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Expert-verified Found in: Page 780 ### 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$$C$$.$${\bf{F}}(x,y) - x{y^2}{\bf{i}} + {x^2}y{\bf{j}}$$.

(a)

The potential function $$f$$ of vector field $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ is$$f(x,y) = \frac{1}{3}{x^3}{y^3}$$

(b)

The value of $$\int_C V f - dr$$ along the curve $$C$$ is $$- 9$$.

See the step by step solution

## Step 1: Given information

The given equation is $${\bf{F}}(x,y) - \left( {y{e^x} + \sin y} \right){\bf{i}} + \left( {{e^x} + x\cos y} \right){\bf{j}}$$

## 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

(a)

Vector field is $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ -

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

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

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

Substitute $${f_x}(x,y){\rm{i}} + {f_y}(x,y){\bf{j}}$$ 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 $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$.

\begin{aligned}{l}{f_x}(x,y) = {x^2}{y^3}(1)\\{f_y}(x,y)& = {x^3}{y^2}(2)\end{aligned}

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

\begin{align} & f(x,y)=\int{\left( {{x}^{3}}{{y}^{3}} \right)}dx\\ & ={{y}^{3}}\int{{{x}^{2}}}dx\\ & ={{y}^{3}}\left( \frac{{{s}^{3}}}{3} \right)+g(y)\left\{ \because \int{{{t}^{2}}}dt=\frac{{{t}^{3}}}{3} \right\}\\ & f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y)(3)\end{align}

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

\begin{align} & {{f}_{y}}(x,y)=\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y) \right) \\ & =\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}} \right)+\frac{\partial }{dy}(g(y)) \\ & =\frac{1}{3}{{x}^{3}}\frac{\partial }{\partial y}\left( {{y}^{3}} \right)+{{g}^{\prime }}(y) \\ & =\frac{1}{3}{{x}^{3}}\left( 3{{y}^{2}} \right)+{{g}^{\prime }}(y)\left\{ \because \frac{\partial }{id}{{t}^{n}}=n{{l}^{n-1}} \right\} \\ & {{f}_{y}}(x,y)={{x}^{3}}{{y}^{2}}+{{g}^{\prime }}(y)(4) \end{align}

Compare the equations (2) and (4).

$${g^\prime }(y) = 0$$

Apply integration on both sides of equation.

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

Here,

$$K$$ is constant.

Substitute $$K$$ for $$g(y)$$ in equation (3),

$$f(x,y) = \frac{1}{3}{x^3}{y^3} + K(5)$$

Consider the value of $$K$$ as 0 .

Substitute 0 for $$K$$ in equation (5),

$$\begin{array}{c}f(x,y) = \frac{1}{3}{x^3}{y^3} + 0\\ = \frac{1}{3}{x^3}{y^3}\end{array}$$

Thus, the potential function $$f$$ of vector field $${\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}$$ is $$f(x,y) = \frac{1}{3}{x^3}{y^3}$$.

## Step 4: Evaluate the given function along 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,

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

Write the expression of curve.

$${\bf{r}}(t) = \left\langle {{t^3} - 2t,{t^3} + 2t} \right\rangle (7)$$

Substitute 0 for $$t$$ in equation$$(7)$$,

\begin{aligned}r(0) & = \left\langle {{0^3} - 2(0),{0^3} + 2(0)} \right\rangle \\ & = \langle 0,0\rangle \end{aligned}

Hence, the initial point of curve $$(a)$$ is $$(0,0)$$.

Substitute 1 for $$t$$ in equation$$(7)$$,

\begin{aligned}{\bf{r}}(1) & = \left\langle {{1^3} - 2(1),{1^3} + 2(1)} \right\rangle \\ & = \langle 1 - 2,1 + 2\rangle \\ & = \langle - 1,3\rangle \end{aligned}

Hence, the terminal point of curve $$(b)$$ is$$( - 1,3)$$.

Find the common value $$\int_C \nabla f \cdot dr$$ by using equation$$(6)$$.

\begin{aligned} & \int_{C}{\nabla }f\cdot dr=f(-1,3)-f(0,0) \\ & =\left[ \frac{1}{3}{{(-1)}^{3}}{{(3)}^{3}} \right]-\left[ \frac{1}{3}{{(0)}^{3}}{{(0)}^{3}} \right]\ \ \ \left\{ \because f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}} \right\} \\ & =-9-0 \\ & =-9 \end{aligned}

Thus, the value of $$\int_C V f - dr$$ along the curve $$C$$ is $$- 9$$. ### Want to see more solutions like these? 