Question

$\sf \red{ Question:- }$
For the following vector field F, decide whether it is conservative or not by computing the appropriate first order partial derivatives. Type in a potential function f (that is, ▽ f = F) with f(0, 0) = 0.
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F (x, y) = (-14x - 2y)i + (-2x + 14y)j
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Answer 1

Given

A vector field F (x, y) = (-14x - 2y)i + (-2x + 14y)j .

To Find

Whether this field is conservative or not.

Concept

A field is called conservative if it is path independent. The curl of such field will be zero. That is ▽ X F = 0. In other words, it can be said that a field  will be conservative if it can be written as gradiant of a scalar field. In my solution, i have used this approach. I tried to find a scalar field whose gradiant is the given field.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\nabla f = ( - 14x - 2y) \hat{i} + ( - 2x + 14y)\hat{j} - - (1)$

We know,

If

$\rm :\longmapsto\:F \: = \: m\hat{i} + n\hat{j} \: where \: m \: and \: n \: are \:$

continuous function of x and y such that first order partial derivative exist, then vector feild F is conservative iff

$\boxed{ \rm{ \frac{\partial m}{\partial y} = \frac{\partial n}{\partial x}}}$

So,

On comparing with given equation, we have

$\red{\rm :\longmapsto\:m = - 14x - 2y}$

and

$\red{\rm :\longmapsto\:n = - 2x + 14y}$

Now,

$\red{\rm :\longmapsto\: \dfrac{\partial m}{\partial y} = - 2}$

and

$\red{\rm :\longmapsto\: \dfrac{\partial n}{\partial x} = - 2}$

$\bf\implies \:\dfrac{\partial m}{\partial y} = \dfrac{\partial n}{\partial x}$

$\bf\implies \:\nabla f \: is \: conservative.$

Now, to find a potential function f(x, y) with f(0,0) = 0.

Given that,

$\rm :\longmapsto\:\nabla f = ( - 14x - 2y) \hat{i} + ( - 2x + 14y)\hat{j} - - (1)$

It implies,

$\red{\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = - 14x - 2y}$

and

$\red{\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = - 2x + 14y}$

Now,

$\red{\rm :\longmapsto\:\dfrac{\partial f}{\partial x} = - 14x - 2y}$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\:f = - 14 \times \dfrac{ {x}^{2} }{2} - 2xy + g(y)$

where, g(y) is a constant of integration.

$\rm :\longmapsto\:f = -7{x}^{2} - 2xy + g(y) - - - (2)$

Now, Differentiate partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = 0 - 2x + \dfrac{\partial g(y)}{\partial y}$

$\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = - 2x + \dfrac{\partial g(y)}{\partial y}$

But,

$\red{\rm :\longmapsto\:\dfrac{\partial f}{\partial y} = - 2x + 14y}$

So, on equating we get

$\rm :\longmapsto\: - 2x + 14y= - 2x + \dfrac{\partial g(y)}{\partial y}$

$\rm :\longmapsto\: 14y= \dfrac{\partial g(y)}{\partial y}$

Now, on integrating both sides w. r. t. y, we get

$\rm :\longmapsto\:g(y) = 14 \times \dfrac{ {y}^{2} }{2} + c$

$\rm :\longmapsto\:g(y) = 7{y}^{2} + c - - - - (3)$

On Substituting equation (3) in equation (2), we get

$\rm :\longmapsto\:f = -7{x}^{2} - 2xy + {7y}^{2} + c - - - (4)$

Now, given that,

$\red{\rm :\longmapsto\:f(0,0) = 0}$

So, on substituting the value, we get

$\rm :\longmapsto\:0 = - 0 - 0 + 0 + c$

$\bf\implies \:c = 0$

On substituting c = 0, in equation (4), we get

$\bf :\longmapsto\:f = -7{x}^{2} - 2xy + {7y}^{2}$

is the required potential function.