Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

(a) Find a function \(f\) such that \({\bf{F}} = \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve \(C\).
\({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\), \(C\) is the line segment from \((1,0, - 2)\) to \((4,6,3)\)

a) The potential function \(f\) of vector field \({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\) is\(f(x,y,z) = xyz + {z^2}\).

b) The value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\) is 77.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Vector field is \({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\) and line segment from \((1,0, - 2)\) to \((4,6,3)\).

Step 2: Relation between the potential function \(f\) and vector field \({\bf{F}}\)

Relation between the potential function \(f\) and vector field \({\bf{F}}\).

\(\nabla f = {\bf{F}}\)

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 potential function \(f\) such that \(\nabla f = {\bf{F}}\)

Vector field is \({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\).

Consider \(\nabla f = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\).

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

Substitute \({f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\) for \(\nabla f\),

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

Compare the equation \({\bf{F}} = {f_x}(x,y,z){\bf{i}} + {f_y}(x,y,z){\bf{j}} + {f_z}(x,y,z){\bf{k}}\) with \({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\).

\(\begin{array}{l}{f_x}(x,y,z) = yz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\{f_y}(x,y,z) = xz\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\{f_z}(x,y,z) = xy + 2z{\rm{ }}\,\,\,\,\,\,{\rm{(3) }}\end{array}\)

Integrate equation (1) with respect to \(x\).

\(\begin{align} f(x,y,z) & =\int{y}zdx \\ & =yz\int{d}x \\ & =yz(x)+g(y,z)\left\{ \because \int{d}t=t \right\} \\ & f(x,y,z)=xyz+g(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4) \end{align}\)

Apply partial differentiation with respect to \(y\) on both sides of equation (4).

\(\begin{align} & {{f}_{y}}(x,y,z)=\frac{\partial }{\partial y}(xyz+g(y,z)) \\ & =\frac{\partial }{\partial y}(xyz)+\frac{\partial }{\partial y}(g(y,z)) \\ & =xz\frac{\partial }{\partial y}(y)+{{g}^{\prime }}(y,z) \\ & =xz(1)+{{g}^{\prime }}(y,z)\left\{ \because \frac{\partial }{\partial t}(t)=1 \right\} \end{align}\)

\({f_y}(x,y,z) = xz + {g^\prime }(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)\)

Compare the equations \((2)\) and (5).

\({g^\prime }(y,z) = 0\)

Apply integration on both sides of equation.

\(\begin{array}{c}\int {{g^\prime }} (y,z)dy = \int 0 dy\\g(y,z) = h(z)\end{array}\)

Here,

\(h(z)\) is function of \(z\).

Substitute \(h(z)\) for \(g(y,z)\) in equation (4),

\(f(x,y,z) = xyz + h(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)\)

Apply partial differentiation with respect to \(z\) on both sides of equation (6).

\(\begin{align} {{f}_{z}}(x,y,z) & =\frac{\partial }{\partial z}(xyz)+\frac{\partial }{\partial z}(h(z)) \\ & =xy\frac{\partial }{\partial z}(z)+{{h}^{\prime }}(z) \\ & =xy(1)+{{h}^{\prime }}(z)\left\{ \because \frac{\partial }{\partial t}(t)=1 \right\} \\ & {{f}_{z}}(x,y,z)=xy+{{h}^{\prime }}(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7) \end{align}\)

Compare the equation (3) and ( 7 ).

\({h^\prime }(z) = 2z\)

Integrate the equation with respect to \(z\).

\(\begin{align} h(z) & =\int{(2z)}dz \\ & =2\left( \frac{{{z}^{2}}}{2} \right)+K\left\{ \because \int{(t)}dt=\frac{{{t}^{2}}}{2} \right\} \\ & ={{z}^{2}}+K \end{align}\)

Here,

\(K\) is constant.

Consider the value of \(K\) as 0 .

Substitute \({z^2} + K\) for \(h(z)\) in equation \((6)\),

\(f(x,y,z) = xyz + {z^2} + K\)

Substitute 0 for \(K\),

\(\begin{array}{c}f(x,y,z) = xyz + {z^2} + 0\\ = xyz + {z^2}\end{array}\)

Thus, The potential function \(f\) of vector field \({\bf{F}}(x,y) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\) is \(f(x,y,z) = xyz + {z^2}\).

Step 4: Determine the value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\)

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,

\(\int_C \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(8)\)

The initial point of curve is \((1,0, - 2)\) so it is lower limit \(a\) and terminal point is \((4,6,3)\) so it is upper limit \(b\).

Find the common value \(\int_C \nabla f \cdot d{\bf{r}}\) by the use of equation (8).

\(\begin{align} & \int_{C}{\nabla }f\cdot d\mathbf{r}=f(4,6,3)-f(1,0,-2) \\ & =\left[ (4)(6)(3)+{{3}^{2}} \right]-\left[ (1)(0)(-2)+{{(-2)}^{2}} \right]\quad \left\{ \because f(x,y,z)=xyz+{{z}^{2}} \right\} \\ & =72+9-0-4 \\ & =77 \end{align}\)

Thus, the value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\) is 77.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

(a) Find a function \(f\) such that \({\bf{F}} = \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve \(C\).
\({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\), \(C\) is the line segment from \((1,0, - 2)\) to \((4,6,3)\)

Step by Step Solution

Step 1: Given data

Step 2: Relation between the potential function \(f\) and vector field \({\bf{F}}\)

Step 3: Find the potential function \(f\) such that \(\nabla f = {\bf{F}}\)

Step 4: Determine the value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\)

Most popular questions for Math Textbooks

Match the vector field F with the plots labelled I–IV. Give reasons for your choices.

\(F(x,y) = \left\langle {y,y + 2} \right\rangle \)

At time \(t = 1\), a particle is located at position \((1,3)\). If it moves in a velocity field \(V(x,y) = \left\langle {xy - {y^2} - 10} \right\rangle \) find its approximate location at time\(t = 1.05\).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

(a) Find a function \(f\) such that \({\bf{F}} = \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve \(C\).\({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\), \(C\) is the line segment from \((1,0, - 2)\) to \((4,6,3)\)

Step by Step Solution

Step 1: Given data

Step 2: Relation between the potential function \(f\) and vector field \({\bf{F}}\)

Step 3: Find the potential function \(f\) such that \(\nabla f = {\bf{F}}\)

Step 4: Determine the value of \(\int_C \nabla f \cdot d{\bf{r}}\) along the curve \(C\)

Most popular questions for Math Textbooks

Match the vector field F with the plots labelled I–IV. Give reasons for your choices.

\(F(x,y) = \left\langle {y,y + 2} \right\rangle \)

At time \(t = 1\), a particle is located at position \((1,3)\). If it moves in a velocity field \(V(x,y) = \left\langle {xy - {y^2} - 10} \right\rangle \) find its approximate location at time\(t = 1.05\).

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(a) Find a function \(f\) such that \({\bf{F}} = \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve \(C\).
\({\bf{F}}(x,y,z) = yz{\bf{i}} + xz{\bf{j}} + (xy + 2z){\bf{k}}\), \(C\) is the line segment from \((1,0, - 2)\) to \((4,6,3)\)