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Expert-verified Found in: Page 781 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # (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 $$C$$.$${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$,

a)The potential function $$f$$ of vector field

$${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$ is $$f(x,y,z) = x\sin y + y\cos z.$$

b) The value of $$\int_C \nabla f \cdot d{\bf{r}}$$ along the curve $$C$$ is $$1 - \frac{\pi }{2}$$.

See the step by step solution

## Step 1: Given data

Vector field is $${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$ and curve $${\bf{r}}(t) = \sin t{\bf{i}} + t{\bf{j}} + 2t{\bf{k}},0 \le t \le \frac{\pi }{2}.$$

## 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 $$\nabla f = {\bf{F}}$$

a)

Vector field is $${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$.

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 $${\bf{F}}$$.

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

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

$${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$.

\begin{aligned}{l}{f_x}(x,y,z) & = \sin y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\{f_y}(x,y,z) & = x\cos y + \cos z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\{f_z}(x,y,z) & = - y\sin z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)\end{aligned}

Integrate equation (1) with respect to x.

\begin{align} f(x,y,z) & =\int{(\sin y)}dx \\ & =\sin y\int{d}x \\ & =\sin y(x)+g(y,z)\left\{ \because \int{d}t=t \right\} \\ f(x,y,z) & =x\sin y+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}(x\sin y+g(y,z)) \\ & =\frac{\partial }{\partial y}(x\sin y)+\frac{\partial }{\partial y}(g(y,z)) \\ & =x\frac{\partial }{\partial y}(\sin y)+{{g}_{y}}(y,z) \\ & =x(\cos y)+{{g}_{y}}(y,z)\left\{ \because \frac{\partial }{\partial t}(\sin t)=1 \right\}\end{align}

$${f_y}(x,y,z) = x\cos y + {g_y}(y,z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

Compare the equations (2) and (5).

$${g_y}(y,z) = \cos z$$

Apply integration on both sides of equation.

\begin{align} \int{{{g}_{y}}}(y,z)dy & =\int{\cos }zdy \\ g(y,z) & =y\cos z+h(z)\quad \left\{ \because \int{d}t=t \right\} \end{align}

Here,

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

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

$$f(x,y,z) = x\sin y + y\cos z + h(z){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(6) }}$$

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

\begin{align} {{f}_{z}}(x,y,z) & =\frac{\partial }{\partial z}(x\sin y)+\frac{\partial }{\partial z}(y\cos z)+\frac{\partial }{\partial z}(h(z)) \\ & =x\sin y\frac{\partial }{\partial z}(1)+y\frac{\partial }{\partial z}(\cos z)+{{h}^{\prime }}(z) \\ & =x\sin y(0)+y(-\sin z)+{{h}^{\prime }}(z)\left\{ \because \frac{\partial }{\partial t}(k)=0,\frac{\partial }{\partial t}(\cos t)=-\sin t \right\} \\ {{f}_{z}}(x,y,z) & =-y\sin z+{{h}^{\prime }}(z)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7) \end{align}

Compare the equation (3) and (7).

$${h^\prime }(z) = 0$$

Integrate equation with respect to $$z$$.

\begin{align} h(z) & =\int{0}dz \\ & =K\left\{ \because \int{0}dt=K \right\} \end{align}

Here,

$$K$$ is constant.

Consider the value of $$K$$ as 0 . Therefore, $$h(z) = 0$$.

Substitute 0 for $$h(z)$$ in equation (6),

\begin{aligned}f(x,y,z) & = x\sin y + y\cos z + 0\\ & = x\sin y + y\cos z\end{aligned}

Thus, The potential function $$f$$ of vector field

$${\bf{F}}(x,y,z) = \sin y{\bf{i}} + (x\cos y + \cos z){\bf{j}} - y\sin z{\bf{k}}$$is $$f(x,y,z) = x\sin y + y\cos z.$$

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

b)

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 point on curve $${\bf{r}}(t) = \sin t{\bf{i}} + t{\bf{j}} + 2t{\bf{k}}$$ is,

$$(x,y,z) = (\sin t,t,2t)\,\,\,\,\,\,\,\,\,\,\,\,\,(9)$$

Substitute 0 for $$t$$,

$$\begin{array}{c}(x,y,z) = (\sin (0),0,2(0))\\ = (0,0,0)\end{array}$$

Hence, the lower limit $$(a)$$ is $$(0,0,0)$$.

Substitute $$\frac{\pi }{2}$$ for $$t$$,

$$\begin{array}{c}(x,y,z) = \left( {\sin \left( {\frac{\pi }{2}} \right),\frac{\pi }{2},2\left( {\frac{\pi }{2}} \right)} \right)\\ = \left( {1,\frac{\pi }{2},\pi } \right)\end{array}$$

Hence, the upper limit $$(b)$$ is $$\left( {1,\frac{\pi }{2},\pi } \right)$$.

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

\begin{aligned} \int_{C}{\nabla }f\cdot d\mathbf{r} & =f\left( 1,\frac{\pi }{2},\pi \right)-f(0,0,0) \\ & =\left\{\begin{aligned}{*{35}{l}} \left. \left( (1)\sin \left( \frac{\pi }{2} \right)+\left( \frac{\pi }{2} \right)\cos (\pi ) \right)- \right\} \\ ((0)\sin (0)+(0)\cos (0)) \\ \end{aligned} \right\}\{\because f(x,y,z)=x\sin y+y\cos z\} \\ & =(1)(1)+\left( \frac{\pi }{2} \right)(-1)-0 \\ & =1-\frac{\pi }{2} \end{aligned}

Thus, the value of $$\int_C \nabla f \cdot d{\bf{r}}$$ along the curve $$C$$ is $$1 - \frac{\pi }{2}$$. ### Want to see more solutions like these? 