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Found in: Page 780

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Determine whether of not $${\bf{F}}$$ is a conservative vector field. If it is, find a function $$f$$ such that$${\bf{F}} = \nabla f$$.$${\bf{F}}(x,y) = \left( {\ln y + 2x{y^2}} \right){\bf{i}} + \left( {3{x^2}{y^2} + x/y} \right){\bf{j}}$$

The vector field $${\bf{F}}(x,y) = \left( {\ln y + 2x{y^3}} \right){\bf{i}} + \left( {3{x^2}{y^2} + \frac{x}{y}} \right){\bf{j}}$$ is a conservative vector field and the potential function $$f$$ is $$f(x,y) = x\ln y + {x^2}{y^3} + K$$.

See the step by step solution

## Step 1: Given information

The given equation is $${\bf{F}}(x,y) - \left( {\ln y + 2x{y^2}} \right){\bf{i}} + \left( {3{x^2}{y^2} + x/y} \right){\bf{j}}$$

## Step 2: Definition of conservative vector field

Consider vector function $${\bf{r}}(t),a \le t \le b$$ with a smooth curve$$C$$. Consider $$f$$ is a differentiable function two or three variables of gradient function $$\nabla f$$ and is continuous on curve$$C$$. Then,

$$\int_c \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))$$

## Step 3: find the function f of conservative vector field

Here,

$$\frac{{{\rm{ }}\partial P{\rm{ }}}}{{{\rm{ }}\partial {\rm{y }}}}$$ is continuous first-order partial derivative of $$P$$, and

$$\frac{{\partial Q}}{{\partial x}}$$ is continuous first-order partial derivative of $$Q$$.

Compare the vector field $${\bf{F}}(x,y) = \left( {\ln y + 2x{y^3}} \right){\bf{i}} + \left( {3{x^2}{y^2} + \frac{x}{y}} \right){\bf{j}}$$ with \begin{aligned}{c}{\bf{F}}(x,y) = P(x,y){\bf{i}} + Q(x,y){\bf{j}}\\P = \ln y + 2x{y^3}\;\;\;{\kern 1pt} \left( 2 \right)\\Q = 3{x^2}{y^2} + \frac{x}{y}\;\;\;{\kern 1pt} \,\left( 3 \right)\end{aligned}.

Apply partial differentiation with respect to $$y$$ on both sides of equation (2).

\begin{aligned}{c}\frac{{\partial P}}{{\partial y}} &= \frac{d}{{dy}}\left( {\ln y + 2x{y^3}} \right)\\ &= \frac{\partial }{{\partial y}}(\ln y) + \frac{\partial }{{\partial y}}\left( {2x{y^3}} \right)\\ &= \frac{1}{y} + 2x\frac{\partial }{{\partial y}}\left( {{y^3}} \right)\\ &= \frac{1}{y} + 6x{y^2}\end{aligned}

Apply partial differentiation with respect to $$x$$ on both sides of equation (3).

\begin{aligned}{c}\frac{{dQ}}{{dx}} &= \frac{\partial }{{dx}}\left( {3{x^2}{y^2} + \frac{x}{y}} \right)\\ &= \frac{\partial }{{dx}}\left( {3{x^2}{y^2}} \right) + \frac{\partial }{{dx}}\left( {\frac{x}{y}} \right)\\ &= {y^2}\frac{\partial }{{dx}}\left( {3{x^2}} \right) + \frac{1}{y}\frac{\partial }{{\partial x}}(x)\\&= 6x{y^2} + \frac{1}{y}\end{aligned}

Substitute $$6x{y^2} + \frac{1}{y}$$ for $$\frac{{\partial P}}{{\partial y}}$$ and $$6x{y^2} + \frac{1}{y}$$ for $$\frac{{\partial Q}}{{dx}}$$ in equation (1),

$$6x{y^2} + \frac{1}{y} = 6x{y^2} + \frac{1}{y}$$

Hence $${\bf{F}}(x,y) = \left( {\ln y + 2x{y^3}} \right){\bf{i}} + \left( {3{x^2}{y^2} + \frac{x}{y}} \right){\bf{j}}$$ is conservative vector field.

## Step 4: find the potential function of conservative vector field

Consider $$\nabla f = {f_x}(x,y){\bf{i}} + {f_Y}(x,y){\bf{j}}$$.

Write the relation between the potential function $$f$$ and vector field $${\bf{F}}$$.

$$\nabla f = {\bf{F}}$$

Substitute $${f_x}(x,y){\rm{i}} + {f_y}(x,y){\rm{j}}$$ for $$\nabla {f_{{\rm{, }}}}$$

$${\bf{F}} = {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}$$

Compare the equation $${\bf{F}}= {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}$$ with $${\bf{F}}(x,y)= \left( {\ln y + 2x{y^3}} \right){\bf{i}} + \left( {3{x^2}{y^2} + \frac{x}{y}} \right){\bf{j}}$$.

$${f_x}(x,y) = \ln y + 2x{y^3}\,\,\,\,\left( 4 \right)$$

$${f_y}(x,y) = 3{x^2}{y^2} + \frac{x}{y}\;\;\;{\kern 1pt} (5)$$

Integrate equation (4) with respect to $$x$$.

\begin{aligned}{c}f(x,y)& = \int {\left( {\ln y + 2x{y^3}} \right)} dx\\& = \int {\ln } ydx + 2{y^3}\\\int x dx& = x\ln y + {x^2}{y^3} + g(y)\\f(x,y) &= x\ln y + {x^2}{y^3} + g(y)\,\,\,\,\,\,\left( 6 \right)\end{aligned}

Apply partial differentiation with respect to $$y$$ on both sides of the above equation.

\begin{aligned}{c}{f_y}(x,y) &= \frac{\partial }{{dy}}\left( {x\ln y + {x^2}{y^3} + g(y)} \right)\\ &= \frac{\partial }{{dy}}(x\ln y) + \frac{\partial }{{\partial y}}\left( {{x^2}{y^3}} \right) + \frac{\partial }{{dy}}(g(y))\\& = x\frac{1}{y} + {x^2}\frac{\partial }{{dy}}\left( {{y^3}} \right) + {g^\prime }(y)\\ &= \frac{x}{y} + 3{x^2}{y^2} + {g^\prime }(y)\end{aligned}

Apply integration on both sides of equation.

\begin{aligned}\int{{{g}^{\prime }}}(y)dy=\int{0}dy\\g(y)=K\left\{\because\int{0}dt=K\right\}\end{aligned}

Substitute $$K$$ for $$g(y)$$ in equation (6),

$$f(x,y) = x\ln y + {x^2}{y^3} + K$$

Thus, the vector field $${\bf{F}}(x,y) = \left( {\ln y + 2x{y^3}} \right){\bf{i}} + \left( {3{x^2}{y^2} + \frac{x}{y}} \right){\bf{j}}$$ is a conservative vector field and the potential function $$f$$ is $$f(x,y) = x\ln y + {x^2}{y^3} + K$$.