Question

Find the curl of the vector and state
its nature at (1,1,-0.2) F= 30 1 + 2xy +
5xz2 k
​

Answer 1

ANSWER

(1,1,-0.2) F = 30 i + 2xy j + 5xz2 k a) √4.01 b) √4.02 c) √4.03

hope this helps you!

mark as brainliest answer.

Answer 2

Concept

A vector may be a quantity or phenomenon that has two independent properties: magnitude and direction.

Given

The given point is $(1,1,-0.2)$ and $F=30i+2xyj+5xz^2 k$.

Find

We have to seek out the curl of the vector and also its nature.

Solution

Firstly, we are going to find the curl of the vector.

$\begin{aligned}\text{Curl }\vec{F}&=\vec{\nabla}\times \vec{F}\\ &=\left|\begin{array}{lll}\hat{i} & \hat{j}& \hat{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ 30& 2xy &5xz^2 \end{array}\right|\\ &=-5z^2 \hat{j}+2y\hat{k} \end{aligned}$

Now, we are going to find the curl of the vector at the purpose  $(1,1,-0.2)$.

$\begin{aligned}\text{curl }\vec{F}&=-5(-0.2)^2 \hat{j}+2(1)\hat{k}\\ &=-0.2\hat{j}+2\hat{k}\end{aligned}$

Further, we'll find the magnitude of the curl of the vector.

$\begin{aligned}|\text{curl }\vec{F}|&=\sqrt{(-0.2)^2+(2)^2}\\ &=\sqrt{4.04}\end{aligned}$

The $\text{curl }\vec{F}\neq 0$, therefore the field is rotational.

Hence, the curl of the vector is $\sqrt{4.04}$ and its nature is rotational.

#SPJ2