Find constants $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ such that the vector field defined by $F=\left(4 x y+a z^{3} ight) i+$ $\left(b x^{2}+3 z ight) j+\left(6 x z^{2}+c y ight) k$ is irrotational. With these values of $a, b$ and $c$, determine a scalar function $\phi(x, y, z)$ such that $F=
abla \phi$.

Question

Find constants $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ such that the vector field defined by $F=\left(4 x y+a z^{3}ight) i+$ $\left(b x^{2}+3 zight) j+\left(6 x z^{2}+c yight) k$ is irrotational. With these values of $a, b$ and $c$, determine a scalar function $\phi(x, y, z)$ such that $F=
abla \phi$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To determine the constants $a$, $b$, and $c$ such that the vector field $F = (4xy + az^3) \mathbf{i} + (bx^2 + 3z) \mathbf{j} + (6xz^2 + cy) \mathbf{k}$ is irrotational, we need to ensure that the curl of $F$ is zero. ‖ ‖ step 2 ⋮ The curl of $F$ is given by $ abla imes F = \left( \frac{\partial F_k}{\partial y} - \frac{\partial F_j}{\partial z} ight) \mathbf{i} + \left( \frac{\partial F_i}{\partial z} - \frac{\partial F_k}{\partial x} ight) \mathbf{j} + \left( \frac{\partial F_j}{\partial x} - \frac{\partial F_i}{\partial y} ight) \mathbf{k}$. ‖ ‖ step 3 ⋮ Calculate $\frac{\partial F_k}{\partial y}$: $\frac{\partial (6xz^2 + cy)}{\partial y} = c$. ‖ ‖ step 4 ⋮ Calculate $\frac{\partial F_j}{\partial z}$: $\frac{\partial (bx^2 + 3z)}{\partial z} = 3$. ‖ ‖ step 5 ⋮ For the $\mathbf{i}$ component to be zero: $c - 3 = 0 \implies c = 3$. ‖ ‖ step 6 ⋮ Calculate $\frac{\partial F_i}{\partial z}$: $\frac{\partial (4xy + az^3)}{\partial z} = 3az^2$. ‖ ‖ step 7 ⋮ Calculate $\frac{\partial F_k}{\partial x}$: $\frac{\partial (6xz^2 + cy)}{\partial x} = 6z^2$. ‖ ‖ step 8 ⋮ For the $\mathbf{j}$ component to be zero: $3az^2 - 6z^2 = 0 \implies a = 2$. ‖ ‖ step 9 ⋮ Calculate $\frac{\partial F_j}{\partial x}$: $\frac{\partial (bx^2 + 3z)}{\partial x} = 2bx$. ‖ ‖ step 10 ⋮ Calculate $\frac{\partial F_i}{\partial y}$: $\frac{\partial (4xy + az^3)}{\partial y} = 4x$. ‖ ‖ step 11 ⋮ For the $\mathbf{k}$ component to be zero: $2bx - 4x = 0 \implies b = 2$. ‖ ‖ step 12 ⋮ With $a = 2$, $b = 2$, and $c = 3$, we now find the scalar function $\phi(x, y, z)$ such that $F = abla \phi$. ‖ ‖ step 13 ⋮ Integrate $F_i = 4xy + 2z^3$ with respect to $x$: $\phi(x, y, z) = 2x^2y + 2xz^3 + g(y, z)$. ‖ ‖ step 14 ⋮ Differentiate $\phi$ with respect to $y$ and set it equal to $F_j$: $\frac{\partial \phi}{\partial y} = 2x^2 + \frac{\partial g}{\partial y} = 2x^2 + 3z \implies \frac{\partial g}{\partial y} = 3z \implies g(y, z) = 3yz + h(z)$. ‖ ‖ step 15 ⋮ Differentiate $\phi$ with respect to $z$ and set it equal to $F_k$: $\frac{\partial \phi}{\partial z} = 6xz^2 + 3y + \frac{\partial h}{\partial z} = 6xz^2 + 3y \implies \frac{\partial h}{\partial z} = 0 \implies h(z) = ext{constant}$. ‖ ‖ step 16 ⋮ Therefore, the scalar function $\phi(x, y, z)$ is $\phi(x, y, z) = 2x^2y + 2xz^3 + 3yz + ext{constant}$. ‖ ∻Answer∻ ⚹ The constants are $a = 2$, $b = 2$, and $c = 3$. The scalar function is $\phi(x, y, z) = 2x^2y + 2xz^3 + 3yz + ext{constant}$. ⚹ ∻Key Concept∻ ⚹Irrotational Vector Field⚹ ∻Explanation∻ ⚹To find the constants that make a vector field irrotational, we set the curl of the vector field to zero and solve for the constants. Then, we find the scalar potential function by integrating the components of the vector field.⚹◊What is the condition for a vector field to be irrotational in terms of its components?⍭ Generate me a similar question◊

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