Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

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Essential Calculus: Early Transcendentals

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Short Answer

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

Step by Step Solution

Step 1: Given information

Step 2: Definition of conservative vector field

Step 3: find the function f of conservative vector field

Step 4: find the potential function of conservative vector field

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Given information

Step 2: Definition of conservative vector field

Step 3: find the function f of conservative vector field

Step 4: find the potential function of conservative vector field

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