Question

Determine if the vector field F is conservative. F = (2xy 16x)i + 2y(x 1) + 9k​

Answer 1

The given force field $\vec F=(2xy^2-16x)\hat i+2y(x^2-1)\hat j+9\hat k$ is conservative

Explanation:

Given vector field

$\vec F=(2xy^2-16x)\hat i+2y(x^2-1)\hat j+9\hat k$

Here

$F_1=2xy^2-16x$

$\frac{\partial F_1}{\partial y}=4xy$

And

$F_2=2y(x^2-1)$

$\frac{\partial F_2}{\partial x}=4xy$

$\since \frac{\partial F_1}{\partial y}-\frac{\partial F_2}{\partial x}=0$

Therefore, the given field is conservative.

Hope this answer is helpful.

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Answer 2

Given that,

Vector field $F=(2xy^2-16x) i+2y(x^2-1)j+9k$

We know that,

If the curl F is zero then the vector field will be conservative.

So, $curl F=\Delta \times F$

We need to calculate the vector field F is conservative

Using formula of curl F

$curl F=\Delta \times F$

Put the value into the formula

$curl F=(\dfrac{\delta}{\delta x}i+\dfrac{\delta}{\delta y}j+\dfrac{\delta}{\delta z})\times((2xy^2-16x) i+2y(x^2-1)j+9k)$

$curl F= i(\dfrac{\delta}{\delta y}(9)-\dfrac{\delta}{\delta z}(2y(x^2-1)))-j(\dfrac{\delta}{\delta x}(9)-\dfrac{\delta}{\delta z}(2xy^2-16x))+k(\dfrac{\delta}{\delta x}(2y(x^2-1)-\dfrac{\delta}{\delta y}(2xy^2-16x))$

$curl F=i(0-0)-j(0-0)+k(4xy-4xy)$

$curl F=0$

Hence, The vector field F is conservative.

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Topic : cross product