If the vector function $\vec{F}=\widehat{a_x}\left(3y-k_{1}z\right)+\widehat{a_y}\left(k_{2}x-2z\right)-\widehat{a_z}\left(k_{3}y+z\right)$ is irrotational, then the values of the constants k₁, k₂ and k₃, respectively, are

Irrotational Vector Function Explained

An irrotational vector function is a fundamental concept in vector calculus and field theory. A vector function, such as the given function $\vec{F}$, is considered irrotational if its curl is equal to zero. This means that the rotational tendency or "circulation" of the field at any point is zero. Mathematically, this condition is expressed as $\nabla \times \vec{F}=0$. When a vector field is irrotational, it can often be expressed as the gradient of a scalar potential function, which simplifies many calculations in physics and engineering, especially in areas like fluid dynamics and electromagnetism.

For the given vector function:

$\vec{F}=\widehat{a_x}\left(3y-k_{1}z\right)+\widehat{a_y}\left(k_{2}x-2z\right)-\widehat{a_z}\left(k_{3}y+z\right)$

We can identify its components as $P$, $Q$, and $R$ along the $\widehat{a_x}$, $\widehat{a_y}$, and $\widehat{a_z}$ directions, respectively:

- $P=3y-k_{1}z$
- $Q=k_{2}x-2z$
- $R=-(k_{3}y+z)=-k_{3}y-z$

Calculating the Curl of the Vector Function

The curl of a vector function $\vec{F}=P\widehat{a_x}+Q\widehat{a_y}+R\widehat{a_z}$ in Cartesian coordinates is given by the determinant:

$\nabla \times \vec{F}=\left|\begin{array}{ccc}\widehat{a_x}& \widehat{a_y}& \widehat{a_z}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ P& Q& R\end{array}\right|$

Expanding this determinant, we get:

$\nabla \times \vec{F}=\widehat{a_x}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)-\widehat{a_y}\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+\widehat{a_z}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)$

Partial Derivatives for Constants

To find the values of the constants $k_1$, $k_2$, and $k_3$, we first need to calculate the partial derivatives of $P$, $Q$, and $R$ with respect to $x$, $y$, and $z$:

- $\frac{\partial P}{\partial y}=\frac{\partial }{\partial y}(3y-k_{1}z)=3$
- $\frac{\partial P}{\partial z}=\frac{\partial }{\partial z}(3y-k_{1}z)=-k_1$
- $\frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}(k_{2}x-2z)=k_2$
- $\frac{\partial Q}{\partial z}=\frac{\partial }{\partial z}(k_{2}x-2z)=-2$
- $\frac{\partial R}{\partial x}=\frac{\partial }{\partial x}(-k_{3}y-z)=0$
- $\frac{\partial R}{\partial y}=\frac{\partial }{\partial y}(-k_{3}y-z)=-k_3$

Determining Constants for Irrotationality

For the vector function $\vec{F}$ to be irrotational, its curl must be identically zero. This means that each component of the curl must be equal to zero.

$\nabla \times \vec{F}=0\widehat{a_x}+0\widehat{a_y}+0\widehat{a_z}$

Equating Components to Zero

Let's set each component of $\nabla \times \vec{F}$ to zero:

**1. $\widehat{a_x}$ component:**

$\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=0$

Substitute the calculated partial derivatives:

$-k_3-(-2)=0$

$-k_3+2=0$

$k_3=2$

**2. $\widehat{a_y}$ component:**

The $\widehat{a_y}$ component in the curl formula has a negative sign: $-\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)=0$. Therefore, we can simply set the expression inside the parenthesis to zero:

$\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}=0$

Substitute the calculated partial derivatives:

$0-(-k_1)=0$

$k_1=0$

**3. $\widehat{a_z}$ component:**

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$

Substitute the calculated partial derivatives:

$k_2-3=0$

$k_2=3$

Final Values of the Constants

Based on the calculations, the values of the constants $k_1$, $k_2$, and $k_3$ for the given vector function to be irrotational are:

- $k_1=0$
- $k_2=3$
- $k_3=2$

Therefore, the constants $k_1$, $k_2$, and $k_3$ are 0.0, 3.0, and 2.0, respectively.