left[ \cos^2(\mu) - 2 \sin^2(\mu) \right] \][/tex]
Removing the dependency on [tex]\( \frac{d\mu}{dt} \)[/tex]:
[tex]\[ a_x = -12 \alpha [\cos^2(\mu) - 2 \sin^2(\mu)] \cos(\mu) \][/tex]
[tex]\[ a_y = \frac{dv_y}{dt} = \frac{d}{dt}[12 \alpha \sin^2(\mu) \cos(\mu)] \][/tex]
Expanding using the product rule:
[tex]\[ a_y = 12 \alpha \left[\sin^2(\mu) \cdot \frac{d}{dt}[\cos(\mu)] + \cos(\mu) \cdot \frac{d}{dt}[\sin^2(\mu)]\right] \][/tex]
Simplifying:
[tex]\[ a_y = 12 \alpha \left[ \sin^2(\mu) (-\sin(\mu) \cdot \frac{d\mu}{dt}) + \cos(\mu) \cdot 2 \sin(\mu) \cos(\mu) \cdot \frac{d\mu}{dt} \right] \][/tex]
[tex]\[ a_y = 12 \alpha \sin(\mu) \cdot \frac{d\mu}{dt} \left[ 2 \cos^2(\mu) - \sin^2(\mu) \right] \][/tex]
[tex]\[ a_y = 12 \alpha [2 \cos^2(\mu) - \sin^2(\mu)] \sin(\mu) \][/tex]
[tex]\[ a_z = \frac{dv_z}{dt} = \frac{d}{dt}[-6 \phi \sin(2 \mu)] \][/tex]
[tex]\[ a_z = -6 \phi \cdot 2 \cos(2 \mu) \cdot \frac{d\mu}{dt} \][/tex]
Simplifying:
[tex]\[ a_z = -12 \phi \cos(2 \mu) \][/tex]
### Part (b): Evaluating the Dot Product [tex]\( \mathbf{r} \cdot \mathbf{v} \)[/tex]
The dot product is given by:
[tex]\[ \mathbf{r} \cdot \mathbf{v} = x v_x + y v_y + z v_z \][/tex]
Substitute the given expressions:
[tex]\[ x = 4 \alpha \cos^3(\mu), \quad y = 4 \alpha \sin^3(\mu), \quad z = 3 \phi \cos(2 \mu) \][/tex]
[tex]\[ v_x = -12 \alpha \cos^2(\mu) \sin(\mu), \quad v_y = 12 \alpha \sin^2(\mu) \cos(\mu), \quad v_z = -6 \phi \sin(2 \mu) \][/tex]
Calculate the individual products:
[tex]\[ x v_x = (4 \alpha \cos^3(\mu)) (-12 \alpha \cos^2(\mu) \sin(\mu)) \][/tex]
[tex]\[ x v_x = -48 \alpha^2 \cos^5(\mu) \sin(\mu) \][/tex]
[tex]\[ y v_y = (4 \alpha \sin^3(\mu)) (12 \alpha \sin^2(\mu) \cos(\mu)) \][/tex]
[tex]\[ y v_y = 48 \alpha^2 \sin^5(\mu) \cos(\mu) \][/tex]
[tex]\[ z v_z = (3 \phi \cos(2 \mu)) (-6 \phi \sin(2 \mu)) \][/tex]
[tex]\[ z v_z = -18 \phi^2 \cos(2 \mu) \sin(2 \mu) \][/tex]
Using trigonometric identities:
[tex]\[ \cos(2 \mu) \sin(2 \mu) = \frac{1}{2} \sin(4 \mu) \][/tex]
[tex]\[ z v_z = -18 \phi^2 \cdot \frac{1}{2} \sin(4 \mu) \][/tex]
[tex]\[ z v_z = -9 \phi^2 \sin(4 \mu) \][/tex]
Thus, the dot product [tex]\( \mathbf{r} \cdot \mathbf{v} \)[/tex] becomes:enzen
[tex]\[ \mathbf{r} \cdot \mathbf{v} = -48 \alpha^2 \cos^5(\mu) \sin(\mu) + 48 \alpha^2 \sin^5(\mu) \cos(\mu) - 9 \phi^2 \sin(4 \mu) \][/tex]
Therefore, the velocity, acceleration, and the dot product vector for the particle are as derived above.
