A particle moves in space along the path `z=ax^(3)+by^(2)` in such a way that `(dx)/(dt)=c=(dy)/(dt)` where `a,b` and `c` are constants. The acceleration of the particle is

Options

A. `(6ac^(2)x+2bc^(2))hatk`
B. `(2ax^(2)+6by^(2))hatk`
C. `(4bc^(2)+3ac^(2))hatk`
D. `(bc^(2)x+2by)hatk`

Answer

To find the acceleration of the particle moving along the path \( z = ax^3 + by^2 \), we will follow these steps:

### Step 1: Understand the given information
We have the path equation:
\[ z = ax^3 + by^2 \]
and the relationships:
\[ \frac{dx}{dt} = c \quad \text{and} \quad \frac{dy}{dt} = c \]
where \( a, b, \) and \( c \) are constants.

### Step 2: Differentiate the path equation with respect to time
To find the velocity in the z-direction, we differentiate \( z \) with respect to \( t \):
\[ \frac{dz}{dt} = \frac{d}{dt}(ax^3 + by^2) \]
Using the chain rule:
\[ \frac{dz}{dt} = a \cdot 3x^2 \frac{dx}{dt} + b \cdot 2y \frac{dy}{dt} \]

### Step 3: Substitute the values of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
Since \( \frac{dx}{dt} = c \) and \( \frac{dy}{dt} = c \), we substitute these into the equation:
\[ \frac{dz}{dt} = a \cdot 3x^2 \cdot c + b \cdot 2y \cdot c \]
Simplifying this gives:
\[ \frac{dz}{dt} = 3acx^2 + 2bc \]

### Step 4: Differentiate again to find acceleration
Now, we differentiate \( \frac{dz}{dt} \) with respect to \( t \) to find the acceleration:
\[ \frac{d^2z}{dt^2} = \frac{d}{dt}(3acx^2 + 2bc) \]
Using the product rule:
\[ \frac{d^2z}{dt^2} = 3ac \cdot 2x \frac{dx}{dt} + 0 \]
Substituting \( \frac{dx}{dt} = c \):
\[ \frac{d^2z}{dt^2} = 6acx \cdot c = 6acx^2 \]

### Step 5: Write the acceleration vector
The acceleration vector in the z-direction is:
\[ \mathbf{a} = \frac{d^2z}{dt^2} \hat{k} = (6acx) \hat{k} \]

### Final Answer
Thus, the acceleration of the particle is:
\[ \mathbf{a} = 6acx \hat{k} \]